Pattern: Misc

[Easy] Simple Moving Average

Problem

  • Given a list, calculate a moving average of a window of size windowSize of numbers.
  • Example:
Input:
a = [3, 1, 10, 3, 5]
windowSize = 2

Output:
2.0
5.5
6.5
4.0

Solution

if windowSize > len(a):
    return -1

# Get the first window.
currSum = sum(a[0:windowSize])
print(currSum/windowSize)

# Slide the window from left to right.
for i in range(len(a) - windowSize):
    currSum += a[windowSize+i]
    currSum -= a[i]
    print(currSum/windowSize)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[2/Medium] Add Two Numbers

Problem

  • You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

  • You may assume the two numbers do not contain any leading zero, except the number 0 itself.

  • Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
  • Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
  • Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
  • Constraints:
    • The number of nodes in each linked list is in the range [1, 100].
    • 0 <= Node.val <= 9
    • It is guaranteed that the list represents a number that does not have leading zeros.
  • See problem on LeetCode.

Solution: Maintain carry and use divmod

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        carry = 0
        res = n = ListNode(0)
        while l1 or l2 or carry:
            if l1:
                carry += l1.val
                l1 = l1.next
            if l2:
                carry += l2.val
                l2 = l2.next
            carry, val = int(carry) // 10, int(carry) % 10 # or divmod(carry, 10)
            n.next = n = ListNode(val) # or the traditional "n.next = ListNode(val); n = n.next"
        return res.next;        
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[7/Medium] Reverse Integer

Problem

  • Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-2^31, 2^31 - 1], then return 0.

  • Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

  • Example 1:

Input: x = 123
Output: 321
  • Example 2:
Input: x = -123
Output: -321
  • Example 3:
Input: x = 120
Output: 21
  • Constraints:
    • -2^31 <= x <= 2^31 - 1
  • See problem on LeetCode.

Solution: Convert to str, reverse str and add sign

class Solution(object):
    def reverse(self, x: int) -> int:
        # extract the sign
        s = (x > 0) - (x < 0) 
        # or cmp(x, 0) which returns negative if x<y, zero if x==y, positive if x>y.
        
        # Change the sign to + for negative numbers since 
        # [::-1] would move the - sign to the end.
        # Note that for positive numbers, this is a no op.
        r = int(str(x * s)[::-1])
        
        # Put the sign back.
        return s * r * (r < 2**31)
Complexity
  • Time: \(O(n + n) = O(n)\)
  • Space: \(O(1)\)

[8/Medium] String to Integer (atoi)

Problem

  • Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++’s atoi function).

  • The algorithm for myAtoi(string s) is as follows:

    1. Read in and ignore any leading whitespace.
    2. Check if the next character (if not already at the end of the string) is ‘-‘ or ‘+’. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
    3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
    4. Convert these digits into an integer (i.e., “123” -> 123, “0032” -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
    5. If the integer is out of the 32-bit signed integer range [-2^31, 2^31 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -2^31 should be clamped to -2^31, and integers greater than 1. 2^31 - 1 should be clamped to 2^31 - 1.
    6. Return the integer as the final result.
  • Note:
    • Only the space character ‘ ‘ is considered a whitespace character.
    • Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.
  • Example 1:
Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
         ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "42" ("42" is read in)
           ^
The parsed integer is 42.
Since 42 is in the range [-2^31, 2^31 - 1], the final result is 42.
  • Example 2:
Input: s = "   -42"
Output: -42
Explanation:
Step 1: "   -42" (leading whitespace is read and ignored)
            ^
Step 2: "   -42" ('-' is read, so the result should be negative)
             ^
Step 3: "   -42" ("42" is read in)
               ^
The parsed integer is -42.
Since -42 is in the range [-2^31, 2^31 - 1], the final result is -42.
  • Example 3:
Input: s = "4193 with words"
Output: 4193
Explanation:
Step 1: "4193 with words" (no characters read because there is no leading whitespace)
         ^
Step 2: "4193 with words" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "4193 with words" ("4193" is read in; reading stops because the next character is a non-digit)
             ^
The parsed integer is 4193.
Since 4193 is in the range [-2^31, 2^31 - 1], the final result is 4193.
  • Constraints:
    • 0 <= s.length <= 200
    • s consists of English letters (lower-case and upper-case), digits (0-9), ' ', '+', '-', and '.'.
  • See problem on LeetCode.

Solution: RegEx

class Solution(object):
    def myAtoi(self, s: str) -> int:
        """
        :type str: str
        :rtype: int
        """
        ###better to do strip before sanity check (although 8ms slower):
        #ls = list(s.strip())
        #if len(ls) == 0 : return 0
        if len(s) == 0: 
            return 0
        ls = list(s.strip())
        
        sign = -1 if ls[0] == '-' else 1
        if ls[0] in ['-','+']: 
            del ls[0]
        
        ret, i = 0, 0
        while i < len(ls) and ls[i].isdigit():
            ret = ret*10 + ord(ls[i]) - ord('0')
            i += 1
        
        return max(-2**31, min(sign * ret,2**31-1))
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)
class Solution:
    def myAtoi(self, s: str) -> int:
        if len(s) == 0: 
            return 0
        
        i = 0
        while i < len(s) and s[i] == ' ':
            i += 1
        if i == len(s): # all are leading whitespace
            return 0
        
        if s[i]=='+':
            sign = 1
            i += 1
        elif s[i]=='-':
            sign = -1
            i += 1
        else:
            sign = 1
        
        val = 0
        while i < len(s) and s[i].isdigit():
            val = val*10 + ord(s[i]) - ord('0')
            i += 1
            
        res = val*sign
        if res < -2**31: return -2**31
        elif res > 2**31-1: return 2**31-1
        else: return res
Complexity
  • Time: \(O(n)\) where \(n = len(s)\)
  • Space: \(O(1)\)

Solution: RegEx

class Solution:
    # @return an integer
    def myAtoi(self, s: str) -> int:
        str = str.strip()
        str = re.findall('(^[\+\-0]*\d+)\D*', str)

        try:
            result = int(''.join(str))
            MAX_INT = 2147483647
            MIN_INT = -2147483648
            
            if result > MAX_INT > 0:
                return MAX_INT
            elif result < MIN_INT < 0:
                return MIN_INT
            else:
                return result
        except:
            return 0
Complexity
  • Time: \(O(n + n) = O(n)\)
  • Space: \(O(1)\)

[9/Easy] Palindrome Number

Problem

  • Given an integer x, return true if x is palindrome integer.

  • An integer is a palindrome when it reads the same backward as forward.

  • For example, 121 is a palindrome while 123 is not.

  • Example 1:

Input: x = 121
Output: true
Explanation: 121 reads as 121 from left to right and from right to left.
  • Example 2:
Input: x = -121
Output: false
Explanation: From left to right, it reads -121. From right to left, it becomes 121-. Therefore it is not a palindrome.
  • Example 3:
Input: x = 10
Output: false
Explanation: Reads 01 from right to left. Therefore it is not a palindrome.
  • Constraints:
    • -2^31 <= x <= 2^31 - 1
  • Follow up: Could you solve it without converting the integer to a string?

  • See problem on LeetCode.

Solution: Convert the number to string and compare it with the reversed string

  • This is the easiest way to check if integer is palindrome.

  • Convert the number to string and compare it with the reversed string.

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False
        
        return str(x) == str(x)[::-1]
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for storing the reversed integer

Solution: Don’t convert the number to string, recreate a new number in reverse order

class Solution:
    def isPalindrome(self, x: int) -> bool:
        if x < 0:
            return False

        inputNum = x
        newNum = 0
        
        # reverse number
        while x:
            newNum = newNum * 10 + x % 10
            x = x // 10
            
        return newNum == inputNum
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for storing the reversed integer

Solution: Instead of reversing the whole integer, let’s convert half of the integer and then check if it’s palindrome

  • Instead of reversing the whole integer, let’s convert half of the integer and then check if it’s palindrome. How do we know when we’ve reached the half? Check if our new half is greater than the original one.

  • For e.g.,

Even case:
If x = 1234, then let's reverse x with a loop. 
Initially, x = 1234, result = 0
Loop iteration #1: x = 123, result = 1
Loop iteration #2: x = 12, result = 12
Odd case:
If x = 15951, then let's reverse x with a loop. 
Initially, x = 15951, result = 0
Loop iteration #1: x = 1595, result = 1
Loop iteration #2: x = 159, result = 15
Loop iteration #3: x = 15, result = 159
  • We see that result > x after 3 loops and we’ve crossed the digits of the integer half way the integer because it’s an integer of odd length. If it’s an even length integer, the loop stops exactly in the middle after iteration 2.

  • Now we can compare x and result if the integer has an even length, or x and result//10 if the integer has an odd length and return True if they match.

class Solution:
    def isPalindrome(self, x: int) -> bool:
        # if x is negative, return False. if x is positive and last digit is 0, that also cannot form a palindrome, return False.
        if x < 0 or (x > 0 and x % 10 == 0):   
            return False

        result = 0
        
        # reverse number till the midpoint
        while x > result:
            result = result * 10 + x % 10
            x = x // 10
            
        return x == result or x == result // 10 # or return True if (x == result or x == result // 10) else False
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for storing the reversed integer

[14/Easy] Longest Common Prefix

Problem

  • Write a function to find the longest common prefix string amongst an array of strings.

  • If there is no common prefix, return an empty string "".

  • Example 1:

Input: strs = ["flower","flow","flight"]
Output: "fl"
  • Example 2:
Input: strs = ["dog","racecar","car"]
Output: ""
Explanation: There is no common prefix among the input strings.
  • Constraints:
    • 1 <= strs.length <= 200
    • 0 <= strs[i].length <= 200
    • strs[i] consists of only lower-case English letters.
  • See problem on LeetCode.

Solution: Chop off the last character every iteration until the prefix fits the string

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        """
        :type strs: List[str]; rtype: str
        """
        # early termination: if strs is None
        if not strs:
           return None

        # early termination: if len(strs) is an iterable but has no strings inside    
        if not len(strs):
            return None
                    
        prefix = strs[0] # or "min(strs, key=len)" or "sorted(strs, key=len)[0]"
        
        for s in strs:
            while s[:len(prefix)] != prefix: # or "while not s.startswith(prefix):"
                # remove a character from "prefix" from the end 
                # until the prefix fits the string
                prefix = prefix[:-1]
                
        return prefix                
Complexity
  • Time: \(O(n^2)\)
  • Space: \(O(1)\)

Solution: Aggregate the characters in each string by their positions and build the prefix incrementally until you hit >1 unique chars

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        """
        :type strs: List[str]; rtype: str
        """
        # the zip() call here aggregates the characters in each string
        # by their positions, so the prefix/initial characters
        # feature first.
        # for example:
        # >>> strs = ["flower","flow","flight"]
        # >>> aggregatedCharsByPos = zip(*strs) # <zip object at location>
        # >>> list(aggregatedCharsByPos)
        # [('f', 'f', 'f'), ('l', 'l', 'l'), ('o', 'o', 'i'), ('w', 'w', 'g')]
        # note that if the length of the iterables are not equal, 
        # zip creates the list of tuples of length equal to the smallest iterable.
        aggregatedCharsByPos, retVal = zip(*strs), ""

        for ch in aggregatedCharsByPos:
            # if there are >1 unique characters at the same position,
            # get out of the loop since you have the longest prefix already  
            if len(set(ch)) > 1: 
                break
                
                # build up return value
            retVal += c[0]
            
        return retVal
Complexity
  • Time: \(O(n + n) = O(n)\)
  • Space: \(O(1)\)

[21/Easy] Merge Two Sorted Lists

Problem

  • You are given the heads of two sorted linked lists list1 and list2.

  • Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

  • Return the head of the merged linked list.

  • Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4]
Output: [1,1,2,3,4,4]
  • Example 2:
Input: list1 = [], list2 = []
Output: []
  • Example 3:
Input: list1 = [], list2 = [0]
Output: [0]
  • Constraints:
    • The number of nodes in both lists is in the range [0, 50].
    • -100 <= Node.val <= 100
    • Both list1 and list2 are sorted in non-decreasing order.
  • See problem on LeetCode.

Solution

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if list1 is None:
            return list2
        elif list2 is None:
            return list1
        elif list1.val <= list2.val:
            list1.next = self.mergeTwoLists(list1.next, list2)
            return list1
        else:
            list2.next = self.mergeTwoLists(list1, list2.next)
            return list2
Complexity
  • Time: \(O(m+n)\) where \(m\) and \(n\) are the lengths of the two lists respectively
  • Space: \(O(1)\)

[31/Medium] Next Permutation

Problem

  • A permutation of an array of integers is an arrangement of its members into a sequence or linear order.
    • For example, for arr = [1,2,3], the following are considered permutations of arr: [1,2,3], [1,3,2], [3,1,2], [2,3,1].
  • The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).
    • For example, the next permutation of arr = [1,2,3] is [1,3,2].
    • Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
    • While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.
  • Given an array of integers nums, find the next permutation of nums.
  • The replacement must be in place and use only constant extra memory.

  • Example 1:
Input: nums = [1,2,3]
Output: [1,3,2]
  • Example 2:
Input: nums = [3,2,1]
Output: [1,2,3]
  • Example 3:
Input: nums = [1,1,5]
Output: [1,5,1]
  • Constraints:
    • 1 <= nums.length <= 100
    • 0 <= nums[i] <= 100
  • See problem on LeetCode.

Solution: Find the first non-increasing element

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        
        # To find next permutation, we'll start from the end.
        i = j = len(nums)-1
        
        # First we'll find the first non-increasing element starting from the end.
        while i > 0 and nums[i-1] >= nums[i]:
            i -= 1
            
        # After completion of the first loop, there will be two cases:
        # 1. "i" becomes zero (this will happen if the given array is sorted in decreasing order). In this case, simply reverse the sequence and return.
        if i == 0:
            nums.reverse()
            return 
        # 2. If "i" is not zero then we'll find the first number greater than nums[i-1] starting from end.
        while nums[j] <= nums[i-1]:
            j -= 1
        
        # Now our i and j pointers are pointing at two different positions:
        # i -> first non-ascending number from end
        # j -> first number greater than nums[i-1]
        
        # We'll swap these two numbers
        nums[i-1], nums[j] = nums[j], nums[i-1]
        
        # We'll reverse the sequence starting from i to end
        nums[i:]= nums[len(nums)-1:i-1:-1]
        
        # We don't need to return anything as we've modified nums in-place
Complexity
  • Time: \(O(3n) = O(n)\)
  • Space: \(O(1)\)

[43/Medium] Multiply Strings

  • Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.

  • Note: You must not use any built-in BigInteger library or convert the inputs to integer directly.

  • Example 1:

Input: num1 = "2", num2 = "3"
Output: "6"
  • Example 2:
Input: num1 = "123", num2 = "456"
Output: "56088"
  • Constraints:
    • 1 <= num1.length, num2.length <= 200
    • num1 and num2 consist of digits only.
    • Both num1 and num2 do not contain any leading zero, except the number 0 itself.
  • See problem on LeetCode.

Solution: Decode numbers, multiply them and encode them together

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        if num1 == '0' or num2 == '0':
            return '0'
        
        def decode(num):
            ans = 0
            for i in num:
                ans = ans*10 + (ord(i) - ord('0'))
            return ans
        
        def encode(s):
            news = ''
            while s:
                a = s % 10
                s = s // 10
                news = chr(ord('0') + a) + news
            return news
        
        return encode(decode(num1) * decode(num2))
  • Same approach; differently structured:
class Solution(object):
    def multiply(self, num1: str, num2: str) -> str:
        """
        :type num1: str
        :type num2: str
        :rtype: str
        """
        res = 0 # output
        carry1 = 1 # carry1 takes care of num1
        
        for i in num1[::-1]: # this goes over num1 from right to left as we do normal multiplication, in the above example first 3 and then 2. 
            carry2 = 1 # takes care of num2
   
            for j in num2[::-1]: # this goes over num2 from right to left as we do normal multiplication, in the above example first 6 and then 3. 
                res += int(i)*int(j)*carry1*carry2 # this is each component calculated separately and added to res
                carry2 *= 10 # after first iteration (number 6 is covered), it goes to the next number from right, which is 3 here, actually 30, right? That's why it multiplies the carry2 by 10. Similar for carry1. 
            
            carry1 *= 10
   
        return str(res)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[68/Hard] Valid Sudoku

Problem

  • Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
    • Each row must contain the digits 1-9 without repetition.
    • Each column must contain the digits 1-9 without repetition.
    • Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition. Note:
    • A Sudoku board (partially filled) could be valid but is not necessarily solvable.
    • Only the filled cells need to be validated according to the mentioned rules.
  • Example 1:

Input: board = 
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output: true
  • Example 2:
Input: board = 
[["8","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
  • Constraints:
    • board.length == 9
    • board[i].length == 9
    • board[i][j] is a digit 1-9 or '.'.
  • See problem on LeetCode.

Solution: Readable

def isValidSudoku(self, board):
    return (self.is_row_valid(board) and
            self.is_col_valid(board) and
            self.is_square_valid(board))

def is_row_valid(self, board):
    for row in board:
        if not self.is_unit_valid(row):
            return False
    return True

def is_col_valid(self, board):
    for col in zip(*board):
        if not self.is_unit_valid(col):
            return False
    return True
    
def is_square_valid(self, board):
    for i in (0, 3, 6):
        for j in (0, 3, 6):
            square = [board[x][y] for x in range(i, i + 3) for y in range(j, j + 3)]
            if not self.is_unit_valid(square):
                return False
    return True
    
def is_unit_valid(self, unit):
    unit = [i for i in unit if i != '.']
    return len(set(unit)) == len(unit)

Solution: One-liner using Counter

def isValidSudoku(self, board):
    return 1 == max(collections.Counter(
        x
        for i, row in enumerate(board)
        for j, c in enumerate(row)
        if c != '.'
        for x in ((c, i), (j, c), (i/3, j/3, c))
    ).values() + [1])
  • The + [1] is only for the empty board, where max would get an empty list and complain. It’s not necessary to get it accepted here, as the empty board isn’t among the test cases, but it’s good to have.

Solution: One-liner using len(set).

def isValidSudoku(self, board):
    seen = sum(([(c, i), (j, c), (i/3, j/3, c)]
                for i, row in enumerate(board)
                for j, c in enumerate(row)
                if c != '.'), [])
    return len(seen) == len(set(seen))

Solution: One-liner using any.

def isValidSudoku(self, board):
    seen = set()
    return not any(x in seen or seen.add(x)
                   for i, row in enumerate(board)
                   for j, c in enumerate(row)
                   if c != '.'
                   for x in ((c, i), (j, c), (i/3, j/3, c)))

Solution: One-liner with iterating a different way

def isValidSudoku(self, board):
    seen = sum(([(c, i), (j, c), (i/3, j/3, c)]
                for i in range(9) for j in range(9)
                for c in [board[i][j]] if c != '.'), [])
    return len(seen) == len(set(seen))

[68/Hard] Text Justification

Problem

  • Given an array of strings words and a width maxWidth, format the text such that each line has exactly maxWidth characters and is fully (left and right) justified.

  • You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ‘ ‘ when necessary so that each line has exactly maxWidth characters.

  • Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line does not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.

  • For the last line of text, it should be left-justified and no extra space is inserted between words.

  • Note:
    • A word is defined as a character sequence consisting of non-space characters only.
    • Each word’s length is guaranteed to be greater than 0 and not exceed maxWidth.
    • The input array words contains at least one word.
  • Example 1:
Input: words = ["This", "is", "an", "example", "of", "text", "justification."], maxWidth = 16
Output:
[
   "This    is    an",
   "example  of text",
   "justification.  "
]
  • Example 2:
Input: words = ["What","must","be","acknowledgment","shall","be"], maxWidth = 16
Output:
[
  "What   must   be",
  "acknowledgment  ",
  "shall be        "
]
Explanation: Note that the last line is "shall be    " instead of "shall     be", because the last line must be left-justified instead of fully-justified.
Note that the second line is also left-justified becase it contains only one word.
  • Example 3:
Input: words = ["Science","is","what","we","understand","well","enough","to","explain","to","a","computer.","Art","is","everything","else","we","do"], maxWidth = 20
Output:
[
  "Science  is  what we",
  "understand      well",
  "enough to explain to",
  "a  computer.  Art is",
  "everything  else  we",
  "do                  "
]
  • Constraints:
    • 1 <= words.length <= 300
    • 1 <= words[i].length <= 20
    • words[i] consists of only English letters and symbols.
    • 1 <= maxWidth <= 100
    • words[i].length <= maxWidth
  • See problem on LeetCode.

Solution: Two loops

class Solution:
    def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
        ans, cur = [], []
        chars = 0
        
        for word in words:
            # if cur is empty or the total chars + total needed spaces + next word fit
            if not cur or (len(word) + chars + len(cur)) <= maxWidth:
                cur.append(word)
                chars += len(word)
            else:
                # place spaces, append the line to the ans, and move on
                line = self.placeSpacesBetween(cur, maxWidth - chars)
                ans.append(line)
                cur.clear()
                cur.append(word)
                chars = len(word)
        
        # left justify any remaining text, which is easy
        if cur:
            extra_spaces = maxWidth - chars - len(cur) + 1
            ans.append(' '.join(cur) + ' ' * extra_spaces)
            
        return ans
    
    
    def placeSpacesBetween(self, words, spaces):
        if len(words) == 1: return words[0] + ' ' * spaces
        
        space_groups = len(words)-1
        spaces_between_words = spaces // space_groups
        extra_spaces = spaces % space_groups
        
        cur = []
        for word in words:
            cur.append(word)
            
            # place the min of remaining spaces or spaces between words plus an extra if available
            cur_extra = min(1, extra_spaces)
            spaces_to_place = min(spaces_between_words + cur_extra, spaces)

            cur.append(' ' * spaces_to_place)
            
            if extra_spaces: extra_spaces -= 1
            spaces -= spaces_to_place
        
        return ''.join(cur)
Complexity
  • Time: \(O(n^2)\)
  • Space: \(O(n)\)

Solution: Two loops; cleaner version

class Solution:
    def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
        i, N, result = 0, len(words), []
        
        while i < N:
            # decide how many words to be put in one line
            oneLine, j, currWidth, positionNum, spaceNum = [words[i]], i + 1, len(words[i]), 0, maxWidth - len(words[i])
            while j < N and currWidth + 1 + len(words[j]) <= maxWidth:
                oneLine.append(words[j])
                currWidth += 1 + len(words[j])
                spaceNum -= len(words[j])
                positionNum, j = positionNum + 1, j + 1
            i = j
            
            # decide the layout of one line
            if i < N and positionNum:
                spaces = [' ' * (spaceNum // positionNum + (k < spaceNum % positionNum)) for k in range(positionNum)] + ['']
            else: # last line or the line only has one word
                spaces = [' '] * positionNum + [' ' * (maxWidth - currWidth)]
                
            result.append(''.join([s for pair in zip(oneLine, spaces) for s in pair]))
        return result
Complexity
  • Time: \(O(n^2)\)
  • Space: \(O(n)\)

Solution: Loop over once and adjust spacing

  • We’ll build the result array line-by-line while iterating over words in the input. Whenever the current line gets too big, we’ll appropriately format it and proceed with the next line until for loop is over. Last but not least, we’ll need to left-justify the last line.
class Solution:
    # Why slots: https://docs.python.org/3/reference/datamodel.html#slots
    # TLDR: 1. faster attribute access. 2. space savings in memory.
    # For letcode problems this can save ~ 0.1MB of memory <insert is something meme>
    __slots__ = ()
    
    def fullJustify(self, words: List[str], maxWidth: int) -> List[str]:
        # Init return array in which, we'll store justified lines
        lines = []
        # current line width
        width = 0
        # current line words
        line = []
        
        for word in words:
            # Gather as many words that will fit under maxWidth restrictions.
            # Line length is a sum of:
            # 1) Current word length
            # 2) Sum of words already in the current line
            # 3) Number of spaces (each word needs to be separated by at least one space)
            if (len(word) + width + len(line)) <= maxWidth:
                width += len(word)
                line.append(word)
                continue
            
            # If the current line only contains one word, fill the remaining string with spaces.
            if len(line) == 1:
                # Use the format function to fill the remaining string with spaces easily and readable.
                # You could also do something like:
                #     line = " ".join(line)
                #     line += " " * (maxWidth - len(line))
                #     lines.append(line)
                lines.append(
                    "{0: <{width}}".format( " ".join(line), width=maxWidth)
                )
            else:
                # Else calculate how many common spaces and extra spaces are there for the current line.
                # Example:
                #  line = ['a', 'computer.', 'Art', 'is']
                # width left in line equals to: maxWidth - width: 20 - 15 = 5
                # len(line) - 1 because to the last word, we aren't adding any spaces
                # Now divmod will give us how many spaces are for all words and how many extra to distribute.
                # divmod(5, 3) = 1, 2
                # This means there should be one common space for each word, and for the first two, add one extra space.
                space, extra = divmod(
                    maxWidth - width,
                    len(line) - 1
                )
                
                i = 0
                # Distribute extra spaces first
                # There cannot be a case where extra spaces count is greater or equal to number words in the current line.
                while extra > 0:
                    line[i] += " "
                    extra -= 1
                    i += 1
                
                # Join line array into a string by common spaces, and append to justified lines.
                lines.append(
                    (" " * space).join(line)
                )
            
            # Create new line array with the current word in iteration, and reset current line width as well.
            line = [word]
            width = len(word)
        
        # Last but not least format last line to be left-justified with no extra space inserted between words.
        # No matter the input, there always be the last line at the end of for loop, which makes things even easier considering the requirement.
        lines.append(
            "{0: <{width}}".format(" ".join(line), width=maxWidth)
        )
        
        return lines
Complexity
  • Time: \(O(n)\) since there is just one for loop, which iterates over words provided as input.
  • Space: \(O(n + l)\) where \(n\) is the length of words, and \(l\) max length of words in one line. Worst case scenario \(l = n\), which will add up to \(O(2n)\) but in asymptotic analysis we don’t care about constants so final complexity is still linear.

[50/Medium] Pow(x, n)

Problem

  • Implement pow(x, n), which calculates x raised to the power n (i.e., x^n).

  • Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000
  • Example 2:
Input: x = 2.10000, n = 3
Output: 9.26100
  • Example 3:
Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
  • Constraints:
    • -100.0 < x < 100.0
    • -2^31 <= n <= 2^31-1
  • See problem on LeetCode.

Solution: Use “**”

class Solution:
    myPow = pow
  • That’s even shorter than the other more obvious “cheat”:
class Solution:
    def myPow(self, x: float, n: int) -> float:
        return x ** n

Solution: Recursion

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if not n:
            return 1
        if n < 0:
            return 1 / self.myPow(x, -n)
        if n % 2:
            return x * self.myPow(x, n-1)
        return self.myPow(x*x, n/2)

Solution: Iteration

class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n < 0:
            x = 1 / x
            n = -n
        pow = 1
        while n:
            if n & 1:
                pow *= x
            x *= x
            n >>= 1
        return pow
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[53/Easy] Maximum Subarray

Problem

  • Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

  • A subarray is a contiguous part of an array.

  • Example 1:

Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: [4,-1,2,1] has the largest sum = 6.
  • Example 2:
Input: nums = [1]
Output: 1
  • Example 3:
Input: nums = [5,4,-1,7,8]
Output: 23
  • Constraints:
    • 1 <= nums.length <= 105
    • -104 <= nums[i] <= 104
  • See problem on LeetCode.

Solution: Kadane’s algorithm

class Solution:
    def maxSubArray(self, A):
        """
        @param A, a list of integers
        @return an integer
        """
        
        # not needed if the question says that the array should have at least one number.
        if not A:
            return 0

        curSum = maxSum = A[0]
        for num in A[1:]:
            # Check if we should abandon the accumulated value so far
            # and start a new run of numbers if the previous sum is negative 
            curSum = max(num, curSum + num) # or curSum = max(curSum, 0) + num
            
            # Update answer if current sum is better than the previous one
            maxSum = max(maxSum, curSum)

        return maxSum
  • Same approach; rephrased longer answer:
class Solution(object):
    def maxSubArray(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        max_so_far = curr_so_far = -float('inf')
        for i in range(len(nums)):
            curr_so_far += nums[i] # Add current number
            
            # Check if we should abandon the accumulated value so far
            # and start a new run of numbers if the previous sum is negative            
            curr_so_far = max(curr_so_far, nums[i])
            
            # Update answer
            max_so_far = max(max_so_far, curr_so_far)
        
        return max_so_far
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[54/Medium] Spiral Matrix

Problem

  • Given an m x n matrix, return all elements of the matrix in spiral order.

  • Example 1:

Input: matrix = [[1,2,3],[4,5,6],[7,8,9]]
Output: [1,2,3,6,9,8,7,4,5]
  • Example 2:

Input: matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]
Output: [1,2,3,4,8,12,11,10,9,5,6,7]
  • Constraints:
    • m == matrix.length
    • n == matrix[i].length
    • 1 <= m, n <= 10
    • -100 <= matrix[i][j] <= 100
  • See problem on LeetCode.
Solution: Recursion
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        """
        :type matrix: List[List[int]]
        :rtype: List[int]
        """
        if len(matrix) == 0:
            return []
        
        new_matrix = []
        for j in range(len(matrix[0])-1, -1, -1):
            new_lst = []
            
            for i in range(1, len(matrix)):
                new_lst.append(matrix[i][j])
            
            new_matrix.append(new_lst)
            
        return matrix[0] + self.spiralOrder(new_matrix)        
Solution: One-liner
  • Take the first row plus the spiral order of the rotated remaining matrix.
def spiralOrder(self, matrix):
    return matrix and [*matrix.pop(0)] + self.spiralOrder([*zip(*matrix)][::-1])
Solution
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        height = len(matrix)
        width = len(matrix[0])
        
        top = 0
        bottom = height - 1
        left = 0
        right = width - 1
        
        ans = []
        while top < bottom and left < right:
            for col in range(left, right):
                ans.append(matrix[top][col])
            
            for row in range(top, bottom):
                ans.append(matrix[row][right])
            
            for col in range(right, left, -1):
                ans.append(matrix[bottom][col])
            
            for row in range(bottom, top, -1):
                ans.append(matrix[row][left])
            
            top += 1
            bottom -= 1
            left += 1
            right -= 1
        
        # If a matrix remain inside it is either a 1xn or a mx1
        # a linear scan will return the same order as spiral for these
        if len(ans) < height*width:
            for row in range(top, bottom+1):
                for col in range(left, right+1):
                    ans.append(matrix[row][col])
        
        return ans
  • Here are 3 examples:
Example 1: row remains in middle
[1,2,3,4]
[5,6,7,8]
[9,0,1,2]
The bottom nested for loops will traverse (in-order): 6,7
Example 2: Single element remains in middle
[1,2,3]
[4,5,6]
[7,8,9]
The bottom nested for loops will traverse (in-order): 5
Example C: Column remains in middle
[1,2,3]
[4,5,6]
[7,8,9]
[0,1,2]
The bottom nested for loops will traverse (in-order): 5,8
  • Same approach; no need to handle edge cases with this solution. We already know the length of our result list should be length of m*n. So, just ignore the redundant elements and return res[:m*n].
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        left, right, top, bottom = 0, n - 1, 0, m - 1
        res = []
        
        while left <= right and top <= bottom:
            # top left to top right
            for col in range(left, right+1):
                res.append(matrix[top][col])
            top += 1
            
            # top right to right bottom
            for row in range(top, bottom+1):
                res.append(matrix[row][right])
            right -= 1
            
            # right bottom to right left
            for col in range(right, left-1, -1):
                res.append(matrix[bottom][col])
            bottom -= 1
            
            # left bottom to top left
            for row in range(bottom, top-1, -1):
                res.append(matrix[row][left])
            left += 1
        
        # just ignore the redundant and return length of m*n
        return res[:m*n]
  • Same approach (but with if-conditions so don’t have to restrict the length of res when returning):
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix), len(matrix[0])
        left, right, top, bottom = 0, n - 1, 0, m - 1
        res = []
        
        while left <= right and top <= bottom:
            # top left to top right
            for col in range(left, right+1):
                res.append(matrix[top][col])
            top += 1
            
            # top right to right bottom
            for row in range(top, bottom+1):
                res.append(matrix[row][right])
            right -= 1
            
            if (top <= bottom):
                # right bottom to right left
                for col in range(right, left-1, -1):
                    res.append(matrix[bottom][col])
                bottom -= 1
            
            if (left <= right):            
                # left bottom to top left
                for row in range(bottom, top-1, -1):
                    res.append(matrix[row][left])
                left += 1
        
        # just ignore the redundant and return length of m*n
        return res
class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m , n = len(matrix), len(matrix[0])
        #refers to row as row can only move up or down
        topBound , bottomBound = 0, m-1
        #refers to cols as cols can only move right or left
        leftBound, rightBound = 0, n-1
        
        res = []
        size = m*n
        
        # we iterate until we have gone over all elements
        while len(res) < size:                    
        # we iterate until we have found the center
        # or while leftBound <= rightBound and topBound <= bottomBound:       
            # runs through the first row straight through each col
            if topBound <= bottomBound:
                for col in range(leftBound, rightBound+1):
                    res.append(matrix[topBound][col])
                topBound += 1
            
            # next runs down the column
            if leftBound <= rightBound:
                for row in range(topBound, bottomBound+1):
                    res.append(matrix[row][rightBound])
                rightBound -= 1
            
            # goes right to left in col
            if topBound <= bottomBound:
                for col in range(rightBound, leftBound -1, -1):
                    res.append(matrix[bottomBound][col])
                bottomBound -= 1
            
            # goes up
            if leftBound <= rightBound:
                for row in range(bottomBound, topBound-1, -1):
                    res.append(matrix[row][leftBound])
                leftBound += 1
            
        return res
  • Note that the if conditions are the “key” concept. This prevents the double counting, especially for cases where the middle most is “horizontal” or “vertical”:
1 2 3 4     or        1 2 3     or     1 2 3 
5 * * 7               4 * 5            4 * 4
7 7 7 7               4 4 4            4 * 4 
                                       4 * 4   
                                       3 3 3  
Complexity
  • Time: \(O(H*W)\) since we traverse all elements in the matrix to be able to return them.
  • Space: \(O(H*W)\) due to the size of the output variable (which is as large as the input matrix). If you don’t consider output variable size in your analysis then: \(O(1)\) space.

[56/Medium] Merge intervals

Problem

  • Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

  • Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
  • Example 2:
Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.
  • Constraints:
    • 1 <= intervals.length <= 104
    • intervals[i].length == 2
    • 0 <= start_i <= end_i <= 104
  • See problem on LeetCode.

Solution: sort first, then compare the start of the current interval with the end of the prior interval

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        out = []
        for i in sorted(intervals, key=lambda i: i[0]):
        # or for i in sorted(intervals, key=lambda i: i.start):        
            if out and i[0] <= out[-1][1]:
            # or if out and i.start <= out[-1].end:
                out[-1][1] = max(out[-1][1], i[1])
                # or out[-1].end = max(out[-1].end, i.end)
            else:
                out += i, # Note that "i," is a tuple (because of the comma), and you can extend a tuple to a list (using "+=").
                # or out += [i]
        return out
Complexity
  • Time: \(O(n\log{n} + n) = O(n\log{n})\)
  • Space: \(O(n)\)

[57/Medium] Insert Interval

Problem

  • You are given an array of non-overlapping intervals intervals where intervals[i] = [start_i, end_i] represent the start and the end of the i-th interval and intervals is sorted in ascending order by start_i. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.
  • Insert newInterval into intervals such that intervals is still sorted in ascending order by start_i and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).
  • Return intervals after the insertion.

  • Example 1:
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
  • Example 2:
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
  • Constraints:
    • 0 <= intervals.length <= 104
    • intervals[i].length == 2
    • 0 <= start_i <= end_i <= 105
    • intervals is sorted by start_i in ascending order.
    • newInterval.length == 2
    • 0 <= start <= end <= 105
  • See problem on LeetCode.

Solution: sort first, then compare the start of the current interval with the end of the prior interval

  • First, append the new interval to the intervals list, then follow the same idea as Merge Intervals.
class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        intervals.append(newInterval)
        out = []
        
        for i in sorted(intervals, key=lambda i: i[0]):
        # or for i in sorted(intervals, key=lambda i: i.start):        
            if out and i[0] <= out[-1][1]:
            # or if out and i.start <= out[-1].end:
                out[-1][1] = max(out[-1][1], i[1])
                # or out[-1].end = max(out[-1].end, i.end)
            else:
                out += i, # Note that "i," is a tuple (because of the comma), and you can extend an iterable to a list (using "+=").
                # or out += [i] or simply out.append(i)
        return out
Complexity
  • Time: \(O(n\log{n} + n) = O(n\log{n})\)
  • Space: \(O(n)\)

Solution: makes use of the property that the intervals were initially sorted according to their start times

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        result = []
        
        for interval in intervals:
            # the start of the new interval is after the end of the current interval, so we can leave the current interval because the new one does not overlap with it
            if newInterval[0] > interval[1]:
                result.append(interval)
                
            # the start of the current interval is after the end of the new interval, so we can add the new interval and update it to the current one
            elif interval[0] > newInterval[1]:
                result.append(newInterval)
                newInterval = interval
                
            # the new interval is in the range of the other interval, we have an overlap, so we must choose the min for start and max for end of interval 
            elif interval[1] >= newInterval[0] or interval[0] <= newInterval[1]:
                newInterval[0] = min(interval[0], newInterval[0])
                newInterval[1] = max(newInterval[1], interval[1])

        result.append(newInterval); 
        return result
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[62/Medium] Unique Paths

Problem

  • There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0]). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1]). The robot can only move either down or right at any point in time.
  • Given the two integers m and n, return the number of possible unique paths that the robot can take to reach the bottom-right corner.
  • The test cases are generated so that the answer will be less than or equal to 2 * 10^9.

  • Example 1:

Input: m = 3, n = 7
Output: 28
  • Example 2:
Input: m = 3, n = 2
Output: 3
Explanation: From the top-left corner, there are a total of 3 ways to reach the bottom-right corner:
1. Right -> Down -> Down
2. Down -> Down -> Right
3. Down -> Right -> Down
  • Constraints:
    • 1 <= m, n <= 100
  • See problem on LeetCode.

Solution: Recursion

  • Since robot can move either down or right, there is only one path to reach the cells in the first row: right->right->...->right.

  • The same is valid for the first column, though the path here is down->down-> ...->down.

  • What about the “inner” cells (m, n)? To such cell one could move either from the cell on the left (m, n - 1), or from the cell above (m - 1, n). That means that the total number of paths to move into (m, n) cell is uniquePaths(m - 1, n) + uniquePaths(m, n - 1).

class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        if m == 1 or n == 1:
            return 1
        
        return self.uniquePaths(m - 1, n) + self.uniquePaths(m, n - 1)

Solution: Dynamic Programming

  • One could rewrite recursive approach into dynamic programming one.
  • Algorithm:
    • Initiate 2D array d[m][n] = number of paths. To start, put number of paths equal to 1 for the first row and the first column. For the simplicity, one could initiate the whole 2D array by ones.
    • Iterate over all “inner” cells: d[col][row] = d[col - 1][row] + d[col][row - 1].
    • Return d[m - 1][n - 1].
class Solution:
    def uniquePaths(self, m: int, n: int) -> int:
        """
        dp[i][j] = number of ways to reach i,j
        dp[0][j] = 1, since this is the top row. To reach any element 
        in the top row, we can only move right, so there is only 1 way.
        Similarly, dp[i][0] = 1.
        dp[i][j] = num ways to reach one left + num ways to reach one above
                 = (number of ways to reach i, j-1) + (num ways to reach i-1, j)
                 = dp[i][j-1] + dp[i-1][j]        
        """
        
        d = [[1] * n for _ in range(m)]

        for col in range(1, m):
            for row in range(1, n):
                # The robot can only move either down or right at any point in time.
                # so to get to [col][row], you have to look at the previous column (left), i.e., [col - 1][row]
                # and the previous row (top), i.e., [col][row - 1]
                d[col][row] = d[col - 1][row] + d[col][row - 1]

        return d[m - 1][n - 1]

Complexity

  • Time: \(\mathcal{O}(n \times m)\)
  • Space: \(\mathcal{O}(n \times m)\)

Solution: Combination

  • Could one do better than \(\mathcal{O}(N \times M)\)? The answer is yes.

  • The problem is a classical combinatorial problem: there are \(h + v\) moves to do from start to finish, \(h = m - 1\) horizontal moves, and \(v = n - 1\) vertical ones. One could choose when to move to the right, i.e. to define \(h\) horizontal moves, and that will fix vertical ones. Or, one could choose when to move down, i.e., to define \(v\) vertical moves, and that will fix horizontal ones.

  • In other words, we’re asked to compute in how many ways one could choose \(p\) elements from \(p + k\) elements. In mathematics, that’s called binomial coefficients
\[C_{h + v}^{h} = C_{h + v}^{v} = \frac{(h + v)!}{h! v!}\]
  • The number of horizontal moves to do is \(h = m - 1\), the number of vertical moves is \(v = n - 1\). That results in a simple formula
\[C_{h + v}^{h} = \frac{(m + n - 2)!}{(m - 1)! (n - 1)!}\]
  • The job is done. Now time complexity will depend on the algorithm to compute factorial function \((m + n - 2)!\). In short, standard computation for k!k! using the definition requires \(\mathcal{O}(k^2 \log k)\) time, and that will be not as good as DP algorithm.
  • The best known algorithm to compute factorial function is done by Peter Borwein. The idea is to express the factorial as a product of prime powers, so that \(k!\) can be computed in \(\mathcal{O}(k (\log k \log \log k)^2)\) time. That’s better than \(\mathcal{O}(k^2)\) and hence beats DP algorithm.
  • The authors prefer not to discuss here various factorial function implementations, and hence provide Python3 solution only, with built-in divide and conquer factorial algorithm. If you’re interested in factorial algorithms, please check out good review on this page.
class Solution:
   def combination_solution():
        ###################################
        # Solution 2: combination         #
        ###################################
        """
        Answer is (m+n-2) choose (m-1)
        Total path length is (m-1) + (n-1).
        Consider the list of moves for a single path:
            [down, right, down, down, right, ...]
        We choose m-1 of these to be down (or choose n-1 of these to be right).
        """
        def choose(n, r):
            return int(math.factorial(n) / (math.factorial(r) * math.factorial(n-r)))
        return choose(m+n-2, m-1)

Complexity

  • Time: \(\mathcal{O}((m + n) (\log (m + n) \log \log (m + n))^2)\)
  • Space: \(\mathcal{O}(1)\)

[65/Hard] Valid number

Problem

  • A valid number can be split up into these components (in order):
    • A decimal number or an integer.
    • (Optional) An e or E, followed by an integer.
  • A decimal number can be split up into these components (in order):
    • (Optional) A sign character (either + or -).
    • One of the following formats:
      • One or more digits, followed by a dot ..
      • One or more digits, followed by a dot ., followed by one or more digits.
      • A dot ., followed by one or more digits.
  • An integer can be split up into these components (in order):
    • (Optional) A sign character (either + or -).
    • One or more digits.
  • For example, all the following are valid numbers: ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"], while the following are not valid numbers: ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"].
  • Given a string s, return true if s is a valid number.

  • Example 1:
Input: s = "0"
Output: true
  • Example 2:
Input: s = "e"
Output: false
  • Example 3:
Input: s = "."
Output: false
  • Constraints:
    • 1 <= s.length <= 20
    • s consists of only English letters (both uppercase and lowercase), digits (0-9), plus +, minus -, or dot ..
  • See problem on LeetCode.

Solution

  • To solve this problem, we should should just make a list of the possible error conditions and then check for each one.
  • The error conditions are:
    • More than one exponent character (e/E), or seeing an e/E when a number has not yet been seen.
    • More than one sign, or a sign appearing after a decimal or number have been seen. This gets reset when passing an e/E.
    • More than one decimal, or a decimal appearing after an e/E has been seen.
    • Any other non-number character appearing.
    • Reaching the end of S without an active number.
  • To help with this process, we can set up some boolean flags for the different things of which we’re keeping track (num, exp, sign, dec). We’ll also need to remember to reset all flags except exp when we find an e/E, as we’re starting a new integer expression.

Solution: Rule-based

  • Method 1:
class Solution:
    def isNumber(self, s: str) -> bool:    
        num, exp, sign, dec = False, False, False, False
        
        for c in s:
            if c >= '0' and c <= '9': num = True     
            elif c == 'e' or c == 'E':
                if exp or not num: return False
                else: exp, num, sign, dec = True, False, False, False
            elif c == '+' or c == '-':
                if sign or num or dec: return False
                else: sign = True
            elif c == '.':
                if dec or exp: return False
                else: dec = True
            else: return False
            
        return num
  • Method 2 (check for seenDigit, seenExponent and seenDot):
class Solution:
    def isNumber(self, s: str) -> bool:
        seenDigit = seenExponent = seenDot = False
        
        for i, c in enumerate(s):
            if c.isdigit():
                seenDigit = True
            elif c in ["+", "-"]:
                if i > 0 and s[i-1] != "e" and s[i-1] != "E":
                    return False
            elif c in ["e", "E"]:
                if seenExponent or not seenDigit:
                    return False
                seenExponent = True
                seenDigit = False
            elif c == ".":
                if seenDot or seenExponent:
                    return False
                seenDot = True
            else:
                return False
        
        return seenDigit
Complexity
  • Time: \(O(n)\) where \(n\) is the number of characters in the input string, i.e., n = len(s).
  • Space: \(O(1)\)

Solution: Use a RegEx

class Solution:
    def isNumber(self, s: str) -> bool:
        # Example:               +-     1 or 1. or 1.2 or .2   e or E +- 1     
        engine = re.compile(r"^[+-]?((\d+\.?\d*)|(\d*\.?\d+))([eE][+-]?\d+)?$")
        return engine.match(s.strip(" ")) # i prefer this over putting more things (\S*) in regex
Complexity
  • Time: \(O(n)\) where \(n\) is the number of characters in the input string, i.e., n = len(s).
  • Space: \(O(1)\)

[67/Easy] Add Binary

Problem

  • Given two binary strings a and b, return their sum as a binary string.

  • Example 1:

Input: a = "11", b = "1"
Output: "100"
  • Example 2:
Input: a = "1010", b = "1011"
Output: "10101"
  • Constraints:
    • 1 <= a.length, b.length <= 104
    • a and b consist only of '0' or '1' characters.
    • Each string does not contain leading zeros except for the zero itself.
  • See problem on LeetCode.

Solution: Pop digits from each number, add them and do

  • The overall idea is to make up the short two strings with 00 to make the two strings have the same length, and then traverse and calculate from the end to get the final result.

  • Let’s understand with an example: Addition of 1 and 1 will lead to carry 1 and print 0, Addition of 1 and 0 give us 1 as carry will lead print 0, Addition of last remaining carry 1 with no body will lead to print 1, so, we get something like “1 0 0” as answer
  • Since the addition will be 3 then print 1 and carry will remain 1.

  • Detailed Explanation:

  • So, first do we understand how do we perform binary addition. Take an example, given two numbers “11” + “1” where “11” is representing “3” & “1” is “1”, in decimal form. Now let’s perform binary addition it’s very similar to the decimal addition that we do. In decimal what we do we add 2 numbers & if it goes beyond 9 we take a carry. And here also we have a number in range 0 - 1, 2 values over here & in Decimal range is 0 - 9, 10 values are there. So, in binary what it means is if result more than 1, there is a carry otherwise no carry. Let me show you in diagram:

  • Per the above diagram, initially carry is 0, then we add 1 + 1 and get 2 which is more than 1, so there is a carry of 1 and result is 0. Now we have carry of 1, again 1 + 1 is 0, and still left with carry of 1. And the last carry one will be return as it is.
  • So, if you see this binary number it is [2^2 * 1 + 2^1 * 0 + 2^0 * 0] and this is the decimal conversion of [1 0 0] which is 4.

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        # output string
        res = ""
        
        # i and j are pointers for string a and b respectively which are used to start adding from right to left
        i, j, carry = len(a) - 1, len(b) - 1, 0
        
        while i >= 0 or j >= 0:
            sum = carry;
            
            # we subtract by '0' to convert the numbers from a char type into an int, so we can perform operations on the numbers
            if i >= 0: sum += ord(a[i]) - ord('0') # ord is use to get the ASCII value of the input character
            if j >= 0: sum += ord(b[j]) - ord('0')
            i, j = i - 1, j - 1
            
            # obtaining a carry depends on the quotient we get by dividing sum / 2 that will be our carry. Carry could be either 1 or 0.
            # if sum is 0, res is 1, then carry would be 0
            # if sum is 1, res is 1, then carry would be 0
            # if sum is 2, res is 0, then carry would be 1
            # if sum is 3, res is 1, then carry would be 1            
            carry = 1 if sum > 1 else 0
            
            # getting remainder and adding it into our result
            res += str(sum % 2)

        # add to res until carry becomes 0
        if carry != 0 : 
            res += str(carry)
            
        # you added the bits from R to L, but need the number from L to R 
        # so reverse the answer and return
        return res[::-1]
Complexity
  • Time: \(O(max(m, n))\) where \(m\) and \(n\) are the length of strings \(a\) and \(b\)
  • Space: \(O(max(m, n))\) due to the res object

Solution: Pop digits from each number, add them and add (sum%2) to result and get carry if

class Solution:
    def addBinary(self, a: str, b: str) -> str:
        carry = 0
        result = []

        a = list(a)
        b = list(b)

        while a or b or carry:
            if a:
                # using pop(), we are reading from the last digit to the first
                carry += int(a.pop())
            if b:
                carry += int(b.pop())

            result += str(carry % 2) # result has to be {0, 1}
            carry = 1 if carry > 1 else 0 # in-place integer division; can also do "carry //= 2" as above

        # you added the bits from R to L, but need the number from L to R 
        # so reverse the answer and return
        return ''.join(result)[::-1]
Complexity
  • Time: \(O(max(m, n))\) where \(m\) and \(n\) are the length of strings \(a\) and \(b\)
  • Space: \(O(max(m, n))\) due to the res object

[88/Easy] Merge Sorted Arrays

Problem

  • You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.

  • Merge nums1 and nums2 into a single array sorted in non-decreasing order.

  • The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

  • Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.
  • Example 2:
Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].
  • Example 3:
Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.
  • Constraints:
    • nums1.length == m + n
    • nums2.length == n
    • 0 <= m, n <= 200
    • 1 <= m + n <= 200
    • -109 <= nums1[i], nums2[j] <= 109
  • Follow up: Can you come up with an algorithm that runs in O(m + n) time?

  • See problem on LeetCode.

Solution: Start at the end of both arrays, get bigger element

class Solution(object):
    def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
        """
        Do not return anything, modify nums1 in-place instead.
        """
        while m > 0 and n > 0:
            # Start at the end of both arrays and get the bigger element.
            if nums1[m - 1] > nums2[n - 1]:
                nums1[m + n - 1] = nums1[m - 1]
                m -= 1
            else:
                nums1[m + n - 1] = nums2[n - 1]
                n -= 1
        nums1[:n] = nums2[:n] # while n > 0:
                              #     nums1[n-1] = nums2[n-1]
                              #     n -= 1
Complexity
  • Time: \(O(m + n + n) = O(m + n)\)
  • Space: \(O(1)\) since nothing new is created (the result is calculated in-place).

[100/Easy] Same Tree

Problem

  • Given the roots of two binary trees p and q, write a function to check if they are the same or not.

  • Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

  • Example 1:

Input: p = [1,2,3], q = [1,2,3]
Output: true
  • Example 2:

Input: p = [1,2], q = [1,null,2]
Output: false
  • Example 3:

Input: p = [1,2,1], q = [1,1,2]
Output: false
  • Constraints:
    • The number of nodes in both trees is in the range [0, 100].
    • -104 <= Node.val <= 104
  • See problem on LeetCode.

Solution: Iterative DFS with Deque

  • Iteration is actually better than recursion in best case scenario, as it stops checking the tree whenever it finds the first mismatching node, but note that in Big O notation they are the same.
from collections import deque
class Solution:
    def isSameTree(self, p, q):
        """
        :type p: TreeNode
        :type q: TreeNode
        :rtype: bool
        """    
        def check(p, q):
            # if both are None
            if not p and not q:
                return True
            # one of p and q is None
            if not q or not p:
                return False
            if p.val != q.val:
                return False
            return True
        
        deq = deque([(p, q),])
        while deq:
            p, q = deq.popleft()
            if not check(p, q):
                return False
            
            if p:
                deq.append((p.left, q.left))
                deq.append((p.right, q.right))
                    
        return True

Solution: Pre-order Recursive DFS

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        # base case: check whether both nodes are null, if so return true.
        if not p and not q:
            return True

        # base case: if one node is null and the other isn't, the trees can't be the same.
        # note that if you've reached here both p and q can't be null (because of the if condition above)
        # but one can be null.
        if not q or not p: # or if not q and p or q and not p:
            return False
                
        # base case: if the node values differ, return false.
        if p.val != q.val:
            return False

        # normal case
        return self.isSameTree(p.right, q.right) and self.isSameTree(p.left, q.left)
Complexity
  • Time: \(O(n)\) where \(n\) is the number of nodes in the binary tree
  • Space: \(O(1)\)

[114/Medium] Flatten Binary Tree to Linked List

Problem

  • Given the root of a binary tree, flatten the tree into a “linked list”:

    • The “linked list” should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
    • The “linked list” should be in the same order as a pre-order traversal of the binary tree.
  • Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]
  • Example 2:
Input: root = []
Output: []
  • Example 3:
Input: root = [0]
Output: [0]
  • Constraints:
    • The number of nodes in the tree is in the range [0, 2000].
    • -0 -100 <= Node.val <= 100
  • See problem on LeetCode.

Solution: Morris Traversal

  • There is actually a way to traverse a binary tree with a space complexity of O(1) while staying at a time complexity of O(N), though it does require modifying the tree’s structure. In this problem that’s specifically being called for, so it’s a valid approach, though it won’t always be appropriate to modify the source binary tree in other situations.

  • The approach is called the Morris traversal. At its heart, it takes advantage of the basic nature of ordered traversals to iterate through and unwind the tree. In a pre-order traversal of a binary tree, each vertex is processed in (node, left, right) order. This means that the entire left subtree could be placed between the node and its right subtree.

  • To do this, however, we’ll first have to locate the last node in the left subtree. This is easy enough, since we know that the last node of a pre-order tree can be found by moving right as many times as possible from its root.

  • So we should be able to move through the binary tree, keeping track of the curent node (curr). Whenever we find a left subtree, we can dispatch a runner to find its last node, then stitch together both ends of the left subtree into the right path of curr, taking heed to sever the left connection at curr.

  • Once that’s done, we can continue to move curr to the right, looking for the next left subtree. When curr can no longer move right, the tree will be successfully flattened.

class Solution:
    def flatten(self, root: TreeNode) -> None:
        curr = root
        while curr:
            if curr.left:
                runner = curr.left
                while runner.right: runner = runner.right
                runner.right, curr.right, curr.left = curr.right, curr.left, None
            curr = curr.right
Complexity
  • Time: \(O(n)\) where \(n\) is the number of nodes in the binary tree
  • Space: \(O(1)\)

Solution: \(O(1)\) Space

  • In order to properly connect the linked list, we’ll need to start at the bottom and work up. This means that we’ll need to move in reverse pre-order traversal order through the binary tree. Since pre-order traversal is normally “node, left, right”, we’ll have to move in the reverse order of “right, left, node”.

  • In order to complete this solution in \(O(1)\) space, we won’t be able to conveniently backtrack via a stack, so the key to this solution will be to retreat all the way back up to the root each time we reach a leaf. This will push the time complexity to \(O(n^2)\).

  • We’ll want to first set up head and curr to keep track of the head of the linked list we’re building and the current node we’re visiting. We’ll know we’re finished once head = root.

  • To follow the reverse pre-order traversal order, we’ll first attempt to go right and then left. Since we’re backtracking to root, however, we’ll eventually run back into the same node that we’ve set as head doing this. To prevent this, we’ll stop before moving to the head node and sever the connection.

  • Now that we can’t run into already-completed territory, we can be confident that any leaf we move to must be the next value for head, so we should connect it to the old head, update head, and reset back to the root.

  • As noted before, once head = root, we’ve finished our traversal and can exit the function.

class Solution:
    def flatten(self, root: TreeNode) -> None:
        head, curr = None, root
        while head != root:
            if curr.right == head: curr.right = None
            if curr.left == head: curr.left = None
            if curr.right: curr = curr.right
            elif curr.left: curr = curr.left
            else: curr.right, head, curr = head, curr, root
Complexity
  • Time: \(O(n^2)\) where \(n\) is the number of nodes in the binary tree, due to repeated backtracking to root
  • Space: \(O(1)\)

Solution: Recursive reversed preorder traversal

  • The trick is to consider the order of the traversal – traverse the graph in reverse preorder: (right, left, node). At any given node, solve the right subtree first. At the end of the right subtree, self.prev points to the start of the right part. Then solve the left subtree. Notice that at the beginning of the left part, self.prev still points to the start of the right part. Finally when the left part is done, self.prev points to the start of the left part.
  • Finally, return the top of the computed solution, which is done using the last three lines. The next node to left-append to the computed solution is the root. First, after root, comes self.prev. Second, root.left = None, which is straightforward. Finally, self.prev is set to root, because self.prev holds the top of the computed tree up to this point.
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution(object):
    def __init__(self):
        self.prev = None
            
    def flatten(self, root: Optional[TreeNode]) -> None:
        if not root:
            return None
            
        self.flatten(root.right)
        self.flatten(root.left)
        
        root.right = self.prev
        root.left = None
        self.prev = root
  • Walkthrough:

          root
          / 
        1 
       / \ 
      3  4
    
    • Let’s see what is happening with this code.
      • Step 1:
        • Node(4).right = None
        • Node(4).left = None
        • prev = Node(4)
      • Step 2:
        • Node(3).right = Node(4) (prev)
        • Node(3).left = None
        • prev = Node(3)->Node(4)
      • Step 3:
        • Node(1).right = prev = Node(3) -> Node(4)
        • Node(1).left = None
        • prev = Node(1)->Node(3)->Node(4) (which is the answer)
      • The answer use self.prev to recode the ordered tree of the right part of current node.
      • Remove the left part of current node.
  • Same approach; rehashed:

    • In order to properly connect the linked list, we’ll need to start at the bottom and work up. This means that we’ll need to move in reverse pre-order traversal order through the binary tree. Since pre-order traversal is normally “node, left, right”, we’ll have to move in the reverse order of “right, left, node”.

    • Binary tree traversal is prime ground for a recursive solution, so let’s define a helper (revPreOrder) for the purpose. We’ll also keep a global variable head to keep track of the head of the linked list as we work our way backwards.

    • Per our reverse pre-order traversal approach, we want to recursively work down the right path first then the left path, if they exist. Once we’ve flattened the left and right paths recursively, head should at this point be equal to the next node after the current one, so we should set it as node.right. We shouldn’t forget to set node.left to None, as well.

    • Once we’re done with the current node, we can update head to node and allow the recursion to complete and move back up to the next layer. Once the recursion stack is exhausted, head will be equal to root again.

    • Lastly, we have to deal with an edge case of an empty root, so we can just make sure to only call the initial recursion on root if root actually is a node. There is no need for a return statement if root is None, because the absence of one will return None automatically to make the test suite happy.

class Solution:
    head = None
    
    def flatten(self, root: TreeNode) -> None:
        def revPreOrder(node: TreeNode) -> None:
            if node.right: revPreOrder(node.right)
            if node.left: revPreOrder(node.left)
            node.left, node.right, self.head = None, self.head, node
        
        # note that there is no need for a return statement if `root` is `None`, 
        # because the absence of one will return `None` automatically to make the test suite happy
        if root: 
            revPreOrder(root)
Complexity
  • Time: \(O(n)\) where \(n\) is the number of nodes in the binary tree
  • Space: \(O(n)\) for the recursion stack, which is as long as the maximum depth of the binary tree, which can go up to \(n\)

[134/Medium] Gas Station

Problem

  • There are n gas stations along a circular route, where the amount of gas at the i-th station is gas[i].

  • You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the i-th station to its next (i + 1)-th station. You begin the journey with an empty tank at one of the gas stations.

  • Given two integer arrays gas and cost, return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

  • Example 1:

Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation:
Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.
  • Example 2:
Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.
  • Constraints:
    • n == gas.length == cost.length
    • 1 <= n <= 105
    • 0 <= gas[i], cost[i] <= 104
  • See problem on LeetCode.

Solution: One loop

  • Intuition:
    • The first idea is to check every single station:
      • Choose the station as starting point.
      • Perform the road trip and check how much gas we have in tank at each station.
    • That means \(\mathcal{O}(N^2)\) time complexity, and for sure one could do better.
  • Let’s notice two things.
    • It’s impossible to perform the road trip if sum(gas) < sum(cost). In this situation the answer is -1.

    • One could compute total amount of gas in the tank total_tank = sum(gas) - sum(cost) during the round trip, and then return -1 if total_tank < 0.

    • It’s impossible to start at a station i if gas[i] - cost[i] < 0, because then there is not enough gas in the tank to travel to i + 1 station.

    • The second fact could be generalized. Let’s introduce curr_tank variable to track the current amount of gas in the tank. If at some station curr_tank is less than 0, that means that one couldn’t reach this station.
    • Next step is to mark this station as a new starting point, and reset curr_tank to zero since one starts with no gas in the tank.
  • Algorithm:
    • Now the algorithm is straightforward:
      1. Initiate total_tank and curr_tank as zero, and choose station 0 as a starting station.
      2. Iterate over all stations:
        • Update total_tank and curr_tank at each step, by adding gas[i] and subtracting cost[i].
        • If curr_tank < 0 at i + 1 station, make i + 1 station a new starting point and reset curr_tank = 0 to start with an empty tank.
      3. Return -1 if total_tank < 0 and starting station otherwise.
class Solution:
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        n = len(gas)
        
        total_tank, curr_tank = 0, 0
        starting_station = 0
        for i in range(n):
            total_tank += gas[i] - cost[i]
            curr_tank += gas[i] - cost[i]
            # If one couldn't get here,
            if curr_tank < 0:
                # Pick up the next station as the starting one.
                starting_station = i + 1
                # Start with an empty tank.
                curr_tank = 0
        
        return starting_station if total_tank >= 0 else -1
Complexity
  • Time: \(O(n)\) since there is only one loop over all stations here.
  • Space: \(O(1)\) since it’s a constant space solution.

[138/Medium] Copy List with Random Pointer

Problem

  • A linked list of length n is given such that each node contains an additional random pointer, which could point to any node in the list, or null.

  • Construct a deep copy of the list. The deep copy should consist of exactly n brand new nodes, where each new node has its value set to the value of its corresponding original node. Both the next and random pointer of the new nodes should point to new nodes in the copied list such that the pointers in the original list and copied list represent the same list state. None of the pointers in the new list should point to nodes in the original list.

  • For example, if there are two nodes X and Y in the original list, where X.random --> Y, then for the corresponding two nodes x and y in the copied list, x.random --> y.

  • Return the head of the copied linked list.

  • The linked list is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:
    • val: an integer representing Node.val
    • random_index: the index of the node (range from 0 to n-1) that the random pointer points to, or null if it does not point to any node.
  • Your code will only be given the head of the original linked list.

  • Example 1:
Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]
  • Example 2:
Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]
  • Example 3:
Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]
  • Constraints:
    • 0 <= n <= 1000
    • -104 <= Node.val <= 104
    • Node.random is null or is pointing to some node in the linked list.
  • See problem on LeetCode.

Solution: Two loops

  • The first while loop builds the mapping of the original object to its copy while the second loop, we basically connect stuff, i.e., next and random using the mapping:
# Definition for a Node.
class Node:
    def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
        self.val = int(x)
        self.next = next
        self.random = random
        
class Solution:
    def copyRandomList(self, head: 'Node') -> 'Node':        
        # build mapping of original node to it's copy
        current = head
        mapping = {}
        while current:
            mapping[current] = Node(current.val)
            current = current.next
            
        # now that we have the complete mapping, we can use it to
        # construct the linked list copy
        current = head
        while current:
            # we have all the copy nodes but they aren't connected i.e. the next
            # & random all point to None. let's connect them here. it's easy
            # since we have a complete node-to-copy mapping
            if current.next:
                mapping[current].next = mapping[current.next]
            if current.random:
                mapping[current].random = mapping[current.random]
            current = current.next
            
        # return the head of the linked list copy
        return mapping[head]
Complexity
  • Time: \(O(n + n) = O(n)\)
  • Space: \(O(n)\)

Solution: Recursion

# Definition for a Node.
class Node:
    def __init__(self, x: int, next: 'Node' = None, random: 'Node' = None):
        self.val = int(x)
        self.next = next
        self.random = random
        
def copyRandomList(self, head: 'Optional[Node]', seen={None: None}) -> 'Optional[Node]':
    # Node could have been seen earlier through either random or next pointers
    if head not in seen:
        seen[head] = Node(head.val)
        seen[head].next = self.copyRandomList(head.next, seen)
        seen[head].random = self.copyRandomList(head.random, seen)
    return seen[head]
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[143/Medium] Reorder List

  • You are given the head of a singly linked-list. The list can be represented as:
L0 -> L1 -> … -> Ln - 1 -> Ln
  • Reorder the list to be on the following form:
L0 -> Ln -> L1 -> Ln - 1 -> L2 -> Ln - 2 -> ...
  • You may not modify the values in the list’s nodes. Only nodes themselves may be changed.

  • Example 1:

Input: head = [1,2,3,4]
Output: [1,4,2,3]
  • Example 2:

Input: head = [1,2,3,4,5]
Output: [1,5,2,4,3]
  • Constraints:
    • The number of nodes in the list is in the range [1, 5 * 104].
    • 1 <= Node.val <= 1000
  • See problem on LeetCode.

Solution: Find middle; reverse linked list; merged sorted linked lists

  • This problem is a combination of these three easy problems:

    • Middle of the Linked List.
    • Reverse Linked List.
    • Merge Two Sorted Lists.
  • Overview:

    • Find a middle node of the linked list. If there are two middle nodes, return the second middle node. Example: for the list 1->2->3->4->5->6, the middle element is 4.
    • Once a middle node has been found, reverse the second part of the list. Example: convert 1->2->3->4->5->6 into 1->2->3 and 6->5->4.
    • Now merge the two sorted lists. Example: merge 1->2->3->4 and 6->5->4 into 1->6->2->5->3->4.

  • Now let’s check each algorithm part in more detail.

  • Find a Middle Node:

    • Let’s use two pointers, slow and fast. While the slow pointer moves one step forward slow = slow.next, the fast pointer moves two steps forward fast = fast.next.next, i.e., fast traverses twice as fast as slow. When the fast pointer reaches the end of the list, the slow pointer should be in the middle.

# find the middle of linked list [Problem 876]
# in 1->2->3->4->5->6 find 4 
slow = fast = head
while fast and fast.next:
    slow = slow.next
    fast = fast.next.next 
  • Reverse the Second Part of the List:
    • Let’s traverse the list starting from the middle node slow and its virtual predecessor None. For each current node, save its neighbors: the previous node prev and the next node tmp = curr.next.
    • While you’re moving along the list, change the node’s next pointer to point to the previous node: curr.next = prev, and shift the current node to the right for the next iteration: prev = curr, curr = tmp.

# reverse the second part of the list [Problem 206]
# convert 1->2->3->4->5->6 into 1->2->3->4 and 6->5->4
# reverse the second half in-place
prev, curr = None, slow
while curr:
    tmp = curr.next
    
    curr.next = prev
    prev = curr
    curr = tmp        
  • There is a more elegant way to do it in Python:
# reverse the second part of the list [Problem 206]
# convert 1->2->3->4->5->6 into 1->2->3->4 and 6->5->4
# reverse the second half in-place
prev, curr = None, slow
while curr:
    curr.next, prev, curr = prev, curr, curr.next      
  • Merge Two Sorted Lists:

    • This algorithm is similar to the one for list reversal.

    • Let’s pick the first node of each list - first and second, and save their successors. While you’re traversing the list, set the first node’s next pointer to point to the second node, and the second node’s next pointer to point to the successor of the first node. For this iteration the job is done, and for the next iteration move to the previously saved nodes’ successors.

# merge two sorted linked lists [Problem 21]
# merge 1->2->3->4 and 6->5->4 into 1->6->2->5->3->4
first, second = head, prev
while second.next:
    tmp = first.next
    first.next = second
    first = tmp
    
    tmp = second.next
    second.next = first
    second = tmp
  • Once again, there is a way to make things simple in Python:
# merge two sorted linked lists [Problem 21]
# merge 1->2->3->4 and 6->5->4 into 1->6->2->5->3->4
first, second = head, prev
while second.next:
    first.next, first = second, first.next
    second.next, second = first, second.next
  • Implementation:
    • Now it’s time to put all the pieces together.
class Solution:
    def reorderList(self, head: ListNode) -> None:
        if not head:
            return 
        
        # find the middle of linked list [Problem 876]
        # in 1->2->3->4->5->6 find 4 
        slow = fast = head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next 
            
        # reverse the second part of the list [Problem 206]
        # convert 1->2->3->4->5->6 into 1->2->3->4 and 6->5->4
        # reverse the second half in-place
        prev, curr = None, slow
        while curr:
            curr.next, prev, curr = prev, curr, curr.next       

        # merge two sorted linked lists [Problem 21]
        # merge 1->2->3->4 and 6->5->4 into 1->6->2->5->3->4
        first, second = head, prev
        while second.next:
            first.next, first = second, first.next
            second.next, second = first, second.next
  • Similar approach; using functions for each of the three sub-tasks:
class Solution:
  def reorderList(self, head: ListNode) -> None:
    def findMid(head: ListNode):
      prev = None
      slow = head
      fast = head

      while fast and fast.next:
        prev = slow
        slow = slow.next
        fast = fast.next.next
      prev.next = None

      return slow

    def reverse(head: ListNode) -> ListNode:
      prev = None
      curr = head

      while curr:
        next = curr.next
        curr.next = prev
        prev = curr
        curr = next

      return prev

    def merge(l1: ListNode, l2: ListNode) -> None:
      while l2:
        next = l1.next
        l1.next = l2
        l1 = l2
        l2 = next

    if not head or not head.next:
      return

    mid = findMid(head)
    reversed = reverse(mid)
    merge(head, reversed)
Complexity
  • Time: \(O(n)\). There are three steps here. To identify the middle node takes \(\mathcal{O}(N)\) time. To reverse the second part of the list, one needs \(N/2\) operations. The final step, to merge two lists, requires \(N/2\) operations as well. In total, that results in \(\mathcal{O}(N)\) time complexity.
  • Space: \(O(1)\), since we do not allocate any additional data structures.

[148/Medium] Sort List

  • Given the head of a linked list, return the list after sorting it in ascending order.

  • Example 1:

Input: head = [4,2,1,3]
Output: [1,2,3,4]
  • Example 2:

Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
  • Example 3:
Input: head = []
Output: []
  • Constraints:
    • The number of nodes in the list is in the range [0, 5 * 10^4].
    • -10^5 <= Node.val <= 10^5
  • Follow up: Can you sort the linked list in \(O(n\log{n})\) time and \(O(1)\) memory (i.e. constant space)?

  • See problem on LeetCode.

Solution: Merge Sort / Divide-and-Conquer

class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        def merge( a, b):            
            if not a:
                return b
            
            elif not b:
                return a
            
            if a.val < b.val:
                a.next = merge(a.next, b)
                return a
                
            else:
                b.next = merge(a, b.next)
                return b
        
        ## base case        
        if head is None:
            # empty node
            return None
        
        elif head.next is None:
            # one node only
            return head
        
        ## general case
        # divide into two halves
        pre, slow, fast = None, head, head
        
        while fast and fast.next:
            pre, slow, fast = slow, slow.next, fast.next.next
        
        pre.next = None

        # sort by divide-and-conquer        
        first_half = self.sortList(head)
        second_half = self.sortList(slow)
        result = merge(first_half, second_half)
        return result
Complexity
  • Time: \(O(n\log{n})\)
  • Space: \(O(1)\)

[158/Hard] Read N Characters Given read4 II - Call Multiple Times

  • Given a file and assume that you can only read the file using a given method read4, implement a method read to read n characters. Your method read may be called multiple times.

  • Method read4:

    • The API read4 reads four consecutive characters from file, then writes those characters into the buffer array buf4.

    • The return value is the number of actual characters read.

    • Note that read4() has its own file pointer, much like FILE *fp in C.

  • Definition of read4:

Parameter:  char[] buf4    
Returns:    int

buf4[] is a destination, not a source. The results from read4 will be copied to buf4[].
  • Below is a high-level example of how read4 works:

File file("abcde"); // File is "abcde", initially file pointer (fp) points to 'a'
char[] buf4 = new char[4]; // Create buffer with enough space to store characters
read4(buf4); // read4 returns 4. Now buf4 = "abcd", fp points to 'e'
read4(buf4); // read4 returns 1. Now buf4 = "e", fp points to end of file
read4(buf4); // read4 returns 0. Now buf4 = "", fp points to end of file
  • Method read:

  • By using the read4 method, implement the method read that reads n characters from file and store it in the buffer array buf. Consider that you cannot manipulate file directly.

  • The return value is the number of actual characters read.

  • Definition of read:

Parameters: char[] buf, int n
Returns:    int

buf[] is a destination, not a source. You will need to write the results to buf[].
  • Note:
    • Consider that you cannot manipulate the file directly. The file is only accessible for read4 but not for read.
    • The read function may be called multiple times.
    • Please remember to RESET your class variables declared in Solution, as static/class variables are persisted across multiple test cases. Please see here for more details.
    • You may assume the destination buffer array, buf, is guaranteed to have enough space for storing n characters.
    • It is guaranteed that in a given test case the same buffer buf is called by read.
  • Example 1:
Input: file = "abc", queries = [1,2,1]
Output: [1,2,0]
Explanation: The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 1); // After calling your read method, buf should contain "a". We read a total of 1 character from the file, so return 1.
sol.read(buf, 2); // Now buf should contain "bc". We read a total of 2 characters from the file, so return 2.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
Assume buf is allocated and guaranteed to have enough space for storing all characters from the file.
  • Example 2:
Input: file = "abc", queries = [4,1]
Output: [3,0]
Explanation: The test case represents the following scenario:
File file("abc");
Solution sol;
sol.read(buf, 4); // After calling your read method, buf should contain "abc". We read a total of 3 characters from the file, so return 3.
sol.read(buf, 1); // We have reached the end of file, no more characters can be read. So return 0.
  • Constraints:
    • 1 <= file.length <= 500
    • file consist of English letters and digits.
    • 1 <= queries.length <= 10
    • 1 <= queries[i] <= 500
  • See problem on LeetCode.

Solution: Queue

class Solution:
# @param buf, Destination buffer (a list of characters)
# @param n,   Maximum number of characters to read (an integer)
# @return     The number of characters read (an integer)
    def __init__(self):
        self.queue = []

    def read(self, buf, n):
        idx = 0
        
        while True:
            buf4 = [""]*4
            l = read4(buf4)
            self.queue.extend(buf4)
            curr = min(len(self.queue), n-idx)
            
            for i in xrange(curr):
                buf[idx] = self.queue.pop(0)
                idx += 1
                
            if curr == 0:
                break 
                
        return idx
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[203/Easy] Remove Linked List Elements

Problem

  • Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head.

  • Example 1:

Input: head = [1,2,6,3,4,5,6], val = 6
Output: [1,2,3,4,5]
  • Example 2:
Input: head = [], val = 1
Output: []
  • Example 3:
Input: head = [7,7,7,7], val = 7
Output: []
  • Constraints:
    • The number of nodes in the list is in the range [0, 104].
    • 1 <= Node.val <= 50
    • 0 <= val <= 50
  • See problem on LeetCode.

Solution: Iteration

  • Edge cases:
    • The linked list is empty, i.e., the head node is None.
    • Multiple nodes with the target value in a row.
    • The head node has the target value.
    • The head node, and any number of nodes immediately after it have the target value.
    • All of the nodes have the target value.
    • The last node has the target value.
  • First we remove all (if any) target nodes from the beginning (we do it because the removing logic is slightly different compared to when the head node is not the node requiring removal).
  • After that we just loop over all nodes, if the next one is one that should be removed, just get it out of the list by moving the next pointer to the next.next node. Otherwise just move along the list.
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def removeElements(self, head, val):
         # empty node
         if not head:
            return None
                  
        # move until the first node is not val                   
        while head is not None and head.val == val:
            head = head.next
        
        # start with non-val    
        current = head
        while current is not None:
            if current.next is not None and current.next.val == val:
                # current_node.next needs removing so we simply replace it with current_node.next.next
                # We know this is always "safe", because current_node.next is definitely not None (the loop condition ensures that), so we can safely access its next
                current.next = current.next.next
            else:
                # we know that current_node.next should be kept, and so we move current_node on to be current_node.next
                current = current.next
        
        return head
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Recursion

  • The idea is to change temp with prev, prev with current, and current with temp.
class Solution:
    def reverseList(self, head: ListNode, prev=None) -> ListNode:
        if not head:
            return prev
            
        temp = head.next
        head.next = prev
        
        return self.reverseList(temp, head)     
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[206/Easy] Reverse Linked List

Problem

  • Given the head of a singly linked list, reverse the list, and return the reversed list.

  • Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
  • Example 2:
Input: head = [1,2]
Output: [2,1]
  • Example 3:
Input: head = []
Output: []
  • Constraints:
    • The number of nodes in the list is the range [0, 5000].
    • -5000 <= Node.val <= 5000
  • Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

  • See problem on LeetCode.

Solution: Iteration

  • The idea is to change temp with prev, prev with current, and current with temp.
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]:
        prev = None
        curr = head
        
        while curr:
            temp = curr.next
            curr.next = prev
            prev = curr
            curr = temp
        
        return prev
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Recursion

  • The idea is to change temp with prev, prev with current, and current with temp.
class Solution:
    def reverseList(self, head: ListNode, prev=None) -> ListNode:
        if not head:
            return prev
            
        temp = head.next
        head.next = prev
        
        return self.reverseList(temp, head)     
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[214/Hard] Shortest Palindrome

Problem

  • You are given a string s. You can convert s to a palindrome by adding characters in front of it.

  • Return the shortest palindrome you can find by performing this transformation.

  • Example 1:

Input: s = "aacecaaa"
Output: "aaacecaaa"
  • Example 2:
Input: s = "abcd"
Output: "dcbabcd"
  • Constraints:
    • 0 <= s.length <= 5 * 104
    • s consists of lowercase English letters only.
  • See problem on LeetCode.
Solution: Reverse string, find the suffix of the overlapping palindrome window, add it as a prefix to the original string
class Solution:
    def shortestPalindrome(self, s):
        # edge cases
        if not s or len(s) == 1 or s == s[::-1]:
            return s    
        
        # reverse string            
        r = s[::-1]
        
        for i in range(len(s) + 1):
            # strip characters from the end of s, i.e., s[:len(s)-i]
            # strip characters from the start of r, i.e., r[:i]
            # find a match which will yield the overlapping palindrome window in s and r
            # (note that the r[i:] being compared is a suffix here)
            # once found, add the prefix r[:i] to s
            if s[:len(s)-i] == r[i:]:
                # add to the start of the string
                return r[:i] + s
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[228/Easy] Summary Ranges

Problem

  • You are given a sorted unique integer array nums.

  • A range [a,b] is the set of all integers from a to b (inclusive).

  • Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

  • Each range [a,b] in the list should be output as:

    • "a->b" if a != b
    • "a" if a == b
  • Example 1:

Input: nums = [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: The ranges are:
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"
  • Example 2:
Input: nums = [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: The ranges are:
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"
  • Constraints:
    • 0 <= nums.length <= 20
    • -2^31 <= nums[i] <= 2^31 - 1
    • All the values of nums are unique.
    • nums is sorted in ascending order.
  • See problem on LeetCode.

Solution: Hold the current range in an extra variable r, format and return them

  • A variation of the last solution, holding the current range in an extra variable r to make things easier. Note that r contains at most two elements, so the in-check takes constant time.
class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        ranges, r = [], []
        
        for n in nums:
            if n-1 not in r:
                r = []
                ranges += r,
            r[1:] = n,
            
        return ['->'.join(map(str, r)) for r in ranges] # or map('->'.join, ranges)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: A tricky short version for Python 2

  • Note that the backtick (``) operator was removed in Python 3 since it is confusingly similar to a single quote. Instead of the backtick, use the equivalent built-in function repr().
def summaryRanges(self, nums):
    ranges = r = []
    for n in nums:
        if `n-1` not in r:
            r = []
            ranges += r,
        r[1:] = `n`,
    return map('->'.join, ranges)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Check if the last number and current number differ by >1; if so, add it to the output

class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        begin, res = 0, []
        strout = lambda b, e: str(b) + "->" + str(e) if b != e else str(b)
        
        # run till "len(nums)+1" to accommodate the last number in the input
        for i in range(1, len(nums)+1):
            # The "i == len(nums)" check is for when we are beyond the last 
            # index of nums (i.e., i = len(nums)), in which case we need to 
            # write the output for the actual last index "nums[i-1]"
            if i == len(nums) or nums[i] - nums[i-1] != 1:
                res.append(strout(nums[begin], nums[i-1]))
                begin = i
        
        return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: Collect the ranges, then format and return them

  • A variation of the first solution, without needing to hold the current range in an extra variable r.
class Solution:
    def summaryRanges(self, nums: List[int]) -> List[str]:
        ranges = []
        
        for n in nums:
            if not ranges or n > ranges[-1][-1] + 1:
                # add a new range list either if ranges is empty ([]) or if the current integer is greater than the previous by atleast 2
                ranges += [],
            # note that due to slicing below, the IndexError is avoided
            # without slicing you get an IndexError: list assignment index out of range 
            ranges[-1][1:] = n,
            # the above statement exercises two unique python features.
            # to understand them, here's some examples:
            # L = [1,2]
            # L[2] = 3
            # Traceback (most recent call last):
            #   File "<stdin>", line 1, in <module>
            # IndexError: list assignment index out of range
            # L[2:] = 3
            #Traceback (most recent call last):
            #  File "<stdin>", line 1, in <module>
            #TypeError: can only assign an iterable
            # L[2:] = 3, # L = [1,2,3]
            
        # note that .join() requires an iterable with strings, so need to cast the range points to str
        return ['->'.join([str(x) for x in r]) for r in ranges] # or ['->'.join(map(str, r)) for r in ranges]
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[238/Medium] Product of Array Except Self

Problem

  • Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].

  • The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.

  • You must write an algorithm that runs in O(n) time and without using the division operation.

  • Example 1:

Input: nums = [1,2,3,4]
Output: [24,12,8,6]
  • Example 2:
Input: nums = [-1,1,0,-3,3]
Output: [0,0,9,0,0]
  • Constraints:
    • 2 <= nums.length <= 105
    • -30 <= nums[i] <= 30
    • The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
  • Follow up: Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.)

  • See problem on LeetCode.

Solution: Brute-force

def productExceptSelf(nums):
    res = []
    
    for i in range(len(nums)):
        product = 1
        for j in range(len(nums)):
            if j == i:
                continue
            product *= nums[j]
        res.append(product)
    
    return res
Complexity
  • Time: \(O(n^2)\)
  • Space: \(O(n)\)

Solution: Two pass with division

  • Note that division is not allowed per the question (this is just for reference).
def productExceptSelf1(nums):
    res = []
    product = 1
    
    for i in range(len(nums)):
        product *= nums[i]
    
    for j in range(len(nums)):
        res.append(int(product/nums[j]))
    
    return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: Two-pass (L->R, R->L); precalculate the prefix and postfix products for each element, then multiply the corresponding prefix and postfix

  • The product of elements except the i-th one is equal to a product of elements on the left side and on the right side of that element. So the idea is to do two passes over the input nums and use an auxiliary list res to store intermediate results.
  • For the first pass, we go from left to right, i.e., from the start to the end of nums, and on each iteration we store the accumulated product in the res at the according index such that the value of the res at the i-th index equal to product of all elements in nums starting from 0 to i-1, i.e., res[i] = product(nums[0:i-1]).
  • For the second pass, we go from right to left, i.e., from the end to the start of nums.
  • We accumulate a product of the prefix and postfix for each element by modifying res such that the value at the index i is equal to the product of the accumulated prefix at this index and the accumulated postfix. In the end, res is the required answer.
class Solution:
    def productExceptSelf(self, nums: List[int]) -> List[int]:
        prefix = 1
        res = []
        
        # left to right        
        for num in nums:
            res.append(prefix)
            prefix *= num
        
        # right to left        
        postfix = 1
        for reverseIndex in range(len(nums)-1, -1,-1): # or for reverseIndex in reversed(range(len(nums))):
            res[reverseIndex] *= postfix
            postfix *= nums[reverseIndex]
        
        return res        
Complexity
  • Time: \(O(n)\) for two passes
  • Space: \(O(n)\)

[252/Easy] Meeting Rooms

Problem

  • Given an array of meeting time intervals where intervals[i] = [start_i, end_i], determine if a person could attend all meetings.

  • Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: false
  • Example 2:
Input: intervals = [[7,10],[2,4]]
Output: true
  • Constraints:
    • 0 <= intervals.length <= 104
    • intervals[i].length == 2
    • 0 <= starti < endi <= 106
  • See problem on LeetCode.

Solution: Detect overlap by comparing start of current interval with end of prior interval

class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort() # or the verbose: intervals.sort(key=lambda x: x[0])
        
        for i in range(1, len(intervals)):
            if intervals[i][0] < intervals[i-1][1]:
                return False
            
        return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Detect overlap by comparing end of current interval with start of next interval

class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort() # or the verbose: intervals.sort(key=lambda x: x[0])
        
        for i in range(len(intervals) - 1):
            if intervals[i][1] > intervals[i+1][0]:
                return False
            
        return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: One-liner using all()

class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
        intervals.sort(key=lambda i: i.start)
        return all(intervals[i-1].end <= intervals[i].start
                   for i in range(1, len(intervals)))
  • Another alternative:
class Solution:
    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:    intervals.sort(key=lambda i: i.start)
        return all(i.end <= j.start for i, j in zip(intervals, intervals[1:]))    
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[253/Easy] Meeting Rooms II

Problem

  • Given an array of meeting time intervals intervals where intervals[i] = [start_i, end_i], return the minimum number of conference rooms required.

  • Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: 2
  • Example 2:
Input: intervals = [[7,10],[2,4]]
Output: 1
  • Constraints:
    • 1 <= intervals.length <= 104
    • 0 <= starti < endi <= 106
  • See problem on LeetCode.

Solution: Min-heap to store the earliest ending time

  • The bottom-line is that we need to know when we need a new room and when we can share the room.
    • We sort the interval time slot by the start time.
    • The definition of overlap: the end time of the current time slot is larger than the begin time of the next time slot, for e.g., [1,11],[10,19], since 11 > 10, there is an overlap.
    • If there is an overlap between the current time slot and the smallest end time in the minimum heap, we need a new room for the meeting.
    • If there is no overlap between the current time and the smallest end time in the minimum heap, we don’t need a new room for the meeting since we can use the the same room of the smallest end time. But we still need the end time in the future to help us decide whether we have an overlap or not.
  • We use a minimum heap to store the \(k\)-largest end time we meet until now.
    • If the current time slot’s begin time is larger than the smallest end time in the minimum heap, i.e., we don’t have overlap, which implies we don’t need a new room, which in turn means that we can share the room with the smallest end time in the minimum heap. However, we need to update the minimum heap by using the replace method.
    • If the current time slot’s begin time is less than the smallest end time in the minimum heap, we have an overlap now, i.e., we need a new room, we to use the heappush method to push the current time slot’s end time into the minimum heap.
  • Note that heapq.heapreplace(heap, item) pops and returns the smallest item from the heap, and also push the new item. The heap size doesn’t change.
import heapq

class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        intervals = sorted(intervals, key=lambda x: x[0]) # Sort by start value
        heap = []
        
        for i in intervals:
            if heap and heap[0] <= i[0]:
                # If the new start time is greater than or equal to the exist end time, means the room has been released, replace the previous time with the new ending time
                # replacing because start time >= end time of heap[0], we can reuse the same meeting room that we used before
                heapq.heapreplace(heap, i[-1])
            else:
                # The room is still in use, add (push a new end time to min heap) a new room
                heapq.heappush(heap, i[-1])
                
        return len(heap)
Complexity
  • Time: $$O(n\log{n})
  • Space: \(O(1)\)

Solution: Greedy Interval Partition

  • This is an interval partition problem based on two heuristics:
    1. We process intervals ordered by starting time and assign each interval to a ‘current vacant’ room.
    2. We only check the room with the earliest ending time for global vacancy. If there is no vacant room, we add one.
  • Since we iterate intervals by starting time, there is no better choice than current interval as rest intervals would all request one more room if current one does. And we also need to track the ending time since we need to determine whether there exists a vacant room at specific time. We only track the earliest ending time as we only check this room for vacancy. If this room is not vacant, there is no need to check the rest and we just add one more room.
  • So combining both heuristics, we can sort starting time and ending time separately. We iterate starting time and use a pointer to track current the earliest ending time. So s is current interval’s starting time and earliestEndingTimeIdx is the pointer to current earliest ending time. If s < ends[earliestEndingTimeIdx], there is no vacant room, we add one more room. Otherwise, we assign current interval to a vacant room and our current earliest ending time should be updated to ends[earliestEndingTimeIdx+1].
class Solution:
    def minMeetingRooms(self, intervals: List[List[int]]) -> int:
        starts = sorted(i[0] for i in intervals)
        ends = sorted(i[1] for i in intervals)
        
        earliestEndingTimeIdx = roomCount = 0
        for s in starts:
            if s < ends[earliestEndingTimeIdx]: 
                # current earliest ending time room unavailable, add a vacant room
                roomCount += 1
            else: 
                # earliest ending room was available and taken; look for the next earliest one
                earliestEndingTimeIdx += 1
            
        return roomCount
Complexity
  • Time: \(O(n\log{n} + n\log{n} + n) = O(n\log{n})\) since sorting takes \(O(n\log{n})\) and one greedy pass takes \(O(n)\)
  • Space: \(O(1)\)

[270/Easy] Closest Binary Search Tree Value

Problem

  • Given the root of a binary search tree and a target value, return the value in the BST that is closest to the target.

  • Example 1:

Input: root = [4,2,5,1,3], target = 3.714286
Output: 4
  • Example 2:
Input: root = [1], target = 4.428571
Output: 1
  • Constraints:
    • The number of nodes in the tree is in the range [1, 104].
    • 0 <= Node.val <= 109
    • -109 <= target <= 109
  • See problem on LeetCode.

Solution: Recursion

  • Closest is either the root’s value (a) or the closest in the appropriate subtree (b).
class Solution(object):
    def closestValue(self, root, target):
        a = root.val
        kid = root.left if target < a else root.right
        if not kid: return a
        b = self.closestValue(kid, target)
        
        return min((b, a), key=lambda x: abs(target - x))
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: Iteration

  • Walk the path down the tree close to the target, return the closest value on the path.
class Solution(object):
    def closestValue(self, root, target):
        path = []
        while root:
            path += root.val,
            root = root.left if target < root.val else root.right
            
        return min(path[::-1], key=lambda x: abs(target - x))
  • Note that the [::-1] is only for handling targets much larger than 32-bit integer range, where different path values x have the same “distance”, i.e., abs(x - target). In such cases, the leaf value is the correct answer. If such large targets aren’t asked, then it’s unnecessary.
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: Iteration with \(O(1)\) space

  • In the code below, root.val is the present root we are currently at during each traversal from the root to the left subtree and/or right subtree; while our closest is what we are using to track the minimum root.val we have seen so far, we are updating our closest anytime we come across (root.val - target) < previous closest.
class Solution(object):
    def closestValue(self, root, target):
        closest = root.val
        while root:
            if abs(root.val - target) < abs(closest - target):
                closest = root.val
            root = root.left if target < root.val else root.right
            
        return closest
  • Same approach; rehashed:
class Solution(object):
    def closestValue(self, root, target):
        closest = root.val
        while root:
            closest = min((root.val, closest), key=lambda x: abs(target - x))
            root = root.left if target < root.val else root.right
            
        return closest
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[277/Medium] Find the Celebrity

Problem

  • Suppose you are at a party with n people labeled from 0 to n - 1 and among them, there may exist one celebrity. The definition of a celebrity is that all the other n - 1 people know the celebrity, but the celebrity does not know any of them.

  • Now you want to find out who the celebrity is or verify that there is not one. The only thing you are allowed to do is ask questions like: “Hi, A. Do you know B?” to get information about whether A knows B. You need to find out the celebrity (or verify there is not one) by asking as few questions as possible (in the asymptotic sense).

  • You are given a helper function bool knows(a, b) that tells you whether A knows B. Implement a function int findCelebrity(n). There will be exactly one celebrity if they are at the party.

  • Return the celebrity’s label if there is a celebrity at the party. If there is no celebrity, return -1.

  • Example 1:

Input: graph = [[1,1,0],[0,1,0],[1,1,1]]
Output: 1
Explanation: There are three persons labeled with 0, 1 and 2. graph[i][j] = 1 means person i knows person j, otherwise graph[i][j] = 0 means person i does not know person j. The celebrity is the person labeled as 1 because both 0 and 2 know him but 1 does not know anybody.
  • Example 2:
Input: graph = [[1,0,1],[1,1,0],[0,1,1]]
Output: -1
Explanation: There is no celebrity.
  • Constraints:
    • n == graph.length
    • n == graph[i].length
    • 2 <= n <= 100
    • graph[i][j] is 0 or 1.
    • graph[i][i] == 1
  • Follow up: If the maximum number of allowed calls to the API knows is 3 * n, could you find a solution without exceeding the maximum number of calls?

  • See problem on LeetCode.

Solution: Three loops; find a candidate, check if they everyone before them, and if everyone knows them

  • Explanation:
    • The first loop is to exclude n - 1 labels that are not possible to be a celebrity.
    • After the first loop, x is the only candidate.
    • The second and third any() statements are to verify x is actually a celebrity by definition.
    • The key part is the first loop. To understand this you can think the knows(a,b) as a a < b comparison, if a knows b then a < b, if a does not know b, a > b. Then if there is a celebrity, he/she must be the - “maximum” of the n people.
    • However, the “maximum” may not be the celebrity in the case of no celebrity at all. Thus we need the second and third any() statements to check if x is actually celebrity by definition.
    • The total calls of knows is thus 3n at most. One small improvement is that in the second loop we only need to check i in the range [0, x). You can figure that out yourself easily.
# The knows API is already defined for you.
# return a bool, whether a knows b
# def knows(a: int, b: int) -> bool:

class Solution:
    def findCelebrity(self, n: int) -> int:
        # Iterate through each person and find a candidate to be the celebrity

        # The resulting candidate must be a celebrity by the following logic
        # If there is a celebrity, they must
        # a) be known by every person before them (thus, end up as "cand" at some point)
        # b) not know anyone after them (thus, never relinquish "cand")        
        cand = 0
        for i in range(n):
            if knows(cand, i):
                cand = i
                
        # We've established that cand does not know anyone after it
        # Let's establish that cand does not know everyone before it                
        if any(knows(cand, i) for i in range(cand)):
            return -1

        # Last thing we don't know:
        # Does everyone know cand?
        if any(not knows(i, cand) for i in range(n)):
            return -1
            
        return cand
  • Same solution; expanded a bit by converting the any() calls to for loops:
class Solution(object):
    def findCelebrity(self, n):
        """
        :type n: int
        :rtype: int
        """

        # Iterate through each person and find a candidate to be the celebrity

        # The resulting candidate must be a celebrity by the following logic
        # If there is a celebrity, they must
        # a) be known by every person before them (thus, end up as "cand" at some point)
        # b) not know anyone after them (thus, never relinquish "cand")
        cand = 0
        for i in range(n):
            if knows(cand, i):
                cand = i

        # We've established that cand does not know anyone after it
        # Let's establish that cand is known by, but does not know everyone before it
        for i in range (0, cand):
            if knows(cand, i):
                return -1
            if not knows(i, cand):
                return -1

        # Last thing we don't know:
        # Does everyone after cand know cand?
        for i in range(cand, n):
            if not knows(i, cand):
                return -1

        return cand
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Two loops

class Solution:
    def findCelebrity(self, n: int) -> int:
        # Iterate through each person and find a candidate to be the celebrity

        # The resulting candidate must be a celebrity by the following logic
        # If there is a celebrity, they must
        # a) be known by every person before them (thus, end up as "cand" at some point)
        # b) not know anyone after them (thus, never relinquish "cand")        
        celebrity = 0
        for i in range(1,n):
            if knows(celebrity,i):
                celebrity = i
                
        # We've established that cand does not know anyone after it
        # Let's establish that cand is known by everyone but cand does not know anyone
            if celebrity != i and (knows(celebrity,i) or not knows(i, celebrity)):
                return -1
                
        return celebrity
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[283/Easy] Move Zeroes

Problem

  • Given an integer array nums, move all 0’s to the end of it while maintaining the relative order of the non-zero elements.

  • Note that you must do this in-place without making a copy of the array.

  • Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]
  • Example 2:
Input: nums = [0]
Output: [0]
  • Constraints:
    • 1 <= nums.length <= 104
    • -2^31 <= nums[i] <= 2^31 - 1
  • See problem on LeetCode.

Solution: Keep track of position of the first zero and the first non-zero value after the first zero, swap them

  • Algorithm:
    • We use zero to keep track of position of the first zero in the list (which changes as we go) and i to keep track of the first non-zero value after the first zero.
    • Each time we have zero point to a zero and i point to the first non-zero after zero, we swap the values that store at i and zero.
    • By doing this, we move zeros towards the end of the list gradually until i reaches the end.
    • And when it does, we are done.
  • Remarks:
    • No return value needed, since we are doing in-place modification.
    • We use nums[i], nums[j] = nums[j], nums[i] to achieve the in-place modification because Python allows you to swaps values in a list using syntax: x, y = y, x.
class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        zero = 0  # records the position of "0"
        for i in range(len(nums)):
            if nums[i] != 0:
                nums[i], nums[zero] = nums[zero], nums[i]
                zero += 1
  • Note that this solution is inefficient when all nums are non zeroes since you are still swapping the variable with itself. To avoid this, check if i and zero are different pointers before swapping:
class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        zero = 0  # records the position of "0"
        for i in range(len(nums)):
            if nums[i] != 0:
                if i != zero:
                    nums[i], nums[zero] = nums[zero], nums[i]
                    
                zero += 1
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: del a zero and append one

class Solution(object):
    def moveZeroes(self, nums: List[int]) -> None:
        i = 0
        end = len(nums)
        
        while i < end:
            if nums[i] == 0:
                del nums[i]
                nums.append(0)
                end -= 1
            else:
                i += 1
  • Same approach; rehashed:
class Solution(object):
    def moveZeroes(self, nums: List[int]) -> None:
        """
        :type nums: List[int]
        :rtype: void Do not return anything, modify nums in-place instead.
        """
        for _ in range(nums.count(0)):
            nums.remove(0)
            nums.append(0)
Complexity
  • Time: \(O(n^2)\) since the del operation takes \(O(n)\) time
  • Space: \(O(1)\)

[289/Medium] Game of Life

Problem

  • According to Wikipedia’s article: “The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970.”

  • The board is made up of an m x n grid of cells, where each cell has an initial state: live (represented by a 1) or dead (represented by a 0). Each cell interacts with its eight neighbors (horizontal, vertical, diagonal) using the following four rules (taken from the above Wikipedia article):

    • Any live cell with fewer than two live neighbors dies as if caused by under-population.
    • Any live cell with two or three live neighbors lives on to the next generation.
    • Any live cell with more than three live neighbors dies, as if by over-population.
    • Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
  • The next state is created by applying the above rules simultaneously to every cell in the current state, where births and deaths occur simultaneously. Given the current state of the m x n grid board, return the next state.

  • Example 1:

Input: board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]]
Output: [[0,0,0],[1,0,1],[0,1,1],[0,1,0]]
  • Example 2:

Input: board = [[1,1],[1,0]]
Output: [[1,1],[1,1]]
  • Constraints:
    • m == board.length
    • n == board[i].length
    • 1 <= m, n <= 25
    • board[i][j] is 0 or 1.
  • Follow up:
    • Could you solve it in-place? Remember that the board needs to be updated simultaneously: You cannot update some cells first and then use their updated values to update other cells.
    • In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches upon the border of the array (i.e., live cells reach the border). How would you address these problems?
  • See problem on LeetCode.

Solution: In-place board modify

  • In-place manipulation. At first glance, I thought it was to be solved by DFS/BFS. However, the key word is: simultaneously, so no DFS/BFS.
  • When the value needs to be updated, we do not just change 0 to 1 / 1 to 0 but we do in increments and decrements of 2. Check the table below:
  previous value state change  current state   current value
  0              no change     dead            0
  1              no change     live            1
  0              changed (+2)  live            2
  1              changed (-2)  dead            -1
  • In other words:
apply by using:
    under-population: < 2
    live to next generation: 2 or 3 
    over-population: > 3
    reproduction: == 3
class Solution:
    def gameOfLife(self, board: List[List[int]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """

        directions = set((x,y) for x in (-1, 0, 1) for y in (-1, 0, 1) if x or y)
        # or directions = [(1,0), (1,-1), (0,-1), (-1,-1), (-1,0), (-1,1), (0,1), (1,1)]
        
        # all new 0's denoted as -1  (1 ==> 0)
        # all new 1's denoted as 2   (0 ==> 1)        
        for i in range(len(board)):
            for j in range(len(board[0])):
                # Live neighbors count
                live = 0
                
                # Check and count neighbors in all directions
                for x, y in directions: 
                    if (i + x < len(board) and i + x >= 0) and (j + y < len(board[0]) and j + y >=0) and abs(board[i + x][j + y]) == 1: # originally 1's
                        live += 1
                
                # Rule 1 or Rule 3
                if board[i][j] == 1 and (live < 2 or live > 3):     
                    board[i][j] = -1
                
                # Rule 4
                if board[i][j] == 0 and live == 3:
                    board[i][j] = 2
        
        for i in range(len(board)):
            for j in range(len(board[0])):
                board[i][j] = 1 if(board[i][j] > 0) else 0
        
        return board
Complexity
  • Time: \(O(m \times n)\) where the board is made up of an m x n grid of cells
  • Space: \(O(1)\)

[Easy] Moving Average with Cold Start

Problem

  • Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

  • Example:

Input:
a = [1, 10, 3, 5]
windowSize = 3

Output:
1.0
5.5
4.66667
6.0

Solution: Deque

from collections import deque

queue = deque()

windowSize = 3

for i in range(len(a)):
    if windowSize > len(queue):
        queue.append(a[i])
        print(sum(queue)/len(queue))
    else:
        queue.popleft()
        queue.append(a[i])
        print(sum(queue)/windowSize)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[346/Easy] Moving Average from Data Stream

Problem

  • Given a stream of integers and a window size, calculate the moving average of all integers in the sliding window.

  • Implement the MovingAverage class:

    • MovingAverage(int size) Initializes the object with the size of the window size.
    • double next(int val) Returns the moving average of the last size values of the stream.
  • Example 1:

Input
["MovingAverage", "next", "next", "next", "next"]
[[3], [1], [10], [3], [5]]
Output
[null, 1.0, 5.5, 4.66667, 6.0]

Explanation
MovingAverage movingAverage = new MovingAverage(3);
movingAverage.next(1); // return 1.0 = 1 / 1
movingAverage.next(10); // return 5.5 = (1 + 10) / 2
movingAverage.next(3); // return 4.66667 = (1 + 10 + 3) / 3
movingAverage.next(5); // return 6.0 = (10 + 3 + 5) / 3
  • Constraints:
    • 1 <= size <= 1000
    • -105 <= val <= 105
    • At most 104 calls will be made to next.
  • See problem on LeetCode.

Solution: Deque

import collections

class MovingAverage(object):

    def __init__(self, size: int):
        """
        Initialize your data structure here.
        :type size: int
        """
        self.queue = collections.deque(maxlen=size)
        

    def next(self, val: int) -> float:
        """
        :type val: int
        :rtype: float
        """
        queue = self.queue
        
        # Appending to a `maxlen` deque will 
        # automatically popleft the leftmost element
        queue.append(val)
        
        return float(sum(queue))/len(queue)
    
# Your MovingAverage object will be instantiated and called as such:
# obj = MovingAverage(size)
# param_1 = obj.next(val)
  • Same approach; more explicit deque sizing (with self.max and self.currLength):
from collections import deque
class MovingAverage:
    def __init__(self, size: int):
        """
        Initialize your data structure here.
        """
        self.currLength = 0
        self.max = size
        self.sum = 0
        self.queue = deque()

    def next(self, val: int) -> float:
        if self.currLength < self.max:
            self.currLength +=1
        else:
            self.sum -= self.queue.popleft()
        
        self.queue.append(val)
        self.sum += val
        
        return self.sum/self.currLength
  • Even more cleaner:
from collections import deque
class MovingAverage:
    def __init__(self, size: int):
        """
        Initialize your data structure here.
        """
        self.size = size
        self.queue = deque()
        self.sum = 0

    def next(self, val: int) -> float:
        if len(self.queue) >= self.size:
            # Remove leftmost element value 
            # from sum and from the deque
            self.sum -= self.queue.popleft() 
    
        self.queue.append(val)
        self.sum += val
        
        return self.sum/len(self.queue)
Complexity
  • Time: \(O(1)\) since queue takes \(O(1)\) to insert.
  • Space: \(O(n)\) where \(n\) is the size of queue. The algorithm needs to maintain two integers: sum and numVal, and a queue. So it uses linear space.

[383/Easy] Ransom Note

Problem

  • Given two strings ransomNote and magazine, return true if ransomNote can be constructed from magazine and false otherwise.

  • Each letter in magazine can only be used once in ransomNote.

  • Example 1:

Input: ransomNote = "a", magazine = "b"
Output: false
  • Example 2:
Input: ransomNote = "aa", magazine = "ab"
Output: false
  • Example 3:
Input: ransomNote = "aa", magazine = "aab"
Output: true
  • Constraints:
    • 1 <= ransomNote.length, magazine.length <= 105
    • ransomNote and magazine consist of lowercase English letters.
  • See problem on LeetCode.

Solution: For each unique char in ransomNote, check if it’s count in ransomNote doesn’t exceed that in magazine

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        for i in set(ransomNote):
            if ransomNote.count(i) > magazine.count(i):
                return False
                
        return True

Solution: Remove ransomNote chars from magazine; confirm if length of ransomNote = 0

class Solution:
    def canConstruct(self, ransomNote: str, magazine: str) -> bool:
        for i in ransomNote:
            if i not in magazine:
                return False

            # str.replace(old, new[, count]) returns a copy of the string with all occurrences 
            # of the substring old replaced by new.
            # If the optional argument count is given, only the first count occurrences are replaced.
            # https://docs.python.org/3/library/stdtypes.html#str.replace
            magazine = magazine.replace(i, "", 1)

            if not len(ransomNote):
                return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[384/Medium] Shuffle an Array

Problem

  • Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.

  • Implement the Solution class:
    • Solution(int[] nums) Initializes the object with the integer array nums.
    • int[] reset() Resets the array to its original configuration and returns it.
    • int[] shuffle() Returns a random shuffling of the array.
  • Example 1:
Input
["Solution", "shuffle", "reset", "shuffle"]
[[[1, 2, 3]], [], [], []]
Output
[null, [3, 1, 2], [1, 2, 3], [1, 3, 2]]
Explanation
Solution solution = new Solution([1, 2, 3]);
solution.shuffle();    // Shuffle the array [1,2,3] and return its result.
                       // Any permutation of [1,2,3] must be equally likely to be returned.
                       // Example: return [3, 1, 2]
solution.reset();      // Resets the array back to its original configuration [1,2,3]. Return [1, 2, 3]
solution.shuffle();    // Returns the random shuffling of array [1,2,3]. Example: return [1, 3, 2]
  • Constraints:

    • 1 <= nums.length <= 50
    • -106 <= nums[i] <= 106
    • All the elements of nums are unique.
    • At most 104 calls in total will be made to reset and shuffle.
  • See problem on LeetCode.

Solution

class Solution:
    def __init__(self, nums: List[int]):
        self.nums = nums
        # keep a backup copy
        self.arr = nums[:]

    def reset(self) -> List[int]:
        # make sure to not directly assign self.arr to self.nums
        # since that will mean we will lose our backup once
        # we start modifying it
        self.nums = self.arr[:]
        return self.nums

    def shuffle(self) -> List[int]:
        # shuffle returns none, so copy and shuffle 
    
        # approach 1 
        # random.shuffle(self.nums)
        # return self.nums
        
        # approach 2
        for i in range(len(self.nums)):
            # Fisher-Yates Algorithm
            # generate an index in [i, len(ans))
            # note that random.range(a, b) returns a random number between [a, b)
            # while random.randint(a, b) returns a random number between [a, b] 
            # i.e., including both end points.
            num = random.randrange(i, len(self.nums)) 
            # swap ans[num] and ans[i]
            self.nums[i], self.nums[num] = self.nums[num], self.nums[i]
            
        return self.nums
        
# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.reset()
# param_2 = obj.shuffle()        

Complexity

  • Time: \(O(n)\)
  • Space: \(O(n)\) for self.nums and the self.arr backup

[398/Medium] Random Pick Index

Problem

  • Given an integer array nums with possible duplicates, randomly output the index of a given target number. You can assume that the given target number must exist in the array.

  • Implement the Solution class:
    • Solution(int[] nums): Initializes the object with the array nums.
    • int pick(int target): Picks a random index i from nums where nums[i] == target. If there are multiple valid i’s, then each index should have an equal probability of returning.
  • Example 1:
Input
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output
[null, 4, 0, 2]

Explanation
Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
  • Constraints:
    • 1 <= nums.length <= 2 * 104
    • -2^31 <= nums[i] <= 2^31 - 1
    • target is an integer from nums.
    • At most 104 calls will be made to pick.
  • See problem on LeetCode.

Solution: Form index of targets in pick and use random.choice() to pick elements from a uniform distribution

import random

class Solution:
    def __init__(self, nums: List[int]):
        self.nums = nums

    def pick(self, target: int) -> int:
        # answer is a list of indices containing the target
        answer = []
        
        for idx, num in enumerate(self.nums):
            if target == num:
                answer.append(idx)
        
        # random.choice() samples uniformly (with PDF = 1/(b-a) where [a,b] are the min and max elements in the data distribution)
        return random.choice(answer)

# Your Solution object will be instantiated and called as such:
# obj = Solution(nums)
# param_1 = obj.pick(target)
Complexity
  • Time: \(O(n)\) for pick()
  • Space: \(O(n)\) for answer

Solution: Compute index of targets in __init__ using and use random.choice() to pick elements from a uniform distribution

import random

class Solution:
    def __init__(self, nums: List[int]):
        self.hash = defaultdict(list)
        for i, num in enumerate(nums):
            self.hash[num].append(i)
        
    def pick(self, target: int) -> int:
        return random.choice(self.hash[target])
Complexity
  • Time: \(O(1)\) for pick()
  • Space: \(O(n)\) for hash

[409/Easy] Longest Palindrome

Problem

  • Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters.
  • Letters are case sensitive, for example, "Aa" is not considered a palindrome here.

  • Example 1:
Input: s = "abccccdd"
Output: 7
Explanation:
One longest palindrome that can be built is "dccaccd", whose length is 7.
  • Example 2:
Input: s = "a"
Output: 1
  • Example 3:
Input: s = "bb"
Output: 2
  • Constraints:
    • 1 <= s.length <= 2000
    • s consists of lowercase and/or uppercase English letters only.
  • See problem on LeetCode.

Solution: Add the number of even characters to the result; if an odd character is found, add only one of it to the result for a string to be palindrome

  • For a string to be palindrome, it needs characters that occur an even number of times. For e.g., aabbaa or aaa.
  • If a character occurs once, i.e., an odd number of times, we can accommodate only one such character. For e.g., aba or aacbcaa. In other words, a second character occurring once will need to be discarded for the string to still be a palindrome.
class Solution:
    def longestPalindrome(self, s: str) -> int:        
        counts = Counter(s)
        res = 0
        foundOdd = False
        
        for val in counts.values():
            if not val%2:
                res += val
            else:
                foundOdd = True
                # add even occurrences of the char to res by subtracting 1 from its count
                res += val-1
        
        # add only one instance of the odd character to res
        return res+1 if foundOdd else res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for counts

[415/Easy] Add Strings

Problem

  • Given two non-negative integers, num1 and num2 represented as string, return the sum of num1 and num2 as a string.
  • You must solve the problem without using any built-in library for handling large integers (such as BigInteger). You must also not convert the inputs to integers directly.
  • Example 1:
Input: num1 = "11", num2 = "123"
Output: "134"
  • Example 2:
Input: num1 = "456", num2 = "77"
Output: "533"
  • Example 3:
Input: num1 = "0", num2 = "0"
Output: "0"
  • Constraints:
    • 1 <= num1.length, num2.length <= 104
    • num1 and num2 consist of only digits.
    • num1 and num2 don't have any leading zeros except for the zero itself.
  • See problem on LeetCode.

Solution: Use a Dictionary

class Solution:
  def addStrings(self, num1: str, num2: str) -> str:
      
      def str2int(num):
          numDict = {'0' : 0, '1' : 1, '2' : 2, '3' : 3, '4' : 4, '5' : 5,
                '6' : 6, '7' : 7, '8' : 8, '9' : 9}
          output = 0
          for d in num:
              output = output * 10 + numDict[d]
          return output
      
      return str(str2int(num1) + str2int(num2)) 
Complexity
  • Time: \(O(n + n) = O(2n) = O(n)\)
  • Space: \(O(1)\)

Solution: Use ord to get unicode

class Solution:
  def addStrings(self, num1: str, num2: str) -> str:
      def str2int(num):
          result  = 0
          for n in num:
              result = result * 10 + ord(n) - ord('0')
          return result
      return str(str2int(num1) + str2int(num2))
Complexity
  • Time: \(O(n + n) = O(2n) = O(n)\)
  • Space: \(O(1)\)

[435/Medium] Non-overlapping Intervals

Problem

  • Given an array of intervals intervals where intervals[i] = [start_i, end_i], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

  • Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
  • Example 2:
Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
  • Example 3:
Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
  • Constraints:
    • 1 <= intervals.length <= 105
    • intervals[i].length == 2
    • -5 * 104 <= starti < endi <= 5 * 104
  • See problem on LeetCode.

Solution: Sort by start time

  • Algorithm:
    • Attend the one with smaller start time first
    • (Greedy) Remove the one with bigger end time if an overlap occurs (because it will always incur more overlappings in the remaining array with ascending order of start time)
  • Example:
[[1,10], [2,3], [3,4], [5,6]]
First, we attend [1,10]
Next, [2,3] overlaps with [1,10] as 2 < 10 (i.e., [2,3] starts before [1,10] ends). Here, we are going to remove [1,10] as it would definitely produce more overlappings as we continue the iteration.
Thirdly, attend [3,4]
Last, attend [5,6]
class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort()
        prev = float("-inf")
        remove = 0
        
        for i in intervals:
            if i[0] >= prev:
                # no overlap case: the current interval starts after prev ended
                # thus, no need to remove, just update prev
                prev = i[1]
            else:
                # overlap case: the current interval starts before prev ended
                # remove the interval with the larger end time 
                # and update prev to hold the smaller end time
                remove += 1
                prev = min(prev, i[1])
                
        return remove
Complexity
  • Time: \(O(n\log{n})\) due to sorting
  • Space: \(O(1)\)

Solution: Sort by end time

  • If we think more clearly, the first solution can be improved slightly (save a line of code) by sorting the intervals with end time. Why? Because the greedy nature remains true even if we the remaining array is not sorted by start time. Meanwhile, if we sort them by end time and if overlap occurs, the interval that comes later must be the one to remove as it has larger end time. Take a look at the following example.

  • Example:

1) sort by start time
Extra procedure: after comparing [1,10] and [2,3], we have to choose which one to remove by comparing the end time.
[[1,10], [2,3], [3,4], [5,6]]

2) sort by end time
Saved procedure: after comparing [5, 6] and [1,10], we immediately can remove the later one which is [1,10] as it must produce more overlaps in the following, i.e. with [6,11] and [9,12].
[[2,3], [3,4], [5,6], [1,10], [6, 11], [9, 12]]
class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: x[1]) # sort based on end times
        prev = float("-inf")
        remove = 0
        
        for i in intervals:
            if i[0] >= prev:
                # no overlap case: the current interval starts after prev ended
                # thus, no need to remove, just update prev                
                prev = i[1]
            else:
                # overlap case: the current interval starts before prev ended                
                remove += 1
                
        return remove
Complexity
  • Time: \(O(n\log{n})\) due to sorting
  • Space: \(O(1)\)

[442/Medium] Find All Duplicates in an Array

Problem

  • Given an integer array nums of length n where all the integers of nums are in the range [1, n] and each integer appears once or twice, return an array of all the integers that appears twice.

  • You must write an algorithm that runs in O(n) time and uses only constant extra space.

  • Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [2,3]
  • Example 2:
Input: nums = [1,1,2]
Output: [1]
  • Example 3:
Input: nums = [1]
Output: []
  • Constraints:
    • n == nums.length
    • 1 <= n <= 105
    • 1 <= nums[i] <= n
    • Each element in nums appears once or twice.
  • See problem on LeetCode.

Solution: Use the given array as a hashmap, add element to itself, divide by two to get duplicates

class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:
        """
            We can't use an extra hashmap because space complexity has to be O(1).
            So let's use the given array as hashmap based on the logic that the all numbers are in the range of 1-n so we can use the index of array as the numbers (index+1 as 0th index will represent 1) keys of hashmap and the values as the number of occurrences of each number.
            We can achieve the above be decrementing each element by 1 to sustain the nth index as max value of array will be n but max index would be n-1.
            Now we will iterate over the array and will treat each element as the correct index of number.
            Everytime we encounter a number, the value at its correct index will be incremented by n.
            At the end we divide each number by n so that we get the count of each number(index of array) as the values of array.
            Now the elements with value == 2 would be the duplicates.
        """
        n = len(nums)
        nums = [num-1 for num in nums]
        print(nums)
        for num in nums:
            nums[num % n] += n
        
        output = []
        for key, num in enumerate(nums):
            if num//n == 2:
                output.append(key+1) 
        # or output = [(key+1) for key, num in enumerate(nums) if num//n == 2]
        return output 
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Cyclic Sort

class Solution:
    def cyclicSort(self, nums):
        i = 0
        while i < len(nums):
            correct = nums[i] - 1
            if nums[correct] != nums[i]:
                nums[correct], nums[i] = nums[i], nums[correct]
            else:
                i += 1

    def findDuplicates(self, nums: List[int]) -> List[int]:
        self.cyclicSort(nums)
        output = []
        for i in range(len(nums)):
            if nums[i] != i+1:
                output.append(nums[i])
        return output
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Marking using the given array as a hashmap

class Solution:
    def findDuplicates(self, nums: List[int]) -> List[int]:
        output = []
        """
            Since we have to solve this problem in constant space complexity so the only way is to use the space of given array and manipulate it.
            Since the elements in the array are in range [1..n], we can consider the elements as idexes also and to be precise (element - 1) as it is a 0-indexed array.
            Since we can't use any extra space so let us mark the elements encountered in the given array only. This can be done by multiplying the element with -1.
            Now for each element, we will treat it as the index and check whether the value present at that index is visited or not. If it is visited then it is a duplicate else we mark that element as visited(multiply by -1).
        """
        for num in nums:
            if nums[abs(num) - 1] < 0:
                output.append(abs(num))
            else:
                nums[abs(num) - 1] *= -1
        return output  
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[443/Medium] String Compression

Problem

  • Given an array of characters chars, compress it using the following algorithm:

  • Begin with an empty string s. For each group of consecutive repeating characters in chars:

    • If the group’s length is 1, append the character to s.
    • Otherwise, append the character followed by the group’s length.
  • The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.

  • After you are done modifying the input array, return the new length of the array.

  • You must write an algorithm that uses only constant extra space.

  • Example 1:

Input: chars = ["a","a","b","b","c","c","c"]
Output: Return 6, and the first 6 characters of the input array should be: ["a","2","b","2","c","3"]
Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
  • Example 2:
Input: chars = ["a"]
Output: Return 1, and the first character of the input array should be: ["a"]
Explanation: The only group is "a", which remains uncompressed since it's a single character.
  • Example 3:
Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output: Return 4, and the first 4 characters of the input array should be: ["a","b","1","2"].
Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
  • Constraints:
    • 1 <= chars.length <= 2000
    • chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.
  • See problem on LeetCode.

Solution: Slow/fast pointers

  • Slow: currentIdx (the idx ready to be modified)
  • Fast: i (traversing the string to find how many times a char is repeated before moving onto the next char)
class Solution:
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        # count: record the times one element had appeared. 
        count = 1
        # currentIdx: the idx ready to be modified.
        currentIdx = 0
        
        for i in range(1, len(chars)+1): 
            # check for valid index and if current char matches the previous char
            # if so, add to count
            if i < len(chars) and chars[i] == chars[i-1]:
                count += 1
            # Look for a new element
            else: 
                # 1) update letter: we modify char[currentIdx] to the char we're writing the count for.
                chars[currentIdx] = chars[i-1] 
                currentIdx += 1
                
                # 2) append count to string after char
                if count == 1:
                    # if the previous element only appear once, skip below.
                    continue
                # add count digit by digit
                for ch in str(count):
                    chars[currentIdx] = ch
                    currentIdx += 1
                    
                # 3) reset count because we're done with the current element
                count = 1
        return currentIdx                            
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) because the input is modified in-place

[523/Medium] Continuous Subarray Sum

Problem

  • Given an integer array nums and an integer k, return true if nums has a continuous subarray of size at least two whose elements sum up to a multiple of k, or false otherwise.

  • An integer x is a multiple of k if there exists an integer n such that x = n * k. 0 is always a multiple of k.

  • Example 1:

Input: nums = [23,2,4,6,7], k = 6
Output: true
Explanation: [2, 4] is a continuous subarray of size 2 whose elements sum up to 6.
  • Example 2:
Input: nums = [23,2,6,4,7], k = 6
Output: true
Explanation: [23, 2, 6, 4, 7] is an continuous subarray of size 5 whose elements sum up to 42.
42 is a multiple of 6 because 42 = 7 * 6 and 7 is an integer.
  • Example 3:
Input: nums = [23,2,6,4,7], k = 13
Output: false
  • Constraints:
    • 1 <= nums.length <= 105
    • 0 <= nums[i] <= 109
    • 0 <= sum(nums[i]) <= 2^31 - 1
    • 1 <= k <= 2^31 - 1
  • See problem on LeetCode.

Solution

  • Same as subarray sum equals k with modification. Basic idea is that, if you get the same remainder again, it means that you’ve encountered some sum which is a multiple of k.
class Solution:
    def checkSubarraySum(self, nums: List[int], k: int) -> bool:
        d = {0: -1}
        prefix_sum = 0
        
        for i, num in enumerate(nums):
            prefix_sum += num
            
            if k != 0:
                prefix_sum = prefix_sum % k
                
            if prefix_sum in d:
                # we know that sum between mp[prefix_sum] + 1 and i is multiple of k, so I don't have to include mp[prefix_sum]
                if i - d[prefix_sum] > 1:
                    return True
            else:
                # if prefix_sum doesn't exists, then add its index, otherwise don't update it, it is preferable to keep the
                # oldest index, so that you can get length of atleast 2.
                d[prefix_sum] = i
            
        return False
  • Explanation with example:
Example:
nums = [23,2,4], k = 6
Lets walk through the code with the example. 
(i=0) : prefix_sum = 23 => 23%6 => (prefix_sum = 5)
(i=1) : prefix_sum = 5+2=7 => 7%6 => (prefix_sum = 1)
(i=2) : prefix_sum = 1+4=5 => 5%6 => (prefix_sum = 5)
  • We have encountered the same prefix_sum(remainder) again which means we have the subarray of prefix_sum%k = 0.
  • But, there’s another aspect to this problem. The subarray must have a minimum size of 2.
  • That is why we check if (i - d[prefix_sum]) > 1.
  • In the above example, this if loop is executed when (i=2) and (mp[prefix_sum]=1).
  • In other words, the same remainder (prefix_sum=5) has been encountered twice and then we check for the respective difference in indices.
Counter example to understand this. Lets take nums = [23,6], k = 6
(i=0) : prefix_sum = 23 => 23%6 => (prefix_sum = 5)
(i=1) : prefix_sum = 5+6=11 => 11%6 => (prefix_sum = 5)
  • So, the same prefix_sum (remainder) has appeared again which means we’ve found the subarray but it is not a subarray of size 2 or more.
  • Because they’ve occurred next to each other, which means that, we have just one element in the subarray which contributes.
  • If you remove 23 from the nums array and keep 6 alone, it will still be a subarray whose prefix_sum%k is 0. But, we want a subarray of size 2 or more.
  • This is the reason why we calculate the index difference and then return True.
  • Also, if k==0, you don’t need to find the remainder. So, you just keep adding prefix_sum and repeat the rest of the process.

[556/Medium] Next Greater Element III

Problem

  • Given a positive integer n, find the smallest integer which has exactly the same digits existing in the integer n and is greater in value than n. If no such positive integer exists, return -1.

  • Note that the returned integer should fit in 32-bit integer, if there is a valid answer but it does not fit in 32-bit integer, return -1.

  • Example 1:

Input: n = 12
Output: 21
  • Example 2:
Input: n = 21
Output: -1
  • Constraints:
    • 1 <= n <= 231 - 1
  • See problem on LeetCode.

Solutions:

  • Traverse the given number from rightmost digit, till you get a smaller digit than the previously traversed digit. For example, if the input number is 721632, stop at 1. If you do not find such a digit, then the output is -1.
  • At the right side of digit 1 find the smallest digit greater than 1. For 721632, the right side of 1 contains 632. ans digit is 2.
  • Swap the two digits 722631.
  • Reverse/sort the right-side digits 631 to yield 722136.
def next_permutation(l):
    """implementing C++ <algorithm> next_permutation()."""
    n = len(l)
    
    if n <= 1:
        return False
    i = n - 1
    
    while True:
        j = i
        i -= 1  
        
        if l[i] < l[j]:
            k = n - 1
        
            while not (l[i] < l[k]):  
                k -= 1

            l[i], l[k] = l[k], l[i]
            l[j:] = reversed(l[j:])
            return True
        
        if i == 0:  
                l.reverse()
                return False 

class Solution:
    def nextGreaterElement(self, n: int) -> int:
        l = list(str(n))
        has_next = next_permutation(l)
        ans = int(''.join(l))
        
        return ans if has_next and ans < 2 ** 31 else -1

Solution: Calculate permutations

  • One string will be a permutation of another string only if both of them contain the same characters with the same frequency. We can consider every possible substring/window in the long string s2 of the same length as that of s1 and check the frequency of occurrence of the characters appearing in the two. If the frequencies of every letter match exactly, then only s1’s permutation can be a substring of s2.
  • Using the sliding window approach:
from itertools import permutations

class Solution:
    def nextGreaterElement(self, n: int) -> int:
        all_nums_of_same_digits = [int(''.join(i)) for i in set(permutations(str(n)))]
        all_nums_of_same_digits.sort()
        i = all_nums_of_same_digits.index(n)
        try:
            return all_nums_of_same_digits[i + 1] if all_nums_of_same_digits[i + 1] <= 2**31-1 else -1
        except:
            return -1

[567/Medium] Permutation in String

Problem

  • Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise. In other words, return true if one of s1’s permutations is the substring of s2.

  • Example 1:

Input: s1 = "ab", s2 = "eidbaooo"
Output: true
Explanation: s2 contains one permutation of s1 ("ba").
  • Example 2:
Input: s1 = "ab", s2 = "eidboaoo"
Output: false
  • Constraints:
    • 1 <= s1.length, s2.length <= 104
    • s1 and s2 consist of lowercase English letters.
  • See problem on LeetCode.

Solution: Use counter

  • One string will be a permutation of another string only if both of them contain the same characters with the same frequency. We can consider every possible substring/window in the long string s2 of the same length as that of s1 and check the frequency of occurrence of the characters appearing in the two. If the frequencies of every letter match exactly, then only s1’s permutation can be a substring of s2.
  • Using the sliding window approach:
class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        window = len(s1)
        counterS1 = Counter(s1)
        
        for i in range(len(s2)-window + 1):
            counterS2 = Counter(s2[i: i+window])
            if counterS1 == counterS2:
                return True
        return False
Complexity
  • Time: \(O(n)\). Single pass.
  • Space: \(O(1)\). Constant space needed.

[665/Medium] Non-decreasing Array

Problem

  • Given an array nums with n integers, your task is to check if it could become non-decreasing by modifying at most one element.

  • We define an array is non-decreasing if nums[i] <= nums[i + 1] holds for every i (0-based) such that (0 <= i <= n - 2).

  • Example 1:

Input: nums = [4,2,3]
Output: true
Explanation: You could modify the first 4 to 1 to get a non-decreasing array.
  • Example 2:
Input: nums = [4,2,1]
Output: false
Explanation: You can't get a non-decreasing array by modify at most one element.
  • Constraints:
    • n == nums.length
    • 1 <= n <= 104
    • -105 <= nums[i] <= 105
  • See problem on LeetCode.

Solution: Greedy

  • Intuition:
    • The one thing that we should focus on here is the fact that the problem description says non-decreasing, which is just another way of saying increasing. In this case, it means the next element must always be greater than or equal to the current element. Thus, in an array, the item on the left must be less than or equal to the item on the right. Since we can only rectify a single violation of this rule, if more than one violation exists, it is impossible to make the array non-decreasing.
    • The first time we encounter such a violation i.e. nums[i - 1] > nums[i], we make a change. You don’t actually have to make a change because the question doesn’t ask us to give the final array in order, it just asks if it could become non-decreasing. Then, we move on with the rest of the array, and continue to check for other violations. If it ever so happens that the rule gets violated again, we return false since this would be the second violation of the rule. If we don’t find a second violation, we can return true since rectifying the first violation made our array non-decreasing.
    • Whenever we encounter a violation at a particular index i, we need to check what modification we can make to make the array sorted. Let’s see this scenario using an example.

    • In the example above, we consider the numbers 4, 5, 3 for deciding on how to fix the violation or. In this case, the correct modification is to change the number 3 to 5. If we change 5 to 3, then we won’t be fixing the violation because the resulting array would be 3, 4, 3, 3, 6, 8.

    • The basic decision making process for fixing a violation is listed below. Without considering the number at the index i - 2, we won’t be able to choose between updating nums[i] or nums[i - 1]. The modification has to fit in with the sorted nature of the array.
      if nums[i - 2] > nums[i] then
          nums[i] = nums[i - 1]
      else
          nums[i - 1] = nums[i]
    
    • Once we make the modification, we expect that the rest of the array will be sorted. If that is not the case, then we return false from our function. Some arrays will have violations in different places e.g. a 10 element array where nums[4] > nums[5] and also nums[8] > nums[9]. This array cannot be sorted by only fixing the violation at nums[5].
    • Additionally, it is possible that a modification in the array leads to another violation that did not exist before. Let’s consider an example where this can happen.

  • Algorithm:
    • We iterate over the array until we reach the end of the array or find a violation.
    • If we reach the end of the array, we know it is sorted and we return true.
    • Otherwise, we found a violation. We consider the nums[i - 2] to fix the violation.
    • If the violation is at the index 1, we won’t have a nums[i - 2] available. In that case we simply set nums[i - 1] equal to nums[i].
    • Otherwise, we check if nums[i - 2] <= nums[i] in which case we set nums[i - 1] equal to nums[i].
    • Finally, if nums[i - 2] > nums[i], then we set nums[i] equal to nums[i - 1].
    • Once we make the modification, we simply iterate over the remaining array. If we find another violation, we return false. Otherwise, we return true once the iteration is complete.
class Solution:
    def checkPossibility(self, nums: List[int]) -> bool:
        num_violations = 0
        
        for i in range(1, len(nums)):
            if nums[i - 1] > nums[i]:
                
                if num_violations == 1:
                    return False
                
                num_violations += 1
                
                if i < 2 or nums[i - 2] <= nums[i]:
                    nums[i - 1] = nums[i]
                else:
                    nums[i] = nums[i - 1]
                    
        return True
Complexity
  • Time: \(O(n)\) considering there are \(n\) elements in the array and we process each element at most once.
  • Space: \(O(1)\) since we don’t use any additional space apart from a couple of variables for executing this algorithm.

[670/Medium] Maximum Swap

Problem

  • You are given an integer num. You can swap two digits at most once to get the maximum valued number.

  • Return the maximum valued number you can get.

  • Example 1:

Input: num = 2736
Output: 7236
Explanation: Swap the number 2 and the number 7.
  • Example 2:
Input: num = 9973
Output: 9973
Explanation: No swap.
  • Constraints:
    • 0 <= num <= 108
  • See problem on LeetCode.

Solution: Find the index of the leftmost increasing sequence

  • Find a index i, where there is a increasing order.
  • On the right side of i, find the max value max_val and its index max_idx.
  • On the left side of i, find the leftmost value and its index left_idx, which is less than max_val.
  • Swap above left_idx and max_idx if necessary.
class Solution:
    def maximumSwap(self, num: int) -> int:
        s = list(str(num))                                  # necessary to convert to list to mutate/swap elements
        n = len(s)
        
        for i in range(n-1):                                # find index where s[i] < s[i+1], meaning a chance to flip
            if s[i] < s[i+1]: break
        else: return num                                    # if nothing is found, return num
        max_idx, max_val = i+1, s[i+1]                      # keep going right, find the maximum value index
        
        for j in range(i+1, n):
            if max_val <= s[j]: max_idx, max_val = j, s[j]
        left_idx = i                                        # going right from i, find leftmost value that is less than max_val
        
        for j in range(i, -1, -1):    
            if s[j] < max_val: left_idx = j
        s[max_idx], s[left_idx] = s[left_idx], s[max_idx]   # swap maximum after i and most left less than max
        
        return int(''.join(s))                              # re-create the integer
Complexity
  • Time: \(O(n)\). Single pass.
  • Space: \(O(1)\). Constant space needed.

Solution: One-pass backward iteration, track max two digits

  • Iterate the list of digits backwards and keep tracking of the max digit, as well as the leftmost digit that is smaller than the max.
class Solution:
    def maximumSwap(self, num: int) -> int:
        # note that the conversion to a list is needed to perform the swap
        # at the end (since a string is immutable/unmodifiable)
        digits = list(str(num))

        p, q, max_i = -1, -1, len(digits)-1
        
        for i in range(max_i, -1, -1):
            # absolute max
            if digits[i] > digits[max_i]:
                max_i = i
            # we need the leftmost value smaller than max (hence backwards loop)
            elif digits[max_i] > digits[i]:
                p, q = i, max_i

        # if p and q are not untouched
        if p > -1 and q > -1:
            # swap digits[p] and digits[q]
            digits[p], digits[q] = digits[q], digits[p]
            
        return int(''.join(digits))
Complexity
  • Time: \(O(n)\). Single pass.
  • Space: \(O(1)\). Constant space needed.

[674/Easy] Longest Continuous Increasing Subsequence/Subarray

Problem

  • Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.

  • A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].

  • Example 1:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.
  • Example 2:
Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.
  • Constraints:
    • 1 <= nums.length <= 104
    • -109 <= nums[i] <= 109
  • See problem on LeetCode.

Solution: Loop over each operation, process it, append outcome to a list

class Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:     
        if not nums:
            return 0
        
        curLen = 1
        maxLen = 1
        
        for i in range(1, len(nums)):
            if nums[i] > nums[i-1]:
                curLen += 1
            else:
                curLen = 1
            maxLen = max(maxLen, curLen)
        return maxLen        
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[681/Medium] Next Closest Time

Problem

  • Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.

  • You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.

  • Example 1:

Input: time = "19:34"
Output: "19:39"
Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later.
It is not 19:33, because this occurs 23 hours and 59 minutes later.
  • Example 2:
Input: time = "23:59"
Output: "22:22"
Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22.
It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.
  • Constraints:
    • time.length == 5
    • time is a valid time in the form "HH:MM".
    • 0 <= HH < 24
    • 0 <= MM < 60
  • See problem on LeetCode.

Solution: Loop over each operation, process it, append outcome to a list

class Solution:
    def nextClosestTime(self, time: str) -> str:
        hour, minute = time.split(':')
        
        # generate sorted permutations of the input digits
        # in both hour and minute
        nums = sorted(set(hour+minute))
        all_time = [a+b for a in nums for b in nums]
        
        # the only change needed is minutes
        # hour remains the same
        i = all_time.index(minute)
        if i+1 < len(all_time) and all_time[i+1] < '60':
            return hour+':'+all_time[i+1]

        # we need to change both the hour and minute
        # change minutes to the smallest value available
        # change hour to the next value available
        # for e.g., 12:59 -> 15:11
        i = all_time.index(hour)
        if i+1 < len(all_time) and all_time[i+1] < '24':
            return all_time[i+1]+':'+all_time[0]
        
        # it's not possible to find another time in the same day
        # so find the smallest in the next day
        # for e.g., 23:59 -> 22:22
        return all_time[0]+':'+all_time[0]
Complexity
  • Time: \(O(1)\)
  • Space: \(O(perm(n+m))\) where \(n\) is the average number of unique digits in the hour and \(m\) is the average number of unique digits in the minute
    • Assuming the worst case of all unique digits in the hour and number, n+m = 4, so perm(4) = 4! = 24 so \(O(24) = O(1)\)

[682/Easy] Baseball Game

Problem

  • You are keeping score for a baseball game with strange rules. The game consists of several rounds, where the scores of past rounds may affect future rounds’ scores.

  • At the beginning of the game, you start with an empty record. You are given a list of strings ops, where ops[i] is the i-th operation you must apply to the record and is one of the following:

    • An integer x - Record a new score of x.
    • "+" - Record a new score that is the sum of the previous two scores. It is guaranteed there will always be two previous scores.
    • "D" - Record a new score that is double the previous score. It is guaranteed there will always be a previous score.
    • "C" - Invalidate the previous score, removing it from the record. It is guaranteed there will always be a previous score.
  • Return the sum of all the scores on the record.

  • Example 1:

Input: ops = ["5","2","C","D","+"]
Output: 30
Explanation:
"5" - Add 5 to the record, record is now [5].
"2" - Add 2 to the record, record is now [5, 2].
"C" - Invalidate and remove the previous score, record is now [5].
"D" - Add 2 * 5 = 10 to the record, record is now [5, 10].
"+" - Add 5 + 10 = 15 to the record, record is now [5, 10, 15].
The total sum is 5 + 10 + 15 = 30.
  • Example 2:
Input: ops = ["5","-2","4","C","D","9","+","+"]
Output: 27
Explanation:
"5" - Add 5 to the record, record is now [5].
"-2" - Add -2 to the record, record is now [5, -2].
"4" - Add 4 to the record, record is now [5, -2, 4].
"C" - Invalidate and remove the previous score, record is now [5, -2].
"D" - Add 2 * -2 = -4 to the record, record is now [5, -2, -4].
"9" - Add 9 to the record, record is now [5, -2, -4, 9].
"+" - Add -4 + 9 = 5 to the record, record is now [5, -2, -4, 9, 5].
"+" - Add 9 + 5 = 14 to the record, record is now [5, -2, -4, 9, 5, 14].
The total sum is 5 + -2 + -4 + 9 + 5 + 14 = 27.
  • Example 3:
Input: ops = ["1"]
Output: 1
  • Constraints:
    • 1 <= ops.length <= 1000
    • ops[i] is "C", "D", "+", or a string representing an integer in the range [-3 * 104, 3 * 104].
    • For operation "+", there will always be at least two previous scores on the record.
    • For operations "C" and "D", there will always be at least one previous score on the record.
  • See problem on LeetCode.

Solution: Loop over each operation, process it, append outcome to a list

class Solution:
    def calPoints(self, ops: List[str]) -> int:
        res = []
        
        for op in ops:
            if op.isdigit():
                res.append(int(op))
            elif op == 'C':
                res = res[:-1]
            elif op == 'D':
                res = int(res[-1])
                res.append(double*2)
            elif op == '+':
                s = int(res[-1] + res[-2])
                res.append(s)
            else:
                res.append(int(op[1:])*-1)
        
        return sum(res)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for res

[683/Hard] K Empty Slots

Problem

  • You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are turned off. We turn on exactly one bulb every day until all bulbs are on after n days.

  • You are given an array bulbs of length n where bulbs[i] = x means that on the (i+1)th day, we will turn on the bulb at position x where i is 0-indexed and x is 1-indexed.

  • Given an integer k, return the minimum day number such that there exists two turned on bulbs that have exactly k bulbs between them that are all turned off. If there isn’t such day, return -1.

  • Example 1:

Input: bulbs = [1,3,2], k = 1
Output: 2
Explanation:
On the first day: bulbs[0] = 1, first bulb is turned on: [1,0,0]
On the second day: bulbs[1] = 3, third bulb is turned on: [1,0,1]
On the third day: bulbs[2] = 2, second bulb is turned on: [1,1,1]
We return 2 because on the second day, there were two on bulbs with one off bulb between them.
  • Example 2:
Input: bulbs = [1,2,3], k = 1
Output: -1
  • Constraints:
    • n == bulbs.length
    • 1 <= n <= 2 * 104
    • 1 <= bulbs[i] <= n
    • bulbs is a permutation of numbers from 1 to n.
    • 0 <= k <= 2 * 104
  • See problem on LeetCode.

Solution


Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for res

[686/Medium] Repeated String Match

Problem

  • Given two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b to be a substring of a after repeating it, return -1.

  • Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc".

  • Example 1:

Input: a = "abcd", b = "cdabcdab"
Output: 3
Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it.
  • Example 2:
Input: a = "a", b = "aa"
Output: 2
  • Constraints:
    • 1 <= a.length, b.length <= 104
    • a and b consist of lowercase English letters.
  • See problem on LeetCode.

Solution:

  • In order for b to be in a, a’s length must be no less than b’s. And hence we need to bottom the multiplier to x (i.e., n // m) in order for a’s length to reach b.
  • On the other side, at most we multiply a by (x + 2) and then we can easily tell whether b can be part of a or not. It’s easy to tell in the first example: a = "abcd", b = "cdabcdab", we need to add 'head' and 'tail' to a, and hence x + 2. If b is not in a * (x + 2), then there’s no need to further expand a.
  • Examples:
    • Consider A = 'abc' and B = 'abcabcabc': len(B)/len(A) = 3. However, you cannot do range(1,3), you will need range(1,4). So this causes us to need at the minimum int(len(B)/len(A)) + 1.
    • Consider A = 'abc' and B = 'abcabcab': Now you get int(len(B)/len(A)) = 2 and we will still need range(1,4) so it will be int(len(B)/len(A)) + 2 because it rounded down. So we will need at least +2 to be safe so far.
    • Finally, consider the worst case scenario with residual characters in the front and beginning. A = 'abc' and B = 'cabcabca': In this case we will need to buffer in front and at the end because of how the string starts and ends (starts with c and ends with a). We get int(len(B)/len(A)) = 2. However we will need 4 copies of A to cover B: ab[cabcabca]bc. This means we need range(1,5) which means we will need int(len(B)/len(A)) + 3. This is definitely the worst case scenario because aside from the start and finish of B both needing extra copies, the rounding down, and the range(N) only going up to N-1 there is no other way to create a worse situation.
  • Here, repeatString is a while findString is b.

  • Similar approach; rehashed:
class Solution:
    def repeatedStringMatch(self, a: str, b: str) -> int:
        m, n = len(a), len(b)
        x = n // m
        inc = [0, 1, 2]
        
        for i in inc:
            if b in a * (x + i):
                return x + i
        
        return -1
class Solution:
    def repeatedStringMatch(self, repeatString: str, findString: str) -> int:
        x = len(findString) // len(repeatString)
        
        if findString in repeatString*x: 
            return a
        elif findString in repeatString*(x+1): 
            return x+1
        elif findString in repeatString*(x+2): 
            return x+2
        
        return -1

[708/Medium] Insert into a Sorted Circular Linked List

Problem

  • Given a Circular Linked List node, which is sorted in ascending order, write a function to insert a value insertVal into the list such that it remains a sorted circular list. The given node can be a reference to any single node in the list and may not necessarily be the smallest value in the circular list.

  • If there are multiple suitable places for insertion, you may choose any place to insert the new value. After the insertion, the circular list should remain sorted.

  • If the list is empty (i.e., the given node is null), you should create a new single circular list and return the reference to that single node. Otherwise, you should return the originally given node.

  • Example 1:

Input: head = [3,4,1], insertVal = 2
Output: [3,4,1,2]
Explanation: In the figure above, there is a sorted circular list of three elements. You are given a reference to the node with value 3, and we need to insert 2 into the list. The new node should be inserted between node 1 and node 3. After the insertion, the list should look like this, and we should still return node 3.

  • Example 2:
Input: head = [], insertVal = 1
Output: [1]
Explanation: The list is empty (given head is null). We create a new single circular list and return the reference to that single node.
  • Example 3:
Input: head = [1], insertVal = 0
Output: [1,0]
  • Constraints:
    • The number of nodes in the list is in the range [0, 5 * 104].
    • -106 <= Node.val, insertVal <= 106

Solution: Iteration (One-pass)

"""
# Definition for a Node.
class Node:
    def __init__(self, val=None, next=None):
        self.val = val
        self.next = next
"""

class Solution:
    def insert(self, head: 'Node', insertVal: int) -> 'Node':
        #edge cases
        if not head:
            node = Node(insertVal)
            node.next = node
            return node
                
        curr = head
        while curr.next != head:
            #insert in the normal position
            if curr.val <= insertVal <= curr.next.val:
                break
            #insert in the link position
            if curr.val > curr.next.val and (curr.val <= insertVal or insertVal <= curr.next.val):
                break
            curr = curr.next
        
        #insert new nodes
        insert_node = Node(insertVal, next=curr.next)
        curr.next = insert_node
    
        return head

[766/Easy] Toeplitz Matrix

Problem

  • Given an m x n matrix, return true if the matrix is Toeplitz. Otherwise, return false.

  • A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements.

  • Example 1:

Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]]
Output: true
Explanation:
In the above grid, the diagonals are:
"[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]".
In each diagonal all elements are the same, so the answer is True.
  • Example 2:

Input: matrix = [[1,2],[2,2]]
Output: false
Explanation:
The diagonal "[1, 2]" has different elements.
  • Constraints:
    • m == matrix.length
    • n == matrix[i].length
    • 1 <= m, n <= 20
    • 0 <= matrix[i][j] <= 99
  • See problem on LeetCode.

Solution: DP

class Solution:
    def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
        d = {}
        
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if i - j in d and d[i-j] != matrix[i][j]:
                    return False
                else: d[i-j] = matrix[i][j]
                    
        return True
Complexity
  • Time: \(O(n*m)\)
  • Space: \(O(n + m - 1)\)

Solution: One-loop

class Solution:
    def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
        for r1, r2 in zip(matrix[:-1], matrix[1:]):
            if r1[:-1] != r2[1:]:
                return False
        return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: One-liner

class Solution:
    def isToeplitzMatrix(self, matrix: List[List[int]]) -> bool:
        return all(r1[:-1] == r2[1:] for r1, r2 in zip(matrix[:-1], matrix[1:]))
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[771/Easy] Jewels and Stones

Problem

  • You’re given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels.
  • Letters are case sensitive, so "a" is considered a different type of stone from "A".

  • Example 1:
Input: jewels = "aA", stones = "aAAbbbb"
Output: 3
  • Example 2:
Input: jewels = "z", stones = "ZZ"
Output: 0
  • Constraints:
    • 1 <= jewels.length, stones.length <= 50
    • jewels and stones consist of only English letters.
    • All the characters of jewels are unique.
  • See problem on LeetCode.

Solution

class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        return sum(s in jewels for s in stones)        
  • Same approach; rehashed:
class Solution:
    def numJewelsInStones(self, jewels: str, stones: str) -> int:
        count = 0
        for i in stones:
            if i in jewels:
                count+=1
   
        return count
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[778/Hard] Swim in Rising Water

Problem

  • You are given an n x n integer matrix grid where each value grid[i][j] represents the elevation at that point (i, j).

  • The rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distances in zero time. Of course, you must stay within the boundaries of the grid during your swim.

  • Return the least time until you can reach the bottom right square (n - 1, n - 1) if you start at the top left square (0, 0).

  • Example 1:

Input: grid = [[0,2],[1,3]]
Output: 3
Explanation:
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.
You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.
  • Example 2:

Input: grid = [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
Explanation: The final route is shown.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.
  • Constraints:
    • n == grid.length
    • n == grid[i].length
    • 1 <= n <= 50
    • 0 <= grid[i][j] < n2
    • Each value grid[i][j] is unique.
  • See problem on LeetCode.

Solution: Dijkstra Algorithm + Min Heap

  • This problem is about finding the shortest path from top-left point, for i.e., index (0,0) to bottom-right point i.e, index (n-1, n-1). Dijkstra’s algorithm can be used to solve the single-source shortest-paths problem on a weighted, directed graph G = (V,E) for the case in which all edge weights are positive.

  • In our case the weight between two nodes is the max of cell value of both nodes. For example if Node1 has a value of 3 and Node2 has a value of 2, then weight between those will be max(3,2) = 3. Also when moving from one vertex/Node to another, instead of adding their weights we will take only the maximum one because travel time is given as 0.

  • We will use Min Heap to implement Dijkstra algorithm. For Min Heap we will use heapq module. We will also create a class to represent a Node.

import heapq

class Node:
    def __init__(self, x, y, val):
        # (x,y) position of node in grid
        self.x = x
        self.y = y
        
        # value of node. Will represent its weight
        self.val = val
        
    # We need this so that we can push Node object into out heap
    def __lt__(self, other):
        return self.val < other.val

# Just a wrapper to use heapq in Object Oriented Programming way
class MinHeap:
    def __init__(self):
        self.heap = []
    def push(self, node):
        heapq.heappush(self.heap, node)
    def pop(self) -> Node:
        return heapq.heappop(self.heap)
  • Now to start solving we will push the node at index (0,0) to Min Heap and start our loop.
  • We will also keep a 2D List to track which nodes we have already visited. You can instead make this a property of the node if you want instead of creating a new List.
  • We will loop until we have not visited the node at (n-1, n-1).
  • We will first pop out the minimum node from Min Heap; update our time taken and push all its unvisited neighbors into the heap.
  • Finally our time taken will be the result.
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        n = len(grid)
        
        # Our 2D List to track which Nodes have been visited
        visited = [[False]*n for _ in range(n)]
        
        # Time taken. Initialised to grid[0][0] because we can't move from
        # there unless time taken is atleast same as this
        # Suppose grid[0][0] is 3, then we cannot move unless t = 3
        t = grid[0][0]
        
        # Our MinHeap with initial element as Node at index (0,0)
        minHeap = MinHeap()
        minHeap.push(Node(0, 0, grid[0][0]))
        
        # Loop until we haven't visited the Node at (n-1, n-1)
        while not visited[n-1][n-1] and minHeap:
            # Get the Node with smallest value
            node = minHeap.pop()
            
            # Update its visited status
            visited[node.x][node.y] = True
            
            # Update the time taken
            t = max(t, node.val)
            
            # Push the neighbors of current node into the heap
            # Make sure to check their visited status first
            if node.y > 0 and not visited[node.x][node.y-1]: # Above Node
                minHeap.push(Node(node.x, node.y-1, grid[node.x][node.y-1]))
                
            if node.x > 0 and not visited[node.x-1][node.y]: # Left Node
                minHeap.push(Node(node.x-1, node.y, grid[node.x-1][node.y]))
                
            if node.y < n-1 and not visited[node.x][node.y+1]: # Below Node
                minHeap.push(Node(node.x, node.y+1, grid[node.x][node.y+1]))
                
            if node.x < n-1 and not visited[node.x+1][node.y]: # Right Node
                minHeap.push(Node(node.x+1, node.y, grid[node.x+1][node.y]))
                
        return t

Solution: Binary Search on Time/Height

  • At time ‘t’, there are t units/height of water, i.e.., each building will have a minimum height of ‘t’ due to rain water. In other words, we will be able to reach a height ‘t’ only at ‘t’ hours, not before that.

  • Also, it is given that if we have a path with height <= t at time ‘t’ from [0,0] to [n-1,m-1], we can reach from start to end in negligible amount of time.

  • So, basically for this problem time is not an issue for traveling, but we need to wait for a certain time to reach a building of height h > t.

Thus, we can perform a binary search on the minimum amount of time we need to wait, such that a path is available between [0,0] and [n-1,m-1] and all the building in the path are of height <= t.

  • Now, What should be the range of our Binary Search?
    • Initially we are at time 0, so left bound is 0. And max amount of time we need to wait can be the max height of the given buildings in the grid (in worst case), so max height of the buildings in the grid is our right bound for binary search.

    • Now, we need to perform binary search, for minimum height for which there is a path from [0,0] to [n-1,m-1].
  • How do we know that a path exists?
    • Simply by performing Depth First Traversal.
    • Here, we will travel in all the four directions (up, down, left, right) by not visiting the node we have visited earlier. And in this traversal we start from [0,0] and if we happen to touch [n-1,m-1], then we can say that a path exists.
    • This is general Depth First Traversal, but we have another constraint that we cannot move to a height > t, so we include a constraint that we can move in any of the four directions, if and only if, the height of the building in that direction is less than ‘t’.
  • Binary Search Idea:
    • If at time ‘t’, we found a path, it means there is a possibility that a path may exist at time < t. So, we again search in the boundary [0, t].
    • If at time ‘t’, we did not find a path, it means definitely there is no path in time <= t. So, we need to search in the boundary [t+1, max height of buildings in grid].
class Solution:
    def swimInWater(self, grid: List[List[int]]) -> int:
        
        def PathSearch(height, i, j, visited):
            if not visited[i][j] and grid[i][j] <= height:
                visited[i][j] = 1
                if i-1 >= 0:
                    PathSearch(height, i-1, j, visited)
                if i+1 < n:
                    PathSearch(height, i+1, j, visited)
                if j-1 >= 0:
                    PathSearch(height, i, j-1, visited)
                if j+1 < n:
                    PathSearch(height, i, j+1, visited)
        
        def BinarySearch(l, r):
            while l < r:
                mid = (l+r)//2
                visited = [[0]*m for _ in range(n)]
                PathSearch(mid, 0, 0, visited)
                if visited[n-1][m-1]:
                    r = mid
                else:
                    l = mid + 1
            return l
        
        
        n, m = len(grid), len(grid[0])
        l, r = 0, max(max(x) for x in grid)
        return BinarySearch(l, r)
  • For this Case, we have a square grid, so N == M.

Complexity

  • Time: \(O(N * N)\) for Depth First Traversal and \(O(log (50 * 50 -1))\) for Binary Search which is constant \(= O(N^2)\)
    • Why 50 * 50 -1? -> As N = 50 and the 0<=grid[i][j]<=N * N - 1. So, instead of calculating the right bound = max height of the buildings in grid, we can also make the right bound as 50 * 50 - 1.
    • For some people, they would not like to consider N * N - 1 as constant. As even though it is small, but it is still variable. (It is a debatable topic).
    • So for them, Time: \(O(N^2) * O(log(N^2)) = O(N^2) * 2 * O(logN) = O(N^2 log N)\).
  • Space: \(O(4 * N * N)\) owing to the Depth First Traversal Recursion Tree + \(O(N * M)\) for the visited array = \(O (N^2)\) overall.

[791/Medium] Custom Sort String

Problem

  • You are given two strings order and s. All the characters of order are unique and were sorted in some custom order previously.
  • Permute the characters of s so that they match the order that order was sorted. More specifically, if a character x occurs before a character y in order, then x should occur before y in the permuted string.
  • Return any permutation of s that satisfies this property.
  • Example 1:
Input: order = "cba", s = "abcd"
Output: "cbad"
Explanation: 
"a", "b", "c" appear in order, so the order of "a", "b", "c" should be "c", "b", and "a". 
Since "d" does not appear in order, it can be at any position in the returned string. "dcba", "cdba", "cbda" are also valid outputs.
  • Example 2:
Input: order = "cbafg", s = "abcd"
Output: "cbad"
  • Constraints:
    • 1 <= order.length <= 26
    • 1 <= s.length <= 200 order and s consist of lowercase English letters.
    • All the characters of order are unique.
  • See problem on LeetCode.

Solution: Build up result by looping over order and checking for chars in string; then add chars not in order to the answer

  • For each character in order, add the number of occurrences of that character in str to ans.
  • Write to ans all other letters from str, that is not in order.
class Solution:
    def customSortString(self, order: str, s: str) -> str:
        ans = ""
        
        for c in order:
            ans += c * s.count(c)
            
        for c in s:
            if c not in order:
                ans += c
        
        return ans
Complexity
  • Time: \(O(m + n)\), where m = length of order and n = length of str.
  • Space: \(O(1)\)

[792/Medium] Number of Matching Subsequences

Problem

  • Given a string s and an array of strings words, return the number of words[i] that is a subsequence of s.
  • A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
    • For example, "ace" is a subsequence of "abcde".
  • Example 1:
Input: s = "abcde", words = ["a","bb","acd","ace"]
Output: 3
Explanation: There are three strings in words that are a subsequence of s: "a", "acd", "ace".
  • Example 2:
Input: s = "dsahjpjauf", words = ["ahjpjau","ja","ahbwzgqnuk","tnmlanowax"]
Output: 2
  • Constraints:
    • 1 <= s.length <= 5 * 104
    • 1 <= words.length <= 5000
    • 1 <= words[i].length <= 50
    • s and words[i] consist of only lowercase English letters.
  • See problem on LeetCode.

Solution: Build up result by looping over order and checking for chars in string; then add chars not in order to the answer

  • For each character in order, add the number of occurrences of that character in str to ans.
  • Write to ans all other letters from str, that is not in order.
class Solution:
    def numMatchingSubseq(self, s: str, words: List[str]) -> int:

        def isSubSeq(word):
            index =- 1
            for char in word:
                # update index, always start index+1 to ensure
                # you don't reuse the prior found character
                # note that str.find() returns the index of first occurrence of the substring if found. If not found, it returns -1.
                # the syntax of find() is: str.find(sub[, start[, end]])
                # str.index() is almost the same as str.find(), the only difference is that str.index() raises an exception if the value is not found.
                index = s.find(char, index+1)
                if index == -1:
                    return False
            return True

        count = 0
        for word in words:
            if isSubSeq(word):
                count+=1

        return count
Complexity
  • Time: \(O(m + n)\), where m = length of order and n = length of str.
  • Space: \(O(1)\)

[796/Easy] Rotate String

Problem

  • Given two strings s and goal, return true if and only if s can become goal after some number of shifts on s.

  • A shift on s consists of moving the leftmost character of s to the rightmost position.
    • For example, if s = "abcde", then it will be "bcdea" after one shift.
  • Example 1:
Input: s = "abcde", goal = "cdeab"
Output: true
  • Example 2:
Input: s = "abcde", goal = "abced"
Output: false
  • Constraints:
    • 1 <= s.length, goal.length <= 100
    • s and goal consist of lowercase English letters.
  • See problem on LeetCode.

Solution

class Solution:
    def rotateString(self, s: str, goal: str) -> bool:
        if len(s) != len(goal):
            return False
        
        return goal in s + s
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[953/Easy] Verifying an Alien Dictionary/Mapping to Normal Order

Problem

  • In an alien language, surprisingly, they also use English lowercase letters, but possibly in a different order. The order of the alphabet is some permutation of lowercase letters.

  • Given a sequence of words written in the alien language, and the order of the alphabet, return true if and only if the given words are sorted lexicographically in this alien language.

  • Example 1:

Input: words = ["hello","leetcode"], order = "hlabcdefgijkmnopqrstuvwxyz"
Output: true
Explanation: As 'h' comes before 'l' in this language, then the sequence is sorted.
  • Example 2:
Input: words = ["word","world","row"], order = "worldabcefghijkmnpqstuvxyz"
Output: false
Explanation: As 'd' comes after 'l' in this language, then words[0] > words[1], hence the sequence is unsorted.
  • Example 3:
Input: words = ["apple","app"], order = "abcdefghijklmnopqrstuvwxyz"
Output: false
Explanation: The first three characters "app" match, and the second string is shorter (in size.) According to lexicographical rules "apple" > "app", because 'l' > '∅', where '∅' is defined as the blank character which is less than any other character ([More info](https://en.wikipedia.org/wiki/Lexicographical_order)).

Solution

  • Method 1:
def isAlienSorted(self, words: List, order: str):
    m = {c: i for i, c in enumerate(order)}
    words = [[m[c] for c in w] for w in words]
    return all(w1 <= w2 for w1, w2 in zip(words, words[1:]))
  • Method 2 (one-liner):
# slow, just for fun
def isAlienSorted(self, words: List, order: str):
    return words == sorted(words, key=lambda w: map(order.index, w))
Complexity
  • Time: \(O(S_1S_2)\) where \(S_1\) and \(S_2\) are the input strings
  • Space: \(O(1)\)

[986/Medium] Interval List Intersections

Problem

  • You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [start_i, end_i] and secondList[j] = [start_j, end_j]. Each list of intervals is pairwise disjoint and in sorted order.

  • Return the intersection of these two interval lists.

  • A closed interval [a, b] (with a <= b) denotes the set of real numbers x with a <= x <= b.

  • The intersection of two closed intervals is a set of real numbers that are either empty or represented as a closed interval. For example, the intersection of [1, 3] and [2, 4] is [2, 3].

  • Example 1:

Input: firstList = [[0,2],[5,10],[13,23],[24,25]], secondList = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
  • Example 2:
Input: firstList = [[1,3],[5,9]], secondList = []
Output: []
  • Constraints:
    • 0 <= firstList.length, secondList.length <= 1000
    • firstList.length + secondList.length >= 1
    • 0 <= start_i < end_i <= 109
    • end_i < start_i+1
    • 0 <= start_j < end_j <= 109
    • end_j < start_j+1
  • See problem on LeetCode.

Solution: Combine and sort, check overlap and add smaller end to result

  • Analysis:
    • Combine and sort the two lists. Yes, this isn’t as efficient as walking through the two lists once with two pointers, but at least you don’t have to worry about tracking and advancing two pointers.
    • Then, iterate through the combined list. Each iteration, if the start of the interval is less than or equal to the highest end (currentEnd) we have encountered, add that interval to the intersection list. Then, check if the end of the interval is higher than the previous highest (currentEnd).
def intervalIntersection(self, A: List[List[int]], B: List[List[int]]) -> List[List[int]]:
    combined = sorted(A + B, key=itemgetter(0))
    intersection = []
    currentEnd = -1

    for start, end in combined:
        if start <= currentEnd:
            intersection.append([start, min(currentEnd, end)])
        currentEnd = max(currentEnd, end)        

    return intersection
Complexity
  • Time: \(O((m+n)*log(m+n))\)
  • Space: \(O(m+n)\)

Solution: Two pointers

  • First part - Finding overlapping range:
    • Let’s see the possible types of overlaps:

    • What’s the condition for two ranges to have an overlapping range? Check out the figures below. If we can have a criss-crossing lock condition satisfied, we have essentially found an overlapping range.

    • After we have made sure that there is an overlapping range, we need to figure out the start and end of the overlapping range. Think of this as trying to squeeze the overlapping range as tight as possible (pushing as far right as possible for start and pushing as far left as possible for end).

  • Second part - Incrementing pointers:
    • The idea behind is to increment the pointer based on the end values of two ranges. Let’s say the current range in A has end value smaller than to equal to end value of the current range in B, that essentially means that you have exhausted that range in A and you should move on to the next range. Let’s try to visually think about this. When you are going through the images, keep track of end values of the ranges and how the little pointer triangles progress.

class Solution:
    def intervalIntersection(self, A: List[List[int]], B: List[List[int]]) -> List[List[int]]:
        i = 0
        j = 0
        
        result = []
        while i < len(A) and j < len(B):
            a_start, a_end = A[i]
            b_start, b_end = B[j]
             if a_start <= b_end and b_start <= a_end:                       # Criss-cross lock
                 result.append([max(a_start, b_start), min(a_end, b_end)])   # Squeezing
                 
             if a_end <= b_end:         # Exhausted this range in A
                 i += 1               # Point to next range in A
             else:                      # Exhausted this range in B
                 j += 1               # Point to next range in B
         return result

[989/Easy] Add to Array-Form of Integer

Problem

  • The array-form of an integer num is an array representing its digits in left to right order.
  • For example, for num = 1321, the array form is [1,3,2,1].
  • Given num, the array-form of an integer, and an integer k, return the array-form of the integer num + k.

  • Example 1:
Input: num = [1,2,0,0], k = 34
Output: [1,2,3,4]
Explanation: 1200 + 34 = 1234
  • Example 2:
Input: num = [2,7,4], k = 181
Output: [4,5,5]
Explanation: 274 + 181 = 455
  • Example 3:
Input: num = [2,1,5], k = 806
Output: [1,0,2,1]
Explanation: 215 + 806 = 1021
  • Constraints:
    • 1 <= num.length <= 104
    • 0 <= num[i] <= 9
    • num does not contain any leading zeros except for the zero itself.
    • 1 <= k <= 104

Solution: Stringify, convert to int and then add

class Solution:
    def addToArrayForm(self, num: List[int], k: int) -> List[int]:
        val = ''.join([str(n) for n in num])

        ans = int(val) + k
        
        return list(str(ans))
Complexity
  • Time: \(O(n + n) = O(n)\)
  • Space: \(O(1)\)

[1108/Easy] Defanging an IP Address

Problem

  • Given a valid (IPv4) IP address, return a defanged version of that IP address.

  • A defanged IP address replaces every period "." with "[.]".

  • Example 1:

Input: address = "1.1.1.1"
Output: "1[.]1[.]1[.]1"
  • Example 2:
Input: address = "255.100.50.0"
Output: "255[.]100[.]50[.]0"
  • Constraints:
    • The given address is a valid IPv4 address.
  • See problem on LeetCode.

Solution

class Solution:
    def defangIPaddr(self, address: str) -> str:
        return address.replace(".","[.]")
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) since str.replace() returns a new string

[1111/Medium] Maximum Nesting Depth of Two Valid Parentheses Strings

Problem

  • A string is a valid parentheses string (denoted VPS) if and only if it consists of "(" and ")" characters only, and:
    • It is the empty string, or
    • It can be written as AB (A concatenated with B), where A and B are VPS’s, or
    • It can be written as (A), where A is a VPS.
  • We can similarly define the nesting depth depth(S) of any VPS S as follows:
    • depth("") = 0
    • depth(A + B) = max(depth(A), depth(B)), where A and B are VPS’s
    • depth("(" + A + ")") = 1 + depth(A), where A is a VPS.`
  • For example, "", "()()", and "()(()())" are VPS’s (with nesting depths 0, 1, and 2), and ")(" and "(()" are not VPS’s.
  • Given a VPS seq, split it into two disjoint subsequences A and B, such that A and B are VPS’s (and A.length + B.length = seq.length).
  • Now choose any such A and B such that max(depth(A), depth(B)) is the minimum possible value.
  • Return an answer array (of length seq.length) that encodes such a choice of A and B: answer[i] = 0 if seq[i] is part of A, else answer[i] = 1. Note that even though multiple answers may exist, you may return any of them.

  • Example 1:
Input: seq = "(()())"
Output: [0,1,1,1,1,0]
  • Example 2:
Input: seq = "()(())()"
Output: [0,0,0,1,1,0,1,1]

Solution:

class Solution:
    def maxDepthAfterSplit(self, seq: str) -> List[int]:
        left = 0
        right = 0
        res = []
        
        for paren in seq:
            if paren == "(":
                res.append(left)
                # alternate between 0 and 1
                left = 1-left
            elif paren == ")":
                res.append(right)
                # alternate between 0 and 1
                right = 1-right
                
        return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for res

[1288/Medium] Remove Covered Intervals

Problem

  • Given an array intervals where intervals[i] = [l_i, r_i] represent the interval [l_i, r_i), remove all intervals that are covered by another interval in the list.

  • The interval [a, b) is covered by the interval [c, d) if and only if c <= a and b <= d.

  • Return the number of remaining intervals.

  • Example 1:

Input: intervals = [[1,4],[3,6],[2,8]]
Output: 2
Explanation: Interval [3,6] is covered by [2,8], therefore it is removed.
  • Example 2:
Input: intervals = [[1,4],[2,3]]
Output: 1
  • Constraints:
    • 1 <= intervals.length <= 1000
    • intervals[i].length == 2
    • 0 <= l_i < r_i <= 105
    • All the given intervals are unique.
  • See problem on LeetCode.

Solution: Sort start point by ascending order, end point by descending order; for interval overlaps, check if the current interval’s end point is less than or equal than previous interval’s end point

  • Greedy Algorithm:
    • The idea of greedy algorithm is to pick the locally optimal move at each step, which would lead to the globally optimal solution.
    • Typical greedy solution has \(\mathcal{O}(n \log n)\) time complexity and consists of two steps:
      • Figure out how to sort the input data. That would take \(\mathcal{O}(n \log n)\) time, and could be done directly by sorting or indirectly by using heap data structure. Usually sorting is better than heap usage because of gain in space.
      • Parse the sorted input in \(\mathcal{O}(n)\) time to construct a solution.
    • In the case of already sorted input, the greedy solution could have \(\mathcal{O}(n)\) time complexity, here is an example.
  • Intuition:
    • Let us figure out how to sort the input. The idea to sort by start point is pretty obvious, because it simplifies further parsing:

    • Let us consider two subsequent intervals after sorting. Since sorting ensures that start1 < start2, it’s sufficient to compare the end boundaries:

      • If end1 < end2, the intervals won’t completely cover one another, though they have some overlapping.
      • If end1 >= end2, the interval 2 is covered by the interval 1.
      • Edge case: How to treat intervals which share the same start point:

        • We’ve missed an important edge case in the previous discussion: what if two intervals share the start point, i.e. start1 == start2?

        • The above algorithm will fail because it cannot distinguish these two situations as follows:

        • One of the intervals is covered by another, but if we sort only by the start point, we would not know which one. Hence, we need to sort by the end point as well.

        If two intervals share the same start point, one has to put the longer interval in front.

        • This way the above algorithm would work fine here as well. Moreover, it can deal with more complex cases, like the one below:
  • Algorithm:
    • Sort in the ascending order by the start point. If two intervals share the same start point, put the longer one to be the first. If the array is sorted by start point in ascending order and end point in descending order, then we just need to check whether the current interval’s end is less or equal than previous end point. If it is, then the current interval is covered by one of previous intervals. Otherwise, the current interval’s starting point should be always greater or equal than the previous end point.
    • Initiate the number of non-covered intervals: count = 0.
    • Iterate over sorted intervals and compare end points.
      • If the current interval is not covered by the previous one end > prev_end, increase the number of non-covered intervals. Assign the current interval to be previous for the next step.
      • Otherwise, the current interval is covered by the previous one. Do nothing.
    • Return count.
    • For e.g., [1, 3], [2, 3], [2, 5], [2, 4]. Sorting the start point in ascending order and end point in descending order, we get, [1, 3], [2, 5], [2, 4], [2, 3].
      • For interval [2, 5], since 5 is not <= 3, there is no overlap.
      • For interval [2, 4], since 4 <= 5, [2, 4] is covered by one of the previous intervals.
      • For interval [2, 3], since 3 <= 4, [2, 3] is covered by one of the previous intervals.
class Solution:
    def removeCoveredIntervals(self, intervals: List[List[int]]) -> int:       
        if len(intervals) == 1:
            return 1
            
        # sort by start point in ascending order and end point in descending order
        # in other words, sort by start point.
        # If two intervals share the same start point
        # put the longer one to be the first.
        intervals.sort(key = lambda x: (x[0], -x[1]))
        count = 0
        
        prev_end = 0
        for _, end in intervals:
            # if current interval is not covered
            # by the previous one
            if end > prev_end:
                count += 1    
                prev_end = end
        
        return count
Complexity
  • Time: \(O(n\log{n})\) since the sorting dominates the complexity of the algorithm.
  • Space: \({O(n)}\) or \(\mathcal{O}(\log{n})\)
    • The space complexity of the sorting algorithm depends on the implementation of each program language.
    • For instance, the sorted() function in Python is implemented with the Timsort algorithm whose space complexity is \(\mathcal{O}(n)\).
    • In Java, the Arrays.sort() is implemented as a variant of quicksort algorithm whose space complexity is \(\mathcal{O}(\log{n})\).

[1344/Medium] Angle Between Hands of a Clock

Problem

  • Given two numbers, hour and minutes, return the smaller angle (in degrees) formed between the hour and the minute hand.

  • Answers within 10^-5 of the actual value will be accepted as correct.

  • Example 1:

Input: hour = 12, minutes = 30
Output: 165
  • Example 2:

Input: hour = 3, minutes = 30
Output: 75
  • Example 3:

Input: hour = 3, minutes = 15
Output: 7.5
  • Constraints:
    • 1 <= hour <= 12
    • 0 <= minutes <= 59
  • See problem on LeetCode.

Solution: Mathematical Solution

  • The minute hand moves by minute_deg_per_min = 360° / 60 = 6° per minute.
  • The hour hand moves by hour_deg_per_hour = 360° / 12 = 30° per hour.
  • The hour hand has an additional movement of hour_deg_per_min = hour_deg_per_hour / 60 = 30° / 60 = 0.5° per minute.

  • Therefore we get the following movements:
hour_hand_deg = (hour * hour_deg_per_hour) + (minutes * hour_deg_per_min) = hour * 30 + minutes * 0.5
minute_hand_deg = minutes * minute_deg_per_min = minutes * 6
  • We need the absolute difference between those two:
diff_deg = |hour_hand_deg - minute_hand_deg| = |hour * 30 + minutes * 0.5 - minutes * 6| = |hour * 30 - minutes * 5.5|
  • As we can easily see when looking at a clock there are two different angles between the hands:
    • The minimum angle on one side is between 0° and 180°.
    • The maximum angle on the other side is between 180° and 360°.
  • We need the minimum angle. If our formula returned a number above 180° we got the maximum angle. We can calculate the minimum angle by subtracting the maximum angle from 360°.
class Solution:
    def angleClock(self, hour: int, minutes: int) -> float:
        diff = abs(30 * hour - 5.5 * minutes)
        return diff if diff <= 180 else 360 - diff
Complexity
  • Time: \(O(1)\)
  • Space: \(O(1)\)

[1389/Easy] Create Target Array in the Given Order

Problem

  • Given two arrays of integers nums and index. Your task is to create target array under the following rules:

    • Initially target array is empty.
    • From left to right read nums[i] and index[i], insert at index index[i] the value nums[i] in target array.
    • Repeat the previous step until there are no elements to read in nums and index.
  • Return the target array.

  • It is guaranteed that the insertion operations will be valid.

  • Example 1:

Input: nums = [0,1,2,3,4], index = [0,1,2,2,1]
Output: [0,4,1,3,2]
Explanation:
nums       index     target
0            0        [0]
1            1        [0,1]
2            2        [0,1,2]
3            2        [0,1,3,2]
4            1        [0,4,1,3,2]
  • Example 2:
Input: nums = [1,2,3,4,0], index = [0,1,2,3,0]
Output: [0,1,2,3,4]
Explanation:
nums       index     target
1            0        [1]
2            1        [1,2]
3            2        [1,2,3]
4            3        [1,2,3,4]
0            0        [0,1,2,3,4]
  • Example 3:
Input: nums = [1], index = [0]
Output: [1]
  • Constraints:
    • 1 <= nums.length, index.length <= 100
    • nums.length == index.length
    • 0 <= nums[i] <= 100
    • 0 <= index[i] <= i
  • See problem on LeetCode.

Solution: Insert into list with .insert()

  • The .insert() method inserts an element to the list at the specified index. The syntax is list.insert(index, element).
class Solution:
    def createTargetArray(self, n: List[int], i: List[int]) -> List[int]:
        t = []
        for num in range(len(n)):
            t.insert(i[num], n[num])
            
        return t
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Iterate using zip, insert into list with .insert()

class Solution:
    def createTargetArray(self, nums: List[int], index: List[int]) -> List[int]:
        res = []
        for i, j in zip(nums, index):
            res.insert(j, i)
            
        return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[1446/Easy] Consecutive Characters

Problem

  • The power of the string is the maximum length of a non-empty substring that contains only one unique character.

  • Given a string s, return the power of s.

  • Example 1:

Input: s = "leetcode"
Output: 2
Explanation: The substring "ee" is of length 2 with the character 'e' only.
  • Example 2:
Input: s = "abbcccddddeeeeedcba"
Output: 5
Explanation: The substring "eeeee" is of length 5 with the character 'e' only.
  • Constraints:
    • 1 <= s.length <= 500
    • s consists of only lowercase English letters.
  • See problem on LeetCode.

Solution:

class Solution:
    def maxPower(self, s: str) -> int:
        ans = count = 1
        
        for i in range(1, len(s)):
            if s[i-1] == s[i]:
                count += 1
                ans = max(ans, count)
            else:
                count = 1
                
        return ans
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[1480/Easy] Running Sum of 1D Array

Problem

  • Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

  • Return the running sum of nums.

  • Example 1:

Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].
  • Example 2:
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].
  • Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]
  • Constraints:
    • 1 <= nums.length <= 1000
    • -10^6 <= nums[i] <= 10^6
  • See problem on LeetCode.

Solution: List slicing

  • On each iteration of the for loop, we append the sum up to that index (sliced list) to a new list.
class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
       res = []
        
        for i in range(len(nums)):
            res.append(sum(nums[:i+1]))
            
        return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: DP

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        dp = [0] * (len(nums)+1)
        
        
        for index in range(1,len(nums)+1):
            dp[index] = dp[index-1]+nums[index-1]
        
        
        return dp[1:]
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

Solution: Add previous to current

def runningSum(self, nums: List[int]) -> List[int]:
        n = len(nums)
        
        for i in range(1, n):
            nums[i] += nums[i-1]
        
        return nums
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[1570/Medium] Dot Product of Two Sparse Vectors

Problem

  • Given two sparse vectors, compute their dot product.

  • Implement class SparseVector:
    • SparseVector(nums) Initializes the object with the vector nums
    • dotProduct(vec) Compute the dot product between the instance of SparseVector and vec
  • A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector.

  • Follow up: What if only one of the vectors is sparse?

  • Example 1:
Input: nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
Output: 8
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8
  • Example 2:
Input: nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]
Output: 0
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0
  • Example 3:
Input: nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
Output: 6
  • Constraints:
    • n == nums1.length == nums2.length
    • 1 <= n <= 10^5
    • 0 <= nums1[i], nums2[i] <= 100
  • See problem on LeetCode.

Solution: Mark non-zero seen and multiply-add (MADD) them

class SparseVector:
    def __init__(self, nums: List[int]):
        self.seen = {}
        for i, n in enumerate(nums):
            if n:
                self.seen[i] = n              

    # Return the dotProduct of two sparse vectors
    def dotProduct(self, vec: 'SparseVector') -> int:
        res = 0
        for j, n in vec.seen.items():
            if j in self.seen:
                res += n * self.seen[j]
        return res
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: One-liner

class SparseVector:
    def __init__(self, A: List[int]):
        self.m = {i: x for i, x in enumerate(A) if x}
    def dotProduct(self, other: 'SparseVector') -> int:
        return sum([x * y for i, x in self.m.items() for j, y in other.m.items() if i == j])
Complexity
  • Time: \(O(n)\)
  • Space: $$O(n)$

[1662/Easy] Check If Two String Arrays are Equivalent

Problem

  • Given two string arrays word1 and word2, return true if the two arrays represent the same string, and false otherwise.

  • A string is represented by an array if the array elements concatenated in order forms the string.

  • Example 1:

Input: word1 = ["ab", "c"], word2 = ["a", "bc"]
Output: true
Explanation:
word1 represents string "ab" + "c" -> "abc"
word2 represents string "a" + "bc" -> "abc"
The strings are the same, so return true.
  • Example 2:
Input: word1 = ["a", "cb"], word2 = ["ab", "c"]
Output: false
  • Example 3:
Input: word1  = ["abc", "d", "defg"], word2 = ["abcddefg"]
Output: true
  • Constraints:
    • 1 <= word1.length, word2.length <= 103
    • 1 <= word1[i].length, word2[i].length <= 103
    • 1 <= sum(word1[i].length), sum(word2[i].length) <= 103
    • word1[i] and word2[i] consist of lowercase letters.
  • See problem on LeetCode.

Solution: Use ''.join()

class Solution:
    def arrayStringsAreEqual(self, word1: List[str], word2: List[str]) -> bool:     
        return "".join(word1) == "".join(word2)
Complexity
  • Time: \(O(max(n,m)\) where \(n\) and \(m\) are the lengths of word1 and word2 respectively
  • Space: $$O(1)$

[1629/Easy] Slowest Key

  • A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

  • You are given a string keysPressed of length n, where keysPressed[i] was the i-th key pressed in the testing sequence, and a sorted list releaseTimes, where releaseTimes[i] was the time the i-th key was released. Both arrays are 0-indexed. The 0th key was pressed at the time 0, and every subsequent key was pressed at the exact time the previous key was released.

  • The tester wants to know the key of the keypress that had the longest duration. The i-th keypress had a duration of releaseTimes[i] - releaseTimes[i - 1], and the 0th keypress had a duration of releaseTimes[0].

  • Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key may not have had the same duration.

  • Return the key of the keypress that had the longest duration. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.

  • Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypresses were as follows:
Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
'c' is lexicographically larger than 'b', so the answer is 'c'.
  • Example 2:
Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output: "a"
Explanation: The keypresses were as follows:
Keypress for 's' had a duration of 12.
Keypress for 'p' had a duration of 23 - 12 = 11.
Keypress for 'u' had a duration of 36 - 23 = 13.
Keypress for 'd' had a duration of 46 - 36 = 10.
Keypress for 'a' had a duration of 62 - 46 = 16.
The longest of these was the keypress for 'a' with duration 16.
  • Constraints:
releaseTimes.length == n
keysPressed.length == n
2 <= n <= 1000
1 <= releaseTimes[i] <= 109
releaseTimes[i] < releaseTimes[i+1]
keysPressed contains only lowercase English letters.

Solution: Max-heap

from heapq import heappush

class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        res, prev = (0, []), 0
        
        for time, letter in zip(releaseTimes, keysPressed):
            if time - prev > res[0]:
                res = (time - prev, [-ord(letter)])
            elif time - prev == res[0]:
                heappush(res[1], -ord(letter))
            prev = time
        
        return chr(-res[1][0])

Solution: Calculate successive keys’ release times and compare with the max duration

  • Using max():
class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        key = [keysPressed[0]]
        max_dur = releaseTimes[0]

        for i in range(1, len(releaseTimes)):
            if releaseTimes[i] - releaseTimes[i - 1] == max_dur:
                key.append(keysPressed[i])

            if releaseTimes[i] - releaseTimes[i - 1] > max_dur:
                max_dur = releaseTimes[i] - releaseTimes[i - 1]
                key = [keysPressed[i]]

        return max(key)
  • Using sorted():
class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        res, prev = (0, []), 0
        
        for time, letter in zip(releaseTimes, keysPressed):
            if time - prev > res[0]:
                res = (time - prev, [letter])
            elif time - prev == res[0]:
                res[1].append(letter)
            prev = time
        
        return sorted(res[1])[-1]
  • Same approach; cleaner:
class Solution:
    def slowestKey(self, releaseTimes: List[int], keysPressed: str) -> str:
        res, prev = (0, ''), 0
        
        for time, letter in zip(releaseTimes, keysPressed):
            if time - prev >= res[0]:
                res = max(res, (time - prev, letter))
            prev = time
        
        return res[1]
Complexity
  • Time: \(O(n)\)
  • Space: $$O(1)$

[1641/Medium] Count Sorted Vowel Strings

Problem

  • Given an integer n, return the number of strings of length n that consist only of vowels (a, e, i, o, u) and are lexicographically sorted.
  • A string s is lexicographically sorted if for all valid i, s[i] is the same as or comes before s[i+1] in the alphabet.

  • Example 1:
Input: n = 1
Output: 5
Explanation: The 5 sorted strings that consist of vowels only are ["a","e","i","o","u"].
  • Example 2:
Input: n = 2
Output: 15
Explanation: The 15 sorted strings that consist of vowels only are
["aa","ae","ai","ao","au","ee","ei","eo","eu","ii","io","iu","oo","ou","uu"].
Note that "ea" is not a valid string since 'e' comes after 'a' in the alphabet.
  • Example 3:
Input: n = 33
Output: 66045
  • Constraints:
    • 1 <= n <= 50
  • See problem on LeetCode.

Solution: Brute Force Using Backtracking

  • Intuition:
    • Backtracking is a general approach that works by identifying potential candidates and incrementally builds the solution using depth first search. Once the required goal is reached, we backtrack to explore another potential candidate.
    • The backtracking solution for the given problem would be, to begin with, the first vowel character, continue using the same vowel until we build a string of length \(n\). Post that, backtrack and use the next character.
    • As we are picking up each vowel character in alphabetically sorted order (first a, then e, then i, and so on), we know that the resultant string is always lexicographically sorted.
    • Example, For n = 3, pick the first vowel a, continue picking up the same vowel until we reach \(n\). The resultant string would be aaa. Now that we have found our first combination, backtrack and pick up the next character e. Our next string would be aae. The process continues until all the vowel characters are explored.
  • Algorithm:
    • As we start with the first vowel a then e and so on, we need a way to determine the current vowel in a recursive function. We use an integer variable vowel for that purpose, where 1 denotes a, 2 denotes e, and so on.
    • We begin with \(n\) positions and first vowel a(1) and use method countVowelStringUtil to recursively compute all the combinations.
    • On every recursion, we decrement n by 1 as we have used up that position and continue passing the same vowel. On the backtrack, we move on to the next vowel.
  • Base Case:
    • If n = 0, we have reached the end of the string and found one combination. Hence, we return 1 and backtrack.
  • We try all the possibilities where adding the new vowel doesn’t break the lexicographic order. For example, if the last character of our actual possibility is ‘e’, we can’t add an ‘a’ after it.
def count(n, last=''):
    if n == 0:
        return 1
    else:
        nb = 0
        for vowel in ['a', 'e', 'i', 'o', 'u']:
            if last <= vowel:
                nb += count(n-1, vowel)
        return nb
Complexity
  • Time: \(\mathcal{O}(n^{5})\). We have to calculate the size of the recursion tree. Let’s analyze the number of nodes generated at each level. The following figure illustrates the number of enumerations at level 2.

  • Space: \(\mathcal{O}(n)\). This space will be used to store the recursion stack. We can’t have more than \(n\) recursive calls on the call stack at any time.

Solution: Recursion + Memoization, Top Down Dynamic Programming

  • If you have solved problems on Dynamic Programming before, it is pretty easy to optimize the Approach 2 using Memoization. In the above approach, the problem can be broken down into subproblems. On drawing the recursion tree, it is evident that the same subproblems are computed multiple times.

  • In the above recursion tree, we could see that subproblems (2, 4), (2, 3) and (1, 2) are computed multiple times and the result would be always the same. The problem has Overlapping Subproblem and can be solved using Dynamic Programming.

  • Algorithm:
    • We could have stored the results of our computation for the first time and used it later. This technique of computing once and returning the stored value is called Memoization.
    • As we must store results for every \(n\) and \(\text{vowels}\), we use a two dimensional array \(\text{memo}\) of size n x 5 and follow the following steps for each recursive call:
      • Check in the memo if we have already calculated results for a given \(n\) and \(\text{vowels}\) and return the result stored in the memo.
      • Save the results of any calculations to \(\text{memo}\).

        Since arrays are 0-indexed, we initialize the array of size (n + 1, 5 + 1) and use the range 1..n and 1..5 for easier understanding and keeping it consistent with the previous approach.

  • Key Insight:
    • The fact that we are building strings from vowels really doesn’t matter. What does matter is that we have 5 elements and we want to know how many combinations (with replacement) we can make.
  • Approach:
    • Each iteration we have 2 choices. Add the current letter to the string or move on to the next letter.
    • Once the string is n characters long, return 1 to signify we made 1 string. And if we reach the 6th letter, then return 0 because there are only 5 vowels to choose from so we cannot complete the string.
    • Finally as the recursive function unwinds sum up how many strings were made.
@functools.lru_cache(None)

class Solution:
    def countVowelStrings(self, n: int) -> int:
        def helper(i, j):
            if i == 5:
                return 0
            if j == 0:
                return 1
            return helper(i+1, j) + helper(i, j-1)
        return helper(0, n)
Complexity
  • Time: \(\mathcal{O}(n)\), as for every \(n\) and \(\text{vowels}\) we calculate result only once, the total number of nodes in the recursion tree would be n x 5, which is roughly equal to \(n\).
  • Space: \(\mathcal{O}(n)\), as we use a 2D array \(\text{memo}\) of size n x 5 and the size of internal call stack would also be \(n\) (As explained in Approach 2), the space complexity would be \(\mathcal{O}(n+n) = \mathcal{O}(n)\).

Solution: One-liner

def countVowelStrings(self, n: int) -> int:
    return len(list(itertools.combinations_with_replacement(range(5), n)))

Solution: Bottom Up Dynamic Programming, Tabulation

  • Intuition:
    • This is another approach to solve the Dynamic Programming problems. We use the iterative approach and store the result of subproblems in bottom-up fashion also known as Tabulation. We store the results of previous computations in tabular format and use those results for the next computations.
  • Algorithm:
    • We maintain a 2D array, dp of size n x 5 where, dp[n][vowels] denotes the total number of combinations for given n and number of \text{vowels}. Using the recurrence relation established in Approach 2, we could iteratively calculate the value for dp[n][vowels] as,
        dp[n][vowels] = dp[n - 1][vowels] + dp[n][vowels - 1]
      
    • As this is the Bottom Up approach to solve the problem, we must initialize the table for the base cases. The base cases are the same as in Approach 2.
      • If n = 1, the number of combinations are always equal to number of vowels. Hence, we initialize all the values of dp[1][vowels] with vowels.
      • If vowels = 1, the number of combinations are always equal to 1. Hence, we initialize all the values of dp[n][1] with 1.
class Solution:
    def countVowelStrings(self, k: int) -> int:        
        dp = [1] * 5
                                
        for _ in range(1, k):
            for i in range(1, 5):                    
                dp[i] = dp[i] + dp[i-1]
        
        return sum(dp)
Complexity
  • Time: \(\mathcal{O}(n)\), as we iterate 5 x n times to fill the dp array, the time complexity is approximately \(\mathcal{O}(5 \cdot n) =\mathcal{O}(n)\).
  • Space: \(\mathcal{O}(n)\), as we use a 2D array of size n x 5, the space complexity would be \(\mathcal{O}(n)\).

Solution: Math

  • Intuition and Algorithm:
    • The problem is a variant of finding Combinations. Mathematically, the problem can be described as, given 5 vowels (let k = 5k=5), we want to find the number of combinations using only nn vowels. Also, we can repeat each of those vowels multiple times.

      In other words, from \(k\) vowels (\(k = 5\)), we can choose \(n\) vowels with repetition. Denoted as \(\left(\!\!{k \choose n}\!\!\right)\), the formulae for Combination with Repetition is given by \(\left(\!\!{k \choose n}\!\!\right) = \frac{(k + n - 1)!}{ (k - 1)! n!}\)

    • We know that the \(k\) value is 5 as there are always 5 vowels to choose from. Substituting \(k\) as 5 in above formulae, \(\left(\!\!{5\choose n}\!\!\right) = \frac{(n+4) \cdot (n+3) \cdot (n+2) \cdot (n+1)}{24}\)
    • The derivation can be illustrated as follows.

class Solution:
    def countVowelStrings(self, k: int) -> int:        
        return (k + 4) * (k + 3) * (k + 2) * (k + 1) // 24
Complexity
  • Time: \(\mathcal{O}(1)\), as the approach runs in constant time.
  • Space: \(\mathcal{O}(1)\), as the approach uses constant extra space.

[1678/Easy] Goal Parser Interpretation

Problem

  • You own a Goal Parser that can interpret a string command. The command consists of an alphabet of “G”, “()” and/or “(al)” in some order. The Goal Parser will interpret “G” as the string “G”, “()” as the string “o”, and “(al)” as the string “al”. The interpreted strings are then concatenated in the original order.

  • Given the string command, return the Goal Parser’s interpretation of command.

  • Example 1:

Input: command = "G()(al)"
Output: "Goal"
Explanation: The Goal Parser interprets the command as follows:
G -> G
() -> o
(al) -> al
The final concatenated result is "Goal".
  • Example 2:
Input: command = "G()()()()(al)"
Output: "Gooooal"
  • Example 3:
Input: command = "(al)G(al)()()G"
Output: "alGalooG"
  • Constraints:
    • 1 <= command.length <= 100
    • command consists of "G", "()", and/or "(al)" in some order.
  • See problem on LeetCode.

Solution: Replace longest substring first, then replace smaller ones

class Solution:
    def interpret(self, command: str) -> str:
        command = command.replace('()','o')
        command = command.replace('(', '')
        command = command.replace(')','')
        
        return command
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) since a new string is created when applying .replace() (as strings are immutable)

[1732/Easy] Find the Highest Altitude

Problem

  • There is a biker going on a road trip. The road trip consists of n + 1 points at different altitudes. The biker starts his trip on point 0 with altitude equal 0.

  • You are given an integer array gain of length n where gain[i] is the net gain in altitude between points i and i + 1 for all (0 <= i < n). Return the highest altitude of a point.

  • Example 1:

Input: gain = [-5,1,5,0,-7]
Output: 1
Explanation: The altitudes are [0,-5,-4,1,1,-6]. The highest is 1.
  • Example 2:
Input: gain = [-4,-3,-2,-1,4,3,2]
Output: 0
Explanation: The altitudes are [0,-4,-7,-9,-10,-6,-3,-1]. The highest is 0.
  • Constraints:
    • n == gain.length
    • 1 <= n <= 100
    • -100 <= gain[i] <= 100
  • See problem on LeetCode.

Solution: DP

class Solution:
    def largestAltitude(self, gain: List[int]) -> int:
        dp = [0] * (len(gain)+1)
        dp[0] = 0
        
        for index in range (1, len(gain)+1):
            dp[index] = dp[index-1] + gain[index-1]
        return max(dp)

Solution: Use the gain list for iteration and append the sum of the i-th element of “res” and the i-th element of the “gain” list; then, find the max of the resulting list

class Solution:
    def largestAltitude(self, gain: List[int]) -> int:
        res = [0]
        
        for i in range(len(gain)):
            res.append(res[i] + gain[i])
            
        return max(res)

[1762/Medium] Buildings With an Ocean View

Problem

  • There are n buildings in a line. You are given an integer array heights of size n that represents the heights of the buildings in the line.
  • The ocean is to the right of the buildings. A building has an ocean view if the building can see the ocean without obstructions. Formally, a building has an ocean view if all the buildings to its right have a smaller height.
  • Return a list of indices (0-indexed) of buildings that have an ocean view, sorted in increasing order.

  • Example 1:
Input: heights = [4,2,3,1]
Output: [0,2,3]
Explanation: Building 1 (0-indexed) does not have an ocean view because building 2 is taller.
  • Example 2:
Input: heights = [4,3,2,1]
Output: [0,1,2,3]
Explanation: All the buildings have an ocean view.
  • Example 3:
Input: heights = [1,3,2,4]
Output: [3]
Explanation: Only building 3 has an ocean view.
  • Constraints:
    • 1 <= heights.length <= 105
    • 1 <= heights[i] <= 109
  • See problem on LeetCode.

Solution: Reverse traversal and check for height violation

class Solution:
    def findBuildings(self, heights: List[int]) -> List[int]:
        retList = []
        retList.append(len(heights)-1)
        maxVal = heights[len(heights)-1]
        
        for index in range(len(heights)-2,-1,-1):
            if(heights[index] > maxVal):
                maxVal = heights[index]
                retList.append(index)
        
        return reversed(retList) # sorted(retList)
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

[1650/Medium] Lowest Common Ancestor of a Binary Tree III

  • Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).

  • Each node will have a reference to its parent node. The definition for Node is below:

class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
}
  • According to the definition of LCA on Wikipedia: “The lowest common ancestor of two nodes p and q in a tree T is the lowest node that has both p and q as descendants (where we allow a node to be a descendant of itself).”

  • Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
  • Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5 since a node can be a descendant of itself according to the LCA definition.
  • Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1
  • Constraints:
    • The number of nodes in the tree is in the range [2, 105].
    • -109 <= Node.val <= 109
    • All Node.val are unique.
    • p != q
    • p and q exist in the tree.
  • See problem on LeetCode.

Solution: Iteration

  • The idea is to store the parents (path) from root to p, and then check q’s path.
"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        path = set()
        while p:
            path.add(p)
            p = p.parent 
        while q not in path:
            q = q.parent 
        return q

Solution: Find depth difference, move pointers to the same level, then move pointers up until they meet

  • We first find the depth of each pointer, and then move each pointer to the same level in the tree. Then, we move the pointers up level by level until they meet.
  • Summary:
    • Find the depth of each pointer.
    • Move the deeper pointer up until it is at the same level as the other pointer.
    • Move each pointer up level-by-level until they meet
  • Example 1:
    Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 8
    Output: 3
    

  • Example 2:
    Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
    Output: 5
    

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def get_depth(self, p):
        # helper function to find the depth of the pointer in the tree
        depth = 0
        while p:
            p = p.parent
            depth += 1
        return depth
    
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        # find the depth of each pointer
        p_depth = self.get_depth(p)
        q_depth = self.get_depth(q)
        
        # Move the lower pointer up so that they are each at the same level. 
        # For the smaller depth (p_depth < q_depth or q_depth < p_depth), 
        # the loop will be skipped and the pointer will stay at the same depth.
        for _ in range(p_depth - q_depth):
            p = p.parent
        for _ in range(q_depth - p_depth):
            q = q.parent
        
        # Now that they are at the same depth, move them up the tree in parallel until they meet
        while p != q:
            p=p.parent
            q=q.parent
        return p
Complexity
  • Time: \(O(h)\)
  • Space: \(O(1)\)

Solution: Recursion + Set

class Solution(object):
    def lowestCommonAncestor(self, p, q):
        pVals = set()
        def traverse_up(root):
            if root == None or root in pVals:
                return root
            pVals.add(root)
            return traverse_up(root.parent)
            
        # This works because `None` in python evaluates to `False`, so doing `None or Node` will return `Node`.    
        return traverse_up(p) or traverse_up(q)

[1887/Medium] Reduction Operations to Make the Array Elements Equal

Problem

  • Given an integer array nums, your goal is to make all elements in nums equal. To complete one operation, follow these steps:
    • Find the largest value in nums. Let its index be i (0-indexed) and its value be largest. If there are multiple elements with the largest value, pick the smallest i.
    • Find the next largest value in nums strictly smaller than largest. Let its value be nextLargest.
    • Reduce nums[i] to nextLargest.
  • Return the number of operations to make all elements in nums equal.

  • Example 1:
Input: nums = [5,1,3]
Output: 3
Explanation: It takes 3 operations to make all elements in nums equal:
1. largest = 5 at index 0. nextLargest = 3. Reduce nums[0] to 3. nums = [3,1,3].
2. largest = 3 at index 0. nextLargest = 1. Reduce nums[0] to 1. nums = [1,1,3].
3. largest = 3 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1].
  • Example 2:
Input: nums = [1,1,1]
Output: 0
Explanation: All elements in nums are already equal.
  • Example 3:
Input: nums = [1,1,2,2,3]
Output: 4
Explanation: It takes 4 operations to make all elements in nums equal:
1. largest = 3 at index 4. nextLargest = 2. Reduce nums[4] to 2. nums = [1,1,2,2,2].
2. largest = 2 at index 2. nextLargest = 1. Reduce nums[2] to 1. nums = [1,1,1,2,2].
3. largest = 2 at index 3. nextLargest = 1. Reduce nums[3] to 1. nums = [1,1,1,1,2].
4. largest = 2 at index 4. nextLargest = 1. Reduce nums[4] to 1. nums = [1,1,1,1,1].
  • Constraints:
    • 1 <= nums.length <= 5 * 104
    • 1 <= nums[i] <= 5 * 104
  • See problem on LeetCode.

Solution: Sort counter in descending order and build result by adding frequencies

  • Algorithm:
    • Go through numbers and their counts in decreasing order of frequencies.
    • We keep track of remaining numbers we need to change in each iteration and add it to our result res.
  • Explanation:
    • sorted(Counter(nums).items(), reverse=True) gives items and their counts in reverse order
    • We are doing items() instead of just taking values() as we want to order the elements in dictionary by the keys and not values.
  • Example:
    • Let’s take a look at [1, 2, 2, 5, 5, 43].
    • We look at 43 and add 1 to to_change (as one of 43 needs to be changed to next number).
    • We look at 5 and add to_change to result (changing 43 to 5), and add two 5’s, which is new_elements to to_change. At this stage, we have to_change = 3.
    • We do so until the end when we are at 1, and we end up adding remaining 2’s to the result. Note that we drop the last new_elements for 1’s (since it doesn’t reflect in res).
class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        res = to_change = 0
        
        for _, new_elements in sorted(Counter(nums).items(), reverse=True):
            res += to_change
            to_change += new_elements
            
        return res
Complexity
  • Time: \(O(n\log{n})\) owing to the sort
  • Space: \(O(n)\) for storing the counts

Solution: Derive count add n-1-i

  • First sort nums, then go from left to right, if a mismatch is found at index i, i.e. nums[i] != nums[i+1], then add n-1-i to the result.

class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        nums.sort()
        n = len(nums)
        count = 0
        
        for i in range(n-1):
            if nums[i] != nums[i+1]:
                count += n-1-i
                
        return count
Complexity
  • Time: \(O(n\log{n})\) owing to the sort + \(O(n)\) owing to nums traversal = \(O(n\log{n})\)
  • Space: \(O(1)\)

Solution: One-liner

  • Same approach as above. First sort nums, then go from left to right, if a mismatch is found at index i, i.e. nums[i-1] != nums[i], then add n-i to the result.
class Solution:
    def reductionOperations(self, nums: List[int]) -> int:
        nums, n = sorted(nums), len(nums)
        return sum([n-i if nums[i-1] != nums[i] else 0 for i in range(1, n)])
Complexity
  • Time: \(O(n\log{n})\) owing to the sort + \(O(n)\) owing to nums traversal = \(O(n\log{n})\)
  • Space: \(O(1)\)

[1911/Medium] Maximum Alternating Subsequence Sum

Problem

  • The alternating sum of a 0-indexed array is defined as the sum of the elements at even indices minus the sum of the elements at odd indices.
  • For example, the alternating sum of [4,2,5,3] is (4 + 5) - (2 + 3) = 4.
  • Given an array nums, return the maximum alternating sum of any subsequence of nums (after reindexing the elements of the subsequence).
  • A subsequence of an array is a new array generated from the original array by deleting some elements (possibly none) without changing the remaining elements’ relative order. For example, [2,7,4] is a subsequence of [4,2,3,7,2,1,4] (the underlined elements), while [2,4,2] is not.
  • Example 1:
Input: nums = [4,2,5,3]
Output: 7
Explanation: It is optimal to choose the subsequence [4,2,5] with alternating sum (4 + 5) - 2 = 7.
  • Example 2:
Input: nums = [5,6,7,8]
Output: 8
Explanation: It is optimal to choose the subsequence [8] with alternating sum 8.
  • Example 3:
Input: nums = [6,2,1,2,4,5]
Output: 10
Explanation: It is optimal to choose the subsequence [6,1,5] with alternating sum (6 + 5) - 1 = 10.
  • Constraints:
    • 1 <= nums.length <= 105
    • 1 <= nums[i] <= 105
  • See problem on LeetCode.

Solution: Add and subtract current number from max count

def maxAlternatingSum(self, A):
    odd = even = 0
    for a in A:
        odd, even = [max(odd, even - a), max(even, odd + a)]
    return even
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[1920/Easy] Build Array from Permutation

Problem

  • Given a zero-based permutation nums (0-indexed), build an array ans of the same length where ans[i] = nums[nums[i]] for each 0 <= i < nums.length and return it.

  • A zero-based permutation nums is an array of distinct integers from 0 to nums.length - 1 (inclusive).

  • Example 1:

Input: nums = [0,2,1,5,3,4]
Output: [0,1,2,4,5,3]
Explanation: The array ans is built as follows: 
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[0], nums[2], nums[1], nums[5], nums[3], nums[4]]
    = [0,1,2,4,5,3]
  • Example 2:
Input: nums = [5,0,1,2,3,4]
Output: [4,5,0,1,2,3]
Explanation: The array ans is built as follows:
ans = [nums[nums[0]], nums[nums[1]], nums[nums[2]], nums[nums[3]], nums[nums[4]], nums[nums[5]]]
    = [nums[5], nums[0], nums[1], nums[2], nums[3], nums[4]]
    = [4,5,0,1,2,3]
  • Constraints:
    • 1 <= nums.length <= 1000
    • 0 <= nums[i] < nums.length
    • The elements in nums are distinct.
  • Follow-up: Can you solve it without using an extra space (i.e., O(1) memory)?

  • See problem on LeetCode.

Solution: Simple list comprehension

class Solution:
    def buildArray(self, nums: List[int]) -> List[int]:
            return [nums[i] for i in nums]
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\)

Solution: Simple list comprehension

def buildArray(self, nums: List[int]) -> List[int]:
        n = len(nums);
        
        for i in range(n):
            nums[i] += n*(nums[nums[i]]%n)
        
        for i in range(n):
            nums[i] = nums[i] // n
        
        return nums
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[2042/Easy] Check if Numbers Are Ascending in a Sentence

Problem

  • A sentence is a list of tokens separated by a single space with no leading or trailing spaces. Every token is either a positive number consisting of digits 0-9 with no leading zeros, or a word consisting of lowercase English letters.
    • For example, "a puppy has 2 eyes 4 legs" is a sentence with seven tokens: "2" and "4" are numbers and the other tokens such as "puppy" are words.
  • Given a string s representing a sentence, you need to check if all the numbers in s are strictly increasing from left to right (i.e., other than the last number, each number is strictly smaller than the number on its right in s).
  • Return True if so, or False otherwise.

  • Example 1:

Input: s = "1 box has 3 blue 4 red 6 green and 12 yellow marbles"
Output: true
Explanation: The numbers in s are: 1, 3, 4, 6, 12.
They are strictly increasing from left to right: 1 < 3 < 4 < 6 < 12.
  • Example 2:
Input: s = "hello world 5 x 5"
Output: false
Explanation: The numbers in s are: 5, 5. They are not strictly increasing.
  • Example 3:

Input: s = "sunset is at 7 51 pm overnight lows will be in the low 50 and 60 s"
Output: false
Explanation: The numbers in s are: 7, 51, 50, 60. They are not strictly increasing.
  • Constraints:
    • 3 <= s.length <= 200
    • s consists of lowercase English letters, spaces, and digits from 0 to 9, inclusive.
    • The number of tokens in s is between 2 and 100, inclusive.
    • The tokens in s are separated by a single space.
    • There are at least two numbers in s.
    • Each number in s is a positive number less than 100, with no leading zeros.
    • s contains no leading or trailing spaces.
  • See problem on LeetCode.

Solution: Split the string, find numbers, check if current number is greater than previous

class Solution:
    def areNumbersAscending(self, s: str) -> bool:
        words = s.split(" ")
        prev = 0
        
        for word in words:
            if word.isnumeric(): # or word.isdigit()
            # https://datagy.io/python-isdigit
                if int(word) <= prev:
                    return False
                prev = int(word)
        
        return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(n)\) for words

Solution: One-liner

class Solution:
    def areNumbersAscending(self, s: str) -> bool:
        num = [int(i) for i in s.split() if i.isnumeric()]
        
        #check if in ascending order and unique
        return True if num == sorted(num) and len(num)==len(set(num)) else False
Complexity
  • Time: \(O(n\log{n})\) due to sorting
  • Space: \(O(n)\)

Solution: Accumulate digits of every number, check if current number is greater than previous

class Solution:
    def areNumbersAscending(self, s: str) -> bool:
        prev = -1
        curr = -1
        temp = []
                
        for ch in s: 
            if ch.isdigit(): # or ord(character) >= 48 and ord(character) <= 57: 
                # accumulate all the digits of the number 
                temp.append(ch)
            # the below else is reached when we've accumulated all the digits number 
            else:
                if len(temp) > 0:
                    curr = int("".join(temp)) 
                    
                    if curr <= prev:
                        return False
                    else:
                        prev = curr
                    
                    temp.clear() # or temp = []
        
        # check the last number
        if len(temp) > 0 and int("".join(temp)) <= prev: 
            return False
        
        return True
Complexity
  • Time: \(O(n)\)
  • Space: \(O(1)\)

[2060/Hard] Check if an Original String Exists Given Two Encoded Strings

Problem

  • An original string, consisting of lowercase English letters, can be encoded by the following steps:

  • Arbitrarily split it into a sequence of some number of non-empty substrings.
  • Arbitrarily choose some elements (possibly none) of the sequence, and replace each with its length (as a numeric string).
  • Concatenate the sequence as the encoded string.
  • For example, one way to encode an original string “abcdefghijklmnop” might be:

  • Split it as a sequence: ["ab", "cdefghijklmn", "o", "p"].
  • Choose the second and third elements to be replaced by their lengths, respectively. The sequence becomes ["ab", "12", "1", "p"].
  • Concatenate the elements of the sequence to get the encoded string: “ab121p”.
  • Given two encoded strings s1 and s2, consisting of lowercase English letters and digits 1-9 (inclusive), return true if there exists an original string that could be encoded as both s1 and s2. Otherwise, return false.

  • Note: The test cases are generated such that the number of consecutive digits in s1 and s2 does not exceed 3.

  • Example 1:
Input: s1 = "internationalization", s2 = "i18n"
Output: true
Explanation: It is possible that "internationalization" was the original string.
- "internationalization" 
  -> Split:       ["internationalization"]
  -> Do not replace any element
  -> Concatenate:  "internationalization", which is s1.
- "internationalization"
  -> Split:       ["i", "nternationalizatio", "n"]
  -> Replace:     ["i", "18",                 "n"]
  -> Concatenate:  "i18n", which is s2
  • Example 2:
Input: s1 = "l123e", s2 = "44"
Output: true
Explanation: It is possible that "leetcode" was the original string.
- "leetcode" 
  -> Split:      ["l", "e", "et", "cod", "e"]
  -> Replace:    ["l", "1", "2",  "3",   "e"]
  -> Concatenate: "l123e", which is s1.
- "leetcode" 
  -> Split:      ["leet", "code"]
  -> Replace:    ["4",    "4"]
  -> Concatenate: "44", which is s2.
  • Example 3:
Input: s1 = "a5b", s2 = "c5b"
Output: false
Explanation: It is impossible.
- The original string encoded as s1 must start with the letter 'a'.
- The original string encoded as s2 must start with the letter 'c'.
  • Constraints:
    • 1 <= s1.length, s2.length <= 40
    • s1 and s2 consist of digits 1-9 (inclusive), and lowercase English letters only.
    • The number of consecutive digits in s1 and s2 does not exceed 3.
  • See problem on LeetCode.

Solution: Top-down DP

  • Preliminaries:
    • To get the ending index of the numeric substring of ‘s’ starting at ‘start’; we use the following code:-
      def getValidPrefixLength(s,start):
          end = start
          while end < len(s) and s[end].isdigit(): end += 1
          return end
    
    • To get all possibles lengths that the numeric substrings ‘s’ could represent we use the following code:-
      @lru_cache(None)
      def possibleLengths(s): 
          """Return all possible lengths represented by numeric string s."""
          lengths = {int(s)}
          for i in range(1, len(s)):
              # add all lengths by splitting numeric string s at index i
              # & sum of first split part(i.e. x) and second split part(i.e. y) gives the answer
              lengths |= {x+y for x in possibleLengths(s[:i]) for y in possibleLengths(s[i:])}
          return length
    
  • Main logic:
    • The main logic is implement a top-down dynamic programming based approach that takes in current pointers of both strings s1 and s2 as well as the amount of prefix length lead that s2 has over s1. We consider six cases in the DP in the following order:
      1. both have reached end,
      2. s1 has not reached end and s1 has numeric prefix,
      3. s2 has not reached end and s2 has numeric prefix,
      4. There is no prefix lead difference between s1 and s2,
      5. s2 leads over s1,
      6. s1 leads over s2.
  • Below is the self-explanatory commented code telling how these cases are implemented:
@lru_cache(None)
def dp(i, j, diff):             
    """Return True if s1[i:] matches s2[j:] with given differences."""

    # If both have reached end return true if none of them are leading
    if i == len(s1) and j == len(s2): return diff == 0

    # s1 has not reached end and s1 starts with a digit
    if i < len(s1) and s1[i].isdigit(): 
        i2 = getValidPrefixLength(s1,i)
        for L in possibleLengths(s1[i:i2]): 
            # substract since lead of s2  decreases by L
            if dp(i2, j, diff-L): return True 

    # s2 has not reached end and s2 starts with a digit
    elif j < len(s2) and s2[j].isdigit(): 
        j2 = getValidPrefixLength(s2,j)
        for L in possibleLengths(s2[j:j2]): 
            # add since lead of s2 increases by L
            if dp(i, j2, diff+L): return True 

    # if none of them have integer prefix or a lead over the other
    elif diff == 0: 
        # if only one of them has reached end or current alphabets are not the same 
        if i == len(s1) or j == len(s2) or s1[i] != s2[j]: return False 
        # skip same alphabets
        return dp(i+1, j+1, 0)

    # if none of them have integer prefix & s2 lead over s1 
    elif diff > 0: 
        # no s1 to balance s2's lead
        if i == len(s1): return False
        # move s1 pointer forward and reduce diff
        return dp(i+1, j, diff-1)

    # if none of them have integer prefix & s1 lead over s2
    else: 
        # no s2 to balance s1's lead
        if j == len(s2): return False 
        # move s2 pointer forward and increase diff
        return dp(i, j+1, diff+1)
  • Finally, we call the function as dp(0, 0, 0). The main code is:
from functools import lru_cache
class Solution:
    def possiblyEquals(self, s1: str, s2: str) -> bool:
        
        def getValidPrefixLength(s,start):
            end = start
            while end < len(s) and s[end].isdigit(): end += 1
            return end
        
        @lru_cache(None)
        def possibleLengths(s): 
            """Return all possible lengths represented by numeric string s."""
            ans = {int(s)}
            for i in range(1, len(s)):
                # add all lengths by splitting numeric string s at i
                ans |= {x+y for x in possibleLengths(s[:i]) for y in possibleLengths(s[i:])}
            return ans
        
        @lru_cache(None)
        def dp(i, j, diff):             
            """Return True if s1[i:] matches s2[j:] with given differences."""
            
            # If both have reached end return true if none of them are leading
            if i == len(s1) and j == len(s2): return diff == 0
            
            # s1 has not reached end and s1 starts with a digit
            if i < len(s1) and s1[i].isdigit(): 
                i2 = getValidPrefixLength(s1,i)
                for L in possibleLengths(s1[i:i2]): 
                    # subtract since lead of s2  decreases by L
                    if dp(i2, j, diff-L): return True 
            
            # s2 has not reached end and s2 starts with a digit
            elif j < len(s2) and s2[j].isdigit(): 
                j2 = getValidPrefixLength(s2,j)
                for L in possibleLengths(s2[j:j2]): 
                    # add since lead of s2 increase by L
                    if dp(i, j2, diff+L): return True 
            
            # if none of them have integer prefix or a lead over the other
            elif diff == 0: 
                # if only one of them has reached end or current alphabets are not the same 
                if i == len(s1) or j == len(s2) or s1[i] != s2[j]: return False 
                # skip same alphabets
                return dp(i+1, j+1, 0)
            
            # if none of them have integer prefix and s2 lead over s1 
            elif diff > 0: 
                # no s1 to balance s2's lead
                if i == len(s1): return False
                # move s1 pointer forward and reduce diff
                return dp(i+1, j, diff-1)
            
            # if none of them have integer prefix and s1 lead over s2
            else: 
                # no s2 to balance s1's lead
                if j == len(s2): return False 
                # move s2 pointer forward and increase diff
                return dp(i, j+1, diff+1)
        
        # start with starts of both s1 and s2 with no lead by any of them
        return dp(0, 0, 0)

[2108/Easy] Find First Palindromic String in the Array

Problem

  • Given an array of strings words, return the first palindromic string in the array. If there is no such string, return an empty string "".

  • A string is palindromic if it reads the same forward and backward.

  • Example 1:

Input: words = ["abc","car","ada","racecar","cool"]
Output: "ada"
Explanation: The first string that is palindromic is "ada".
Note that "racecar" is also palindromic, but it is not the first.

Example 2:

Input: words = ["notapalindrome","racecar"]
Output: "racecar"
Explanation: The first and only string that is palindromic is "racecar".
  • Example 3:
Input: words = ["def","ghi"]
Output: ""
Explanation: There are no palindromic strings, so the empty string is returned.
  • Constraints:
    • 1 <= words.length <= 100
    • 1 <= words[i].length <= 100
    • words[i] consists only of lowercase English letters.
  • See problem on LeetCode.

Solution

class Solution:
    def firstPalindrome(self, words: List[str]) -> str:
        for word in words:
            if self.isPalindrome(word):
                return word
        return ""
    
    def isPalindrome(self, word) -> bool:
        l, r = 0, len(word)-1
        while l < r:
            if word[l] != word[r]:
                return False
            l += 1
            r -= 1
        return True
        # or return word == word[::-1] (less efficient)
Complexity
  • Time: \(O(nm)\) where \(n\) are the number of words in the input and \(m\) is the average length of each word.
  • Space: \(O(n)\) for word[::-1]