## Problem

• Given a sorted array $$nums$$, remove the duplicates in-place such that each element appears only once and returns the new length.
• Do not allocate extra space for another array, you must do this by modifying the input array in-place with $$O(1)$$ extra memory.
• See problem on LeetCode.

## Algorithm

• Since the array is already sorted, we can keep two pointers $$i$$ and $$j$$, where $$i$$ is the slow-runner while $$j$$ is the fast-runner. As long as $$nums[i] = nums[j]$$, we increment $$j$$ to skip the duplicate.
• When we encounter $$nums[j] \neq nums[i]$$, the duplicate run has ended so we must copy its value to $$nums[i + 1]$$. $$i$$ is then incremented and we repeat the same process again until $$j$$ reaches the end of array.

## Code

class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
"""
@param a list of integers
@return an integer
"""

# Handle the special case that A is empty
if not nums: # or "if not len(nums)" or "if len(A) == 0"
return 0

i = 0
for j in range(1, len(nums)):
# Not a duplicate item
if nums[i] != nums[j]:
i += 1
nums[i] = nums[j]

return i + 1


## Complexity analysis

• Time complexity: $$O(n)$$. Assume that $$n$$ is the length of array. Each of $$i$$ and $$j$$ traverses at most $$n$$ steps.
• Space complexity: $$O(1)$$.