Remove Duplicates from Sorted Array

Intuition

Because the array is sorted, every duplicate group is contiguous, so a value is new only when it differs from the last one we kept. The slow pointer always sits on the final element of the deduplicated prefix, while the fast pointer races ahead. Whenever the fast pointer sees a value that differs from the slow pointer's value, we grow the prefix by one and copy that value in. This overwrites duplicates without ever needing extra storage.

Algorithm walkthrough

1. 1Handle the trivial empty case, then start slow at index 0 as the first kept element.
2. 2Move fast from index 1 to the end of the array.
3. 3When nums[fast] differs from nums[slow], advance slow and copy nums[fast] into that slot.
4. 4Leave fast running so it keeps skipping over equal duplicates automatically.
5. 5Return slow + 1, the number of unique values now sitting at the front.

Implementation

from typing import List

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        if not nums:
            return 0
        slow = 0
        for fast in range(1, len(nums)):
            if nums[fast] != nums[slow]:
                slow += 1
                nums[slow] = nums[fast]
        return slow + 1