We use a slow pointer (l) and a fast pointer (r). The slow pointer tracks the position of the last unique element, while the fast pointer iterates through the array.

1. If len(nums) == 0, return 0.
2. Initialize l = 1.
3. For r from 1 to len(nums)-1:
   • If nums[r] != nums[r-1]:
     • nums[l] = nums[r].
     • l += 1.
4. Return l.

• Time Complexity: O(N).
• Space Complexity: O(1).

def remove_duplicates(nums):
    if not nums: return 0
    
    l = 1
    for r in range(1, len(nums)):
        if nums[r] != nums[r - 1]:
            nums[l] = nums[r]
            l += 1
            
    return l