I’m considering below solution:
def remove_duplicates(arr): i = 0 while i < len(arr)-1: if arr[i] == arr[i+1]: arr.pop(i) else: i += 1 return len(arr)
Poping from an intermediate address is O(k) as the array must shift all values to the left afterwards. The provided solution has a better time complexity due to that. If you started from the last indexes and used pop from end, this would be a better solution.