Space efficient solution for Leetcode 80

Leetcode 80 - Medium - Remove duplicates from sorted array #2

"""
Basic solution

Time: O(nlogn) 
- The main loop iterates through the original list (n + k) times
- The inner loop iterates through the tail of the list (k * log(n))
- In the worst case, k would be n/3, which means:
  - main loop reduces to n + n/3 -> 4/3 * n -> n
  - inner loop reduces to n/3 * log(n) -> log(n)
Space: O(n)
- We are editing in-place without creating any new lists or callables
"""
class Solution(object):
    def removeDuplicates(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        k = 0
        i = 2
        while i < len(nums) - k:      
        # As we find more triplicates, the final value in the list begins to duplicate
        # E.g. ..., 3] -> ..., 3, 3] -> ..., 3, 3, 3] -> etc
        # So we need to ignore the last k indices of the list when working out if we've finished
            if nums[i] == nums[i-1] == nums[i-2]:
                k += 1
                for j in range(i+1, len(nums)):
                    nums[j-1] = nums[j]
                i -= 1      # Recheck this index after we've shifted everything left
            i += 1
        return len(nums) - k