[LeetCode] 80. Remove Duplicates from Sorted Array II

Problem : https://leetcode.com/problems/remove-duplicates-from-sorted-array-ii/

Uses 2 pointers approach. Pointer j moves 1 step per time. Pointer 'i' moves only when encounter valid character.

Time Complexity : O ( N )

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        i, j = 1, 1
        count = 1
        
        while j < len(nums):
            if nums[j] == nums[j-1]:
                count += 1
                
                if count <= 2:
                    # keep 2 duplicate  in maximum
                    nums[i] = nums[j]
                    i += 1
                    j += 1
                else:
                    # skip nums[j] as there are already two duplicated numbers
                    j += 1
            else:
                count = 1
                nums[i] = nums[j]
                i += 1
                j += 1
        
        return i

A much simpler solution of overwritting unwanted letters.

Because it is a sorted array, we only need to compare if current letter is same to letter on last 2nd postion of the wanted array.

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        i = j = 2
        
        while j < len(nums):
            if nums[i-2] == nums[j]:
                j += 1
            else:
                nums[i] = nums[j]
                i += 1
                j += 1
        
        return i