LeetCode 80 Remove Duplicates from Sorted Array II

Follow up for "Remove Duplicates": What if duplicates are allowed at most twice?

For example,
Given sorted array A = [1, 1, 1, 2, 2, 3], Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3.
It doesn't matter what you leave beyond the new length.

Diffculty
Medium

Similar Problems
[LeetCode 26] Remove Duplicates

Analysis

Similar to the problem Remove Duplicates, the only difference is that when A[end-1] == A[end] == A[i], A[i] is the duplicate number and we should skip it. So we actually only need to compare A[end-1] and A[i], since when A[end-1] == A[i], there must be A[end] = A[end-1] since it is a sorted array.

A = [1, 1, 1, 2, 2, 3]
           ^     ^
           end

Solutions

1. Cpp solution

class Solution {
public:
    int removeDuplicates(vector<int>& A) {
        const int n = A.size();
        if(n < 3) return n;
        int end = 2;
        for(int i = 2; i < n; i++) {
            if(A[i] != A[end - 2]) 
                A[end++] = A[i];
        }
        return end;
    }
};

2. Go solution

func removeDuplicates(nums []int) int {
    n := len(nums)
    if n < 3 {
        return n
    }

    end := 2

    for i := 2; i < n; i++ {
        if nums[end-2] != nums[i] {
            nums[end] = nums[i]
            end++
        }
    }
    return end
}

Reference