Problem

Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this in place with constant memory.

For example,
Given input array nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn’t matter what you leave beyond the new length.

Analysis

We can use an integer (validIdx) to keep track of current valid length, then we traverse the array
– when nums[i] has been repeated, continue
– otherwise, write it to the array, and advance validIdx

Time complexity: O(n)
Space complexity: O(n)

Solution

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if(nums.size() == 0) return 0;
        int validIdx = 0;
        int curr = nums[0];
        for(int i=0; i<nums.size(); i++){
            //if repeat continue
            if(nums[i] == curr){
                continue;
            }
            //otherwise write the array, and move index
            else{
                nums[validIdx] = curr;
                curr = nums[i];
                validIdx=validIdx+1;
            }
        }
        nums[validIdx] = curr;
        return (validIdx+1);
    }
};