Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums. Return k.

Example

  • Input: nums = [0,0,1,1,1,2,2,3,3,4]
  • Output: 5, nums = [0,1,2,3,4,,,,,_]
  • Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively. It does not matter what you leave beyond the returned k (hence they are underscores).

Solution (C++)

The idea is to have two indeces. One index a which iterates through the whole list, “looking for the next unique element” and one in-place index b, which defines the last unique element. As the list is sorted, we can check for the next unequal element to the element in index b (as previous elements won’t appear again like in a a b a).

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int b = 0; // inplace index, points to last unique element

        // a starts at 1, since the 0-th element is always unique
        for (int a = 1; a < nums.size(); a++){
            if(nums[b] != nums[a]){ // compare last unique to new
                nums[b+1] = nums[a]; // set new unique element 
                b++;
            }
        }

        // ! increment by 1, as this should be the number of unique elements
        return b+1; 
    }
};

or

Simplified code, where we loop through each element, and only increment b, if we found a new unique element.

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {

        int b=0;
        for (auto ele : nums){
            if (b == 0 || nums[b-1] != ele){
                nums[b] = ele;
                b++;
            }
        }
        
        return b;
    }
};