Algorithm

Since we need to store only unique elements, we can use the set data structure. We can insert all the elements of the array in the set irrespective of their frequency as set only allows one occurence of each element.

Declare a set and insert all the elements of the array into the set.
The number of unique elements in array is equal to size of the set.
Traverse the set and fill the first k indices with elements in set.

Code

#include <bits/stdc++.h>
using namespace std;

// Solution class containing removeDuplicates method
class Solution {
public:
    // Removes duplicates using unordered_set and returns count of unique elements
    int removeDuplicates(vector<int>& nums) {
        // Unordered set to store elements we have already seen
        unordered_set<int> seen;

        // Index where the next unique element will be written
        int index = 0;

        // Loop over each element in the array
        for (int num : nums) {
            // If num is not in seen, it's unique
            if (seen.find(num) == seen.end()) {
                // Add this num to the set of seen numbers
                seen.insert(num);

                // Overwrite nums[index] with this unique num
                nums[index] = num;

                // Move index forward
                index++;
            }
        }
        // Return count of unique elements
        return index;
    }
};

int main() {
    vector<int> nums = {0,0,1