Hackerrank Divisible Sum Pairs (C++)

C++ :

int divisibleSumPairs(int n, int k, vector<int> ar) {
    int noOfPairs = 0, sum = 0;

    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            sum = ar[i] + ar[j];
            if(sum % k == 0){
                noOfPairs++;
            }
        } 
    }

    return noOfPairs;
}

Explanation:

This is about adding two numbers at ar[i] and ar[j], such that the sum is divisible by k. n is the ar.size(), which is actually no need to be provided, but since it is, let’s use it.

For the inner for loop, j is (i + 1), so that double summing doesn’t happen (eg. ar[2] + ar[4] and ar[4] and ar[2], since i < j is a requirement for the question. The rest is relatively straightforward.