def divisibleSumPairs(n, k, ar): count = 0 for i in range(n): for j in range(i + 1, n): if (ar[i] + ar[j]) % k == 0: count += 1 return count

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Best Solution in O(n) without data structure Overhead: Lang: Java

import java.io.IOException;

public class Main{ public static void main(String[] args) throws IOException { Reader reader = new Reader(); int n = reader.readInt(); int k = reader.readInt(); int ret = 0; int[] remainders = new int[k];

    //O(n)
    for (int i = 0; i < n; i++) {
        int rIndex = reader.readInt() % k;
        //Special Case
        if(rIndex == 0) {
            ret += remainders[0];
        }
        //Where k is even
        else if (rIndex == k - rIndex) {
            ret += remainders[k / 2];
        }

        //general case
        else {
            ret += remainders[k - rIndex];
        }

        remainders[rIndex]++;
    }

    System.out.println(ret);
}

}

Here is my c++ solution, you can whatch the explanation here: https://youtu.be/aDv1iUSgnd8

Solution 1 : O(n^2)

int divisibleSumPairs(int n, int k, vector<int> ar) {
    int result = 0;
    for(int i = 0; i < ar.size() - 1; i++){
        for(int j = i + 1; j < ar.size(); j++){
            if( (ar[i] + ar[j]) % k == 0) result++;
        }
    }
    return result;
}

Solution 2 : O(n)

int divisibleSumPairs(int n, int k, vector<int> ar) {
    map<int, int> mp;
    for(int i = 0; i < ar.size(); i++){
        mp[ar[i] % k]++;
    }
    int result = 0;
     for(int i = 0; i <= k/2; i++){
        if(i == 0 || i*2 == k) result += (mp[i] * (mp[i] - 1)) / 2;
        else result += mp[i] * mp[k - i]; 
    }
    return result;
}