This approach greatly improves performance while keeping the logic simple and clear. It’s a great exercise to learn about optimizing loops and using modular properties in programming challenges. If you want to explore or attempt this problem yourself, click here to open it on HackerRank.

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

condition should be i <= j not i < j inshallah, correct ?

O(n) solution in JavaScript

function divisibleSumPairs(n, k, ar) {
    let pairCount = 0;
    let remainders = new Array(k).fill(0);
    
    for(let i = 0; i < n; i++){
        let num = ar[i];
        let remainder = num % k;
        let compliment = (k - remainder)% k;
        
        pairCount += remainders[compliment];
        remainders[remainder]++;
    }
    return pairCount;
}

Is there any solution but O(n^2) ?