def divisibleSumPairs(n, k, ar): freq = [0] * k count = 0

for num in ar:
    mod = num % k
    complement = (k - num) % k
    count += freq[complement]
    freq[mod] += 1

return count

My solution in python:

def divisibleSumPairs(n, k, arr):

return sum([1 for i in range(n) for j in range (i + 1 , n) if i < j and (arr[i] + arr[j]) % k == 0])

JAVA Solution

public static int divisibleSumPairs(int n, int k, List ar) {

    int count = 0;

    for(int i = 0; i<n; i++){

        for(int j = i+1; j<n; j++){

            if((ar.get(i)+ ar.get(j)) % k == 0){
                count++;
            }
        }
    }

    return count;

}

C# var remainderCounts = new int[k]; var count = 0;

foreach (var num in ar) { var remainder = num % k; var complement = (k - remainder) % k; count += remainderCounts[complement]; remainderCounts[remainder]++; }

return count;

Do not need the n in the function and using itertools to get all unique pairs:

import os
from itertools import combinations

def divisibleSumPairs(n, k, ar):
    valid_pair = 0
    all_pairs = combinations(ar, 2)
    for pair in all_pairs:
        pair_sum = sum(pair)
        if pair_sum % k == 0:
            valid_pair += 1
    return valid_pair

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')
    
    first_multiple_input = input().rstrip().split()
    n = int(first_multiple_input[0])
    k = int(first_multiple_input[1])
    ar = list(map(int, input().rstrip().split()))
    result = divisibleSumPairs(n, k, ar)

    fptr.write(str(result) + '\n')
    fptr.close()