For Python3 Platform

I wrote the code from scratch just to get more practice and time complexity of my code is O(n+k) and not O(n^2) which is way more efficient

def divisibleSumPairs(n, k, ar):
    freq = [0] * k
    count = 0
    
    for ele in ar:
        remainder = ele % k
        needed = (k-remainder) % k
        
        count += freq[needed]
        freq[remainder] += 1
    
    return count

n, k = map(int, input().split())
ar = list(map(int, input().split()))

result = divisibleSumPairs(n, k, ar)

print(result)

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

Input

n, k = map(int, input().split()) ar = list(map(int, input().split()))

Output

print(divisibleSumPairs(n, k, ar))

Mi solución en Python 3:

def divisibleSumPairs(n, k, ar):
    # Write your code here
    contador= 0
    for i in range(n):
        for j in range(i+1,n):
            if(ar[i]+ ar[j])%k ==0:
                contador +=1
    return contador

//Java solution` public static int divisibleSumPairs(int n, int k, List ar) { int pairs = 0; // include constraints if (k > 0 && 2 <= n && n <= 100) { for (int i = 0; i < n; i++) { for (int j = 1+i; j < n; j++) { int sum = ar.get(i) + ar.get(j); if (sum % k == 0) { pairs++; } } } }

    return pairs;

}

`

//Java solution` public static int divisibleSumPairs(int n, int k, List ar) { int pairs = 0; // include constraints if (k > 0 && 2 <= n && n <= 100) { for (int i = 0; i < n; i++) { for (int j = 1+i; j < n; j++) { int sum = ar.get(i) + ar.get(j); if (sum % k == 0) { pairs++; } } } }

    return pairs;

}

`