Count pairs in array whose sum is divisible by K

Difficulty Level : Medium
Last Updated : 07 Jun, 2021

Given an array A[] and positive integer K, the task is to count the total number of pairs in the array whose sum is divisible by K.
Note: This question is a generalized version of this

Examples:

Input : A[] = {2, 2, 1, 7, 5, 3}, K = 4
Output : 5
Explanation : 
There are five pairs possible whose sum
is divisible by '4' i.e., (2, 2), 
(1, 7), (7, 5), (1, 3) and (5, 3)

Input : A[] = {5, 9, 36, 74, 52, 31, 42}, K = 3
Output : 7

Naive Approach: The simplest approach is to iterate through every pair of the array but using two nested for loops and count those pairs whose sum is divisible by ‘K’. The time complexity of this approach is O(N²).

Efficient Approach: An efficient approach is to use Hashing technique. We will separate elements into buckets depending on their (value mod K). When a number is divided by K then the remainder may be 0, 1, 2, up to (k-1). So take an array to say freq[] of size K (initialized with Zero) and increase the value of freq[A[i]%K] so that we can calculate the number of values giving remainder j on division with K.

C++

// C++ Program to count pairs
// whose sum divisible by 'K'
#include <bits/stdc++.h>
using namespace std;
 
// Program to count pairs whose sum divisible
// by 'K'
int countKdivPairs(int A[], int n, int K)
{
    // Create a frequency array to count
    // occurrences of all remainders when
    // divided by K
    int freq[K] = { 0 };
 
    // Count occurrences of all remainders
    for (int i = 0; i < n; i++)
        ++freq[A[i] % K];
 
    // If both pairs are divisible by 'K'
    int sum = freq[0] * (freq[0] - 1) / 2;
 
    // count for all i and (k-i)
    // freq pairs
    for (int i = 1; i <= K / 2 && i != (K - i); i++)
        sum += freq[i] * freq[K - i];
    // If K is even
    if (K % 2 == 0)
        sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
    return sum;
}
 
// Driver code
int main()
{
 
    int A[] = { 2, 2, 1, 7, 5, 3 };
    int n = sizeof(A) / sizeof(A[0]);
    int K = 4;
    cout << countKdivPairs(A, n, K);
 
    return 0;
}

Java

// Java program to count pairs
// whose sum divisible by 'K'
import java.util.*;
 
class Count {
    public static int countKdivPairs(int A[], int n, int K)
    {
        // Create a frequency array to count
        // occurrences of all remainders when
        // divided by K
        int freq[] = new int[K];
 
        // Count occurrences of all remainders
        for (int i = 0; i < n; i++)
            ++freq[A[i] % K];
 
        // If both pairs are divisible by 'K'
        int sum = freq[0] * (freq[0] - 1) / 2;
 
        // count for all i and (k-i)
        // freq pairs
        for (int i = 1; i <= K / 2 && i != (K - i); i++)
            sum += freq[i] * freq[K - i];
        // If K is even
        if (K % 2 == 0)
            sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
        return sum;
    }
    public static void main(String[] args)
    {
        int A[] = { 2, 2, 1, 7, 5, 3 };
        int n = 6;
        int K = 4;
        System.out.print(countKdivPairs(A, n, K));
    }
}

Python3

# Python3 code to count pairs whose
# sum is divisible by 'K'
 
# Function to count pairs whose
# sum is divisible by 'K'
def countKdivPairs(A, n, K):
     
    # Create a frequency array to count
    # occurrences of all remainders when
    # divided by K
    freq = [0] * K
     
    # Count occurrences of all remainders
    for i in range(n):
        freq[A[i] % K]+= 1
         
    # If both pairs are divisible by 'K'
    sum = freq[0] * (freq[0] - 1) / 2;
     
    # count for all i and (k-i)
    # freq pairs
    i = 1
    while(i <= K//2 and i != (K - i) ):
        sum += freq[i] * freq[K-i]
        i+= 1
 
    # If K is even
    if( K % 2 == 0 ):
        sum += (freq[K//2] * (freq[K//2]-1)/2);
     
    return int(sum)
 
# Driver code
A = [2, 2, 1, 7, 5, 3]
n = len(A)
K = 4
print(countKdivPairs(A, n, K))

C#

// C# program to count pairs
// whose sum divisible by 'K'
using System;
 
class Count
{
    public static int countKdivPairs(int []A, int n, int K)
    {
        // Create a frequency array to count
        // occurrences of all remainders when
        // divided by K
        int []freq = new int[K];
 
        // Count occurrences of all remainders
        for (int i = 0; i < n; i++)
            ++freq[A[i] % K];
 
        // If both pairs are divisible by 'K'
        int sum = freq[0] * (freq[0] - 1) / 2;
 
        // count for all i and (k-i)
        // freq pairs
        for (int i = 1; i <= K / 2 && i != (K - i); i++)
            sum += freq[i] * freq[K - i];
             
        // If K is even
        if (K % 2 == 0)
            sum += (freq[K / 2] * (freq[K / 2] - 1) / 2);
        return sum;
    }
     
    // Driver code
    static public void Main ()
    {
        int []A = { 2, 2, 1, 7, 5, 3 };
        int n = 6;
        int K = 4;
        Console.WriteLine(countKdivPairs(A, n, K));
    }
}
 
// This code is contributed by akt_mit.

PHP

<?php
// PHP Program to count pairs
// whose sum divisible by 'K'
 
// Program to count pairs whose sum
// divisible by 'K'
function countKdivPairs($A, $n, $K)
{
     
    // Create a frequency array to count
    // occurrences of all remainders when
    // divided by K
    $freq = array_fill(0, $K, 0);
 
    // Count occurrences of all remainders
    for ($i = 0; $i < $n; $i++)
        ++$freq[$A[$i] % $K];
 
    // If both pairs are divisible by 'K'
    $sum = (int)($freq[0] * ($freq[0] - 1) / 2);
 
    // count for all i and (k-i)
    // freq pairs
    for ($i = 1; $i <= $K / 2 &&
                 $i != ($K - $i); $i++)
        $sum += $freq[$i] * $freq[$K - $i];
         
    // If K is even
    if ($K % 2 == 0)
        $sum += (int)($freq[(int)($K / 2)] *
                     ($freq[(int)($K / 2)] - 1) / 2);
    return $sum;
}
 
// Driver code
$A = array( 2, 2, 1, 7, 5, 3 );
$n = count($A);
$K = 4;
echo countKdivPairs($A, $n, $K);
 
// This code is contributed by mits
?>

Javascript

<script>
    // Javascript program to count pairs whose sum divisible by 'K'
     
    function countKdivPairs(A, n, K)
    {
        // Create a frequency array to count
        // occurrences of all remainders when
        // divided by K
        let freq = new Array(K);
        freq.fill(0);
  
        // Count occurrences of all remainders
        for (let i = 0; i < n; i++)
            ++freq[A[i] % K];
  
        // If both pairs are divisible by 'K'
        let sum = freq[0] * parseInt((freq[0] - 1) / 2, 10);
  
        // count for all i and (k-i)
        // freq pairs
        for (let i = 1; i <= K / 2 && i != (K - i); i++)
            sum += freq[i] * freq[K - i];
              
        // If K is even
        if (K % 2 == 0)
            sum += parseInt(freq[parseInt(K / 2, 10)] * (freq[parseInt(K / 2, 10)] - 1) / 2, 10);
        return sum;
    }
     
    let A = [ 2, 2, 1, 7, 5, 3 ];
    let n = 6;
    let K = 4;
    document.write(countKdivPairs(A, n, K));
         
</script>

Output :

Time complexity: O(N)
Auxiliary space: O(K)

https://www.youtube.com/watch?v=5UJvXcSUyT0&list=PLM68oyaqFM7Q-sv3gA5xbzfgVkoQ0xDrW&index=16

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