Count pairs in array whose sum is divisible by K

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

Examples:

Input : A[] = {2, 2, 1, 7, 5, 3}, K = 4
Output : 5
Explaination : 
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’. 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, upto (k-1). So take an array 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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));
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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))

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right



Output :

 5

Time complexity: O(N)
Auxiliary space: O(1)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t, Mithun Kumar