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Count pairs in array whose sum is divisible by K

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  • Difficulty Level : Medium
  • Last Updated : 25 Dec, 2022
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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).

 Below is the implementation of the above approach:

C++




// C++ Program to count pairs
// whose sum divisible by 'K'
#include <bits/stdc++.h>
using namespace std;
 
int countKdivPairs(int A[], int n, int K)
{
    // variable for storing answer
    int count = 0;
 
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
 
            // if pair sum is divisible
            if ((A[i] + A[j]) % K == 0)
 
                // Increment count
                count++;
        }
    }
 
    return count;
}
 
// Driver code
int main()
{
 
    int A[] = { 2, 2, 1, 7, 5, 3 };
    int n = sizeof(A) / sizeof(A[0]);
    int K = 4;
 
    // Function call
    cout << countKdivPairs(A, n, K);
 
    return 0;
}


Java




/*package whatever //do not write package name here */
import java.io.*;
class GFG {
 
  static int countKdivPairs(int[] A, int n, int K)
  {
 
    // variable for storing answer
    int count = 0;
 
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
 
        // if pair sum is divisible
        if ((A[i] + A[j]) % K == 0)
 
          // Increment count
          count++;
      }
    }
 
    return count;
  }
 
  public static void main (String[] args)
  {
    int[] A = { 2, 2, 1, 7, 5, 3 };
    int n = A.length;
    int K = 4;
 
    // Function call
    System.out.println(countKdivPairs(A, n, K));
  }
}
 
// This code is contributed by utkarshshirode02


C#




// C# Program to count pairs
// whose sum divisible by 'K'
using System;
class GFG {
 
  static int countKdivPairs(int[] A, int n, int K)
  {
     
    // variable for storing answer
    int count = 0;
 
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
 
        // if pair sum is divisible
        if ((A[i] + A[j]) % K == 0)
 
          // Increment count
          count++;
      }
    }
 
    return count;
  }
 
  // Driver code
  public static void Main()
  {
 
    int[] A = { 2, 2, 1, 7, 5, 3 };
    int n = A.Length;
    int K = 4;
 
    // Function call
    Console.Write(countKdivPairs(A, n, K));
  }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




// Javascript Program to count pairs
// whose sum divisible by 'K'
 
function countKdivPairs(A, n, K)
{
    // variable for storing answer
    let count = 0
 
    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
 
            // if pair sum is divisible
            if ((A[i] + A[j]) % K == 0)
 
                // Increment count
                count++
        }
    }
 
    return count
}
 
// Driver code
let A = [ 2, 2, 1, 7, 5, 3 ]
let n = A.length
let K = 4
 
// Function call
console.log(countKdivPairs(A, n, K))
 
// This code is contributed by Samim Hossain Mondal.


Python3




# Python Program to count pairs
# whose sum divisible by 'K'
def countKdivPairs(A, n, K):
    count = 0
    for i in range(0, n):
        for j in range(i+1, n):
            if((A[i]+A[j]) % K==0):
                # Increment count
                count += 1
    return count
# Driver Code
 
A=[ 2, 2, 1, 7, 5, 3 ]
n = len(A)
K = 4
#Function call
print(countKdivPairs(A, n, K))
    


Output

5

Time complexity: O(N2), for using two nested loops.
Auxiliary Space: O(1), as constant space is used.

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

5

Time complexity: O(N) 
Auxiliary space: O(K), since K extra space has been taken.

https://www.youtube.com/watch?v=5UJvXcSUyT0


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