# Count pairs in array whose sum is divisible by K

• Difficulty Level : Medium
• Last Updated : 22 Jun, 2022

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 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 codeint 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 codeA = [2, 2, 1, 7, 5, 3]n = len(A)K = 4print(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.



## Javascript



Output :

5

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

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