Related Articles
Count number of pairs in array having sum divisible by K | SET 2
• Difficulty Level : Hard
• Last Updated : 22 Sep, 2020

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.
Examples:

Input : A[] = {2, 2, 1, 7, 5, 3}, K = 4
Output :
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 :

Approach: In the previous post, an approach using hashing is discussed. In this article, another approach using hashing is discussed.

Analyzing the statement we can say we need to make pairs of (a, b) such that:

```     (a + b) % K = 0
=>    a%K + b%K = 0
=>    a%K + b%K = K%K
=>    b%K = K%K - a%K
=>     b%K = (K - a%K) % K.     {Range of a%K => [0,K-1]}
```

The idea is a can be paired with (K — a%K) % K. Now we have to find the same for each a present in the given array.

The algorithm would create a hash-map:
Keys: possible remainders for value%K i.e 0 to K-1
Values: count of values with value%K = key

The stepwise algorithm is:

1. Find x = arr[i]%k.
2. This array element can be paired with array elements having mod value k-x. This frequency count of array elements is stored in hash. So add that count to answer.
3. Increment count for x in hash.
4. In case the value of x is zero, then it can be paired only with elements having 0 mod value.

Below is the implementation of the above approach:

## 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 };`   `    ``// To store count of pairs.` `    ``int` `ans = 0;`   `    ``// Traverse the array, compute the remainder` `    ``// and add k-remainder value hash count to ans` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `rem = A[i] % K;` `      `  `        ``// Count number of ( A[i], (K - rem)%K ) pairs` `          ``ans += freq[(K - rem) % K];`   `        ``// Increment count of remainder in hash map` `        ``freq[rem]++;` `    ``}`   `    ``return` `ans;` `}`   `// 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'` `class` `GFG ` `{`   `  ``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];`   `      ``// To store count of pairs.` `      ``int` `ans = ``0``;`   `      ``// Traverse the array, compute the remainder` `      ``// and add k-remainder value hash count to ans` `      ``for` `(``int` `i = ``0``; i < n; i++)` `      ``{` `          ``int` `rem = A[i] % K;`   `          ``// Count number of ( A[i], (K - rem)%K ) pairs` `          ``ans += freq[(K - rem) % K];`   `          ``// Increment count of remainder in hash map` `          ``freq[rem]++;` `      ``}`   `      ``return` `ans;` `  ``}`   `// Driver code` `  ``public` `static` `void` `main(String[] args) ` `  ``{` `      ``int` `A[] = { ``2``, ``2``, ``1``, ``7``, ``5``, ``3` `};` `      ``int` `n = A.length;` `      ``int` `K = ``4``;` `      ``System.out.println(countKdivPairs(A, n, K));` `  ``}` `}`   `// This code is contributed by Princi Singh, Yadvendra Naveen`

## Python3

 `# Python Program to count pairs whose sum divisible` `# by 'K'` `def` `countKdivPairs(A, n, K):` `    `  `    ``# Create a frequency array to count` `    ``# occurrences of all remainders when` `    ``# divided by K` `    ``freq ``=` `[``0` `for` `i ``in` `range``(K)]`   `    ``# To store count of pairs.` `    ``ans ``=` `0`   `    ``# Traverse the array, compute the remainder` `    ``# and add k-remainder value hash count to ans` `    ``for` `i ``in` `range``(n):` `        ``rem ``=` `A[i] ``%` `K` `        `  `        ``# Count number of ( A[i], (K - rem)%K ) pairs` `        ``ans ``+``=` `freq[(K ``-` `rem) ``%` `K]` `        `  `        ``# Increment count of remainder in hash map` `        ``freq[rem] ``+``=` `1`   `    ``return` `ans`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``A ``=` `[``2``, ``2``, ``1``, ``7``, ``5``, ``3``]` `    ``n ``=` `len``(A)` `    ``K ``=` `4` `    ``print``(countKdivPairs(A, n, K))`   `# This code is contributed by` `# Surendra_Gangwar, Yadvendra Naveen`

## C#

 `// C# Program to count pairs` `// whose sum divisible by 'K'` `using` `System;`   `class` `GFG ` `{`   `// Program to count pairs whose sum divisible` `// by 'K'` `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];`   `    ``// To store count of pairs.` `    ``int` `ans = 0;`   `    ``// Traverse the array, compute the remainder` `    ``// and add k-remainder value hash count to ans` `    ``for` `(``int` `i = 0; i < n; i++)` `    ``{` `        ``int` `rem = A[i] % K;`   `        ``// Count number of ( A[i], (K - rem)%K ) pairs` `          ``ans += freq[(K - rem) % K];`   `        ``// Increment count of remainder in hash map` `        ``freq[rem]++;` `    ``}`   `    ``return` `ans;` `}`   `// Driver code` `public` `static` `void` `Main(String[] args) ` `{` `    ``int` `[]A = { 2, 2, 1, 7, 5, 3 };` `    ``int` `n = A.Length;` `    ``int` `K = 4;` `    ``Console.WriteLine(countKdivPairs(A, n, K));` `}` `}`   `// This code contributed by Rajput-Ji, Yadvendra Naveen`

Output:

```5

```

Time Complexity: O(N)
Auxiliary Space: O(K)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up
Recommended Articles
Page :