# 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
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
```

## Recommended: Please solve it on PRACTICE first, before moving on to the solution.

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++

 `// 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 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

 ` `

Output :

``` 5
```

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

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

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.