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 = 4Output :5Explanation :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 = 3Output :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(N^{2}).

**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 <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*

*filter_none*

## 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)); ` ` ` `} ` `} ` |

*chevron_right*

*filter_none*

## 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)) ` |

*chevron_right*

*filter_none*

## 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. ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**Output :**

5

**Time complexity: **O(N)

**Auxiliary space: **O(1)

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.

## Recommended Posts:

- Count pairs in array whose sum is divisible by 4
- Count of pairs from 1 to a and 1 to b whose sum is divisible by N
- Count pairs in an array whose absolute difference is divisible by K
- Check if an array can be divided into pairs whose sum is divisible by k
- Count the number of pairs (i, j) such that either arr[i] is divisible by arr[j] or arr[j] is divisible by arr[i]
- Number of pairs from the first N natural numbers whose sum is divisible by K
- Count number of pairs in array having sum divisible by K | SET 2
- Count pairs in a sorted array whose sum is less than x
- Count number of distinct pairs whose sum exists in the given array
- Count of pairs in an array whose sum is a perfect square
- Count of pairs in an Array whose sum is a Perfect Cube
- Count of pairs in an Array whose sum is Prime
- Find pairs in array whose sum does not exist in Array
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count of all even numbers in the range [L, R] whose sum of digits is divisible by 3
- Count pairs of numbers from 1 to N with Product divisible by their Sum
- Count pairs from 1 to N such that their Sum is divisible by their XOR
- Count pairs (a, b) whose sum of cubes is N (a^3 + b^3 = N)
- Count pairs from two sorted arrays whose sum is equal to a given value x
- Count pairs (a, b) whose sum of squares is N (a^2 + b^2 = N)

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.