# Count pairs in array whose sum is divisible by K

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 = 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’. The 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, 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)

https://www.youtube.com/watch?v=5UJvXcSUyT0&list=PLM68oyaqFM7Q-sv3gA5xbzfgVkoQ0xDrW&index=16

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. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.