Count pairs in array whose sum is divisible by K

• Difficulty Level : Medium
• Last Updated : 25 Dec, 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).

Below is the implementation of the above approach:

C++

 `// C++ Program to count pairs` `// whose sum divisible by 'K'` `#include ` `using` `namespace` `std;`   `int` `countKdivPairs(``int` `A[], ``int` `n, ``int` `K)` `{` `    ``// variable for storing answer` `    ``int` `count = 0;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``for` `(``int` `j = i + 1; j < n; j++) {`   `            ``// if pair sum is divisible` `            ``if` `((A[i] + A[j]) % K == 0)`   `                ``// Increment count` `                ``count++;` `        ``}` `    ``}`   `    ``return` `count;` `}`   `// Driver code` `int` `main()` `{`   `    ``int` `A[] = { 2, 2, 1, 7, 5, 3 };` `    ``int` `n = ``sizeof``(A) / ``sizeof``(A[0]);` `    ``int` `K = 4;`   `    ``// Function call` `    ``cout << countKdivPairs(A, n, K);`   `    ``return` `0;` `}`

Java

 `/*package whatever //do not write package name here */` `import` `java.io.*;` `class` `GFG {`   `  ``static` `int` `countKdivPairs(``int``[] A, ``int` `n, ``int` `K)` `  ``{`   `    ``// variable for storing answer` `    ``int` `count = ``0``;`   `    ``for` `(``int` `i = ``0``; i < n; i++) {` `      ``for` `(``int` `j = i + ``1``; j < n; j++) {`   `        ``// if pair sum is divisible` `        ``if` `((A[i] + A[j]) % K == ``0``)`   `          ``// Increment count` `          ``count++;` `      ``}` `    ``}`   `    ``return` `count;` `  ``}`   `  ``public` `static` `void` `main (String[] args) ` `  ``{` `    ``int``[] A = { ``2``, ``2``, ``1``, ``7``, ``5``, ``3` `};` `    ``int` `n = A.length;` `    ``int` `K = ``4``;`   `    ``// Function call` `    ``System.out.println(countKdivPairs(A, n, K));` `  ``}` `}`   `// This code is contributed by utkarshshirode02`

C#

 `// C# Program to count pairs` `// whose sum divisible by 'K'` `using` `System;` `class` `GFG {`   `  ``static` `int` `countKdivPairs(``int``[] A, ``int` `n, ``int` `K)` `  ``{` `    `  `    ``// variable for storing answer` `    ``int` `count = 0;`   `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``for` `(``int` `j = i + 1; j < n; j++) {`   `        ``// if pair sum is divisible` `        ``if` `((A[i] + A[j]) % K == 0)`   `          ``// Increment count` `          ``count++;` `      ``}` `    ``}`   `    ``return` `count;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `Main()` `  ``{`   `    ``int``[] A = { 2, 2, 1, 7, 5, 3 };` `    ``int` `n = A.Length;` `    ``int` `K = 4;`   `    ``// Function call` `    ``Console.Write(countKdivPairs(A, n, K));` `  ``}` `}`   `// This code is contributed by Samim Hossain Mondal.`

Javascript

 `// Javascript Program to count pairs` `// whose sum divisible by 'K'`   `function` `countKdivPairs(A, n, K)` `{` `    ``// variable for storing answer` `    ``let count = 0`   `    ``for` `(let i = 0; i < n; i++) {` `        ``for` `(let j = i + 1; j < n; j++) {`   `            ``// if pair sum is divisible` `            ``if` `((A[i] + A[j]) % K == 0)`   `                ``// Increment count` `                ``count++` `        ``}` `    ``}`   `    ``return` `count` `}`   `// Driver code` `let A = [ 2, 2, 1, 7, 5, 3 ]` `let n = A.length` `let K = 4`   `// Function call` `console.log(countKdivPairs(A, n, K))`   `// This code is contributed by Samim Hossain Mondal.`

Python3

 `# Python Program to count pairs` `# whose sum divisible by 'K'` `def` `countKdivPairs(A, n, K):` `    ``count ``=` `0` `    ``for` `i ``in` `range``(``0``, n):` `        ``for` `j ``in` `range``(i``+``1``, n):` `            ``if``((A[i]``+``A[j]) ``%` `K``=``=``0``):` `                ``# Increment count` `                ``count ``+``=` `1` `    ``return` `count` `# Driver Code`   `A``=``[ ``2``, ``2``, ``1``, ``7``, ``5``, ``3` `]` `n ``=` `len``(A)` `K ``=` `4` `#Function call` `print``(countKdivPairs(A, n, K))` `   `

Output

`5`

Time complexity: O(N2), for using two nested loops.
Auxiliary Space: O(1), as constant space is used.

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 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

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Javascript

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Output

`5`

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