# Check if an array can be divided into pairs whose sum is divisible by k

- Difficulty Level : Medium
- Last Updated : 08 Jul, 2022

Given an array of integers and a number k, write a function that returns true if the given array can be divided into pairs such that the sum of every pair is divisible by k.

**Examples:**

Input: arr[] = {9, 7, 5, 3},

k = 6Output: True

We can divide the array into (9, 3) and

(7, 5). Sum of both of these pairs

is a multiple of 6.

Input: arr[] = {92, 75, 65, 48, 45, 35},

k = 10Output: True

We can divide the array into (92, 48), (75, 65).

and (45, 35). The sum of all these pairs is a

multiple of 10.

Input: arr[] = {91, 74, 66, 48}, k = 10Output: False

A **Simple Solution** is to iterate through every element arr[i]. Find if there is another not yet visited element that has a remainder like **(k – arr[i]%k)**. If there is no such element, return false. If a pair is found, then mark both elements as visited. The **time complexity** of this solution is O(n^{2} and it requires O(n) extra space.

An **Efficient Solution** is to use Hashing.

1) If length of given array is odd, return false. An odd length array cannot be divided into pairs. 2) Traverse input array and count occurrences of all remainders (use (arr[i] % k)+k)%k for handling the case of negative integers as well). freq[((arr[i] % k)+k)%k]++ 3) Traverse input array again. a) Find the remainder of the current element. b) If remainder divides k into two halves, then there must be even occurrences of it as it forms pair with itself only. c) If the remainder is 0, then there must be even occurrences. d) Else, number of occurrences of current the remainder must be equal to a number of occurrences of "k - current remainder".

An **efficient approach **is to use hashing (unordered_map in C++ and HashMap in Java).

The below image is a dry run of the above approach:

Below is the implementation of the above approach:

## C++

`// A C++ program to check if arr[0..n-1] can be divided` `// in pairs such that every pair is divisible by k.` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Returns true if arr[0..n-1] can be divided into pairs` `// with sum divisible by k.` `bool` `canPairs(` `int` `arr[], ` `int` `n, ` `int` `k)` `{` ` ` `// An odd length array cannot be divided into pairs` ` ` `if` `(n & 1)` ` ` `return` `false` `;` ` ` `// Create a frequency array to count occurrences` ` ` `// of all remainders when divided by k.` ` ` `unordered_map<` `int` `, ` `int` `> freq;` ` ` `// Count occurrences of all remainders` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `freq[((arr[i] % k) + k) % k]++;` ` ` `// Traverse input array and use freq[] to decide` ` ` `// if given array can be divided in pairs` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `// Remainder of current element` ` ` `int` `rem = ((arr[i] % k) + k) % k;` ` ` `// If remainder with current element divides` ` ` `// k into two halves.` ` ` `if` `(2 * rem == k) {` ` ` `// Then there must be even occurrences of` ` ` `// such remainder` ` ` `if` `(freq[rem] % 2 != 0)` ` ` `return` `false` `;` ` ` `}` ` ` `// If remainder is 0, then there must be even` ` ` `// number of elements with 0 remainder` ` ` `else` `if` `(rem == 0) {` ` ` `if` `(freq[rem] & 1)` ` ` `return` `false` `;` ` ` `}` ` ` `// Else number of occurrences of remainder` ` ` `// must be equal to number of occurrences of` ` ` `// k - remainder` ` ` `else` `if` `(freq[rem] != freq[k - rem])` ` ` `return` `false` `;` ` ` `}` ` ` `return` `true` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 92, 75, 65, 48, 45, 35 };` ` ` `int` `k = 10;` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function call` ` ` `canPairs(arr, n, k) ? cout << ` `"True"` `: cout << ` `"False"` `;` ` ` `return` `0;` `}` |

## Java

`// JAVA program to check if arr[0..n-1] can be divided` `// in pairs such that every pair is divisible by k.` `import` `java.util.HashMap;` `public` `class` `Divisiblepair {` ` ` `// Returns true if arr[0..n-1] can be divided into pairs` ` ` `// with sum divisible by k.` ` ` `static` `boolean` `canPairs(` `int` `ar[], ` `int` `k)` ` ` `{` ` ` `// An odd length array cannot be divided into pairs` ` ` `if` `(ar.length % ` `2` `== ` `1` `)` ` ` `return` `false` `;` ` ` `// Create a frequency array to count occurrences` ` ` `// of all remainders when divided by k.` ` ` `HashMap<Integer, Integer> hm = ` `new` `HashMap<>();` ` ` `// Count occurrences of all remainders` ` ` `for` `(` `int` `i = ` `0` `; i < ar.length; i++) {` ` ` `int` `rem = ((ar[i] % k) + k) % k;` ` ` `if` `(!hm.containsKey(rem)) {` ` ` `hm.put(rem, ` `0` `);` ` ` `}` ` ` `hm.put(rem, hm.get(rem) + ` `1` `);` ` ` `}` ` ` `// Traverse input array and use freq[] to decide` ` ` `// if given array can be divided in pairs` ` ` `for` `(` `int` `i = ` `0` `; i < ar.length; i++) {` ` ` `// Remainder of current element` ` ` `int` `rem = ((ar[i] % k) + k) % k;` ` ` `// If remainder with current element divides` ` ` `// k into two halves.` ` ` `if` `(` `2` `* rem == k) {` ` ` `// Then there must be even occurrences of` ` ` `// such remainder` ` ` `if` `(hm.get(rem) % ` `2` `== ` `1` `)` ` ` `return` `false` `;` ` ` `}` ` ` `// If remainder is 0, then there must be two` ` ` `// elements with 0 remainder` ` ` `else` `if` `(rem == ` `0` `) {` ` ` `// Then there must be even occurrences of` ` ` `// such remainder` ` ` `if` `(hm.get(rem) % ` `2` `== ` `1` `)` ` ` `return` `false` `;` ` ` `}` ` ` `// Else number of occurrences of remainder` ` ` `// must be equal to number of occurrences of` ` ` `// k - remainder` ` ` `else` `{` ` ` `if` `(hm.get(k - rem) != hm.get(rem))` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `return` `true` `;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `92` `, ` `75` `, ` `65` `, ` `48` `, ` `45` `, ` `35` `};` ` ` `int` `k = ` `10` `;` ` ` `// Function call` ` ` `boolean` `ans = canPairs(arr, k);` ` ` `if` `(ans)` ` ` `System.out.println(` `"True"` `);` ` ` `else` ` ` `System.out.println(` `"False"` `);` ` ` `}` `}` `// This code is contributed by Rishabh Mahrsee` |

## Python3

`# Python3 program to check if` `# arr[0..n-1] can be divided` `# in pairs such that every` `# pair is divisible by k.` `from` `collections ` `import` `defaultdict` `# Returns true if arr[0..n-1] can be` `# divided into pairs with sum` `# divisible by k.` `def` `canPairs(arr, n, k):` ` ` `# An odd length array cannot` ` ` `# be divided into pairs` ` ` `if` `(n & ` `1` `):` ` ` `return` `0` ` ` `# Create a frequency array to` ` ` `# count occurrences of all` ` ` `# remainders when divided by k.` ` ` `freq ` `=` `defaultdict(` `lambda` `: ` `0` `)` ` ` `# Count occurrences of all remainders` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `freq[((arr[i] ` `%` `k) ` `+` `k) ` `%` `k] ` `+` `=` `1` ` ` `# Traverse input array and use` ` ` `# freq[] to decide if given array` ` ` `# can be divided in pairs` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `# Remainder of current element` ` ` `rem ` `=` `((arr[i] ` `%` `k) ` `+` `k) ` `%` `k` ` ` `# If remainder with current element` ` ` `# divides k into two halves.` ` ` `if` `(` `2` `*` `rem ` `=` `=` `k):` ` ` `# Then there must be even occurrences` ` ` `# of such remainder` ` ` `if` `(freq[rem] ` `%` `2` `!` `=` `0` `):` ` ` `return` `0` ` ` `# If remainder is 0, then there` ` ` `# must be two elements with 0 remainder` ` ` `else` `if` `(rem ` `=` `=` `0` `):` ` ` `if` `(freq[rem] & ` `1` `):` ` ` `return` `0` ` ` `# Else number of occurrences of` ` ` `# remainder must be equal to` ` ` `# number of occurrences of` ` ` `# k - remainder` ` ` `else` `if` `(freq[rem] !` `=` `freq[k ` `-` `rem]):` ` ` `return` `0` ` ` `return` `1` `# Driver code` `arr ` `=` `[` `92` `, ` `75` `, ` `65` `, ` `48` `, ` `45` `, ` `35` `]` `k ` `=` `10` `n ` `=` `len` `(arr)` `# Function call` `if` `(canPairs(arr, n, k)):` ` ` `print` `(` `"True"` `)` `else` `:` ` ` `print` `(` `"False"` `)` `# This code is contributed by Stream_Cipher` |

## C#

`// C# program to check if arr[0..n-1]` `// can be divided in pairs such that` `// every pair is divisible by k.` `using` `System.Collections.Generic;` `using` `System;` `class` `GFG {` ` ` `// Returns true if arr[0..n-1] can be` ` ` `// divided into pairs with sum` ` ` `// divisible by k.` ` ` `static` `bool` `canPairs(` `int` `[] ar, ` `int` `k)` ` ` `{` ` ` `// An odd length array cannot` ` ` `// be divided into pairs` ` ` `if` `(ar.Length % 2 == 1)` ` ` `return` `false` `;` ` ` `// Create a frequency array to count` ` ` `// occurrences of all remainders when` ` ` `// divided by k.` ` ` `Dictionary<Double, ` `int` `> hm` ` ` `= ` `new` `Dictionary<Double, ` `int` `>();` ` ` `// Count occurrences of all remainders` ` ` `for` `(` `int` `i = 0; i < ar.Length; i++) {` ` ` `int` `rem = ((ar[i] % k) + k) % k;` ` ` `if` `(!hm.ContainsKey(rem)) {` ` ` `hm[rem] = 0;` ` ` `}` ` ` `hm[rem]++;` ` ` `}` ` ` `// Traverse input array and use freq[]` ` ` `// to decide if given array can be` ` ` `// divided in pairs` ` ` `for` `(` `int` `i = 0; i < ar.Length; i++) {` ` ` `// Remainder of current element` ` ` `int` `rem = ((ar[i] % k) + k) % k;` ` ` `// If remainder with current element` ` ` `// divides k into two halves.` ` ` `if` `(2 * rem == k) {` ` ` `// Then there must be even occurrences` ` ` `// of such remainder` ` ` `if` `(hm[rem] % 2 == 1)` ` ` `return` `false` `;` ` ` `}` ` ` `// If remainder is 0, then there` ` ` `// must be two elements with 0` ` ` `// remainder` ` ` `else` `if` `(rem == 0) {` ` ` `// Then there must be even occurrences` ` ` `// of such remainder` ` ` `if` `(hm[rem] % 2 == 1)` ` ` `return` `false` `;` ` ` `}` ` ` `// Else number of occurrences of remainder` ` ` `// must be equal to number of occurrences of` ` ` `// k - remainder` ` ` `else` `{` ` ` `if` `(hm[k - rem] != hm[rem])` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `return` `true` `;` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[] arr = { 92, 75, 65, 48, 45, 35 };` ` ` `int` `k = 10;` ` ` `// Function call` ` ` `bool` `ans = canPairs(arr, k);` ` ` `if` `(ans)` ` ` `Console.WriteLine(` `"True"` `);` ` ` `else` ` ` `Console.WriteLine(` `"False"` `);` ` ` `}` `}` `// This code is contributed by Stream_Cipher` |

## Javascript

`<script>` `// Javascript program to check if arr[0..n-1] can be divided` `// in pairs such that every pair is divisible by k.` ` ` `// Returns true if arr[0..n-1] can be divided into pairs` ` ` `// with sum divisible by k.` ` ` `function` `canPairs(ar, k)` ` ` `{` ` ` `// An odd length array cannot be divided into pairs` ` ` `if` `(ar.length % 2 == 1)` ` ` `return` `false` `;` ` ` ` ` `// Create a frequency array to count occurrences` ` ` `// of all remainders when divided by k.` ` ` `let hm = ` `new` `Map();` ` ` ` ` `// Count occurrences of all remainders` ` ` `for` `(let i = 0; i < ar.length; i++) {` ` ` `let rem = ((ar[i] % k) + k) % k;` ` ` `if` `(!hm.has(rem)) {` ` ` `hm.set(rem, 0);` ` ` `}` ` ` `hm.set(rem, hm.get(rem) + 1);` ` ` `}` ` ` ` ` `// Traverse input array and use freq[] to decide` ` ` `// if given array can be divided in pairs` ` ` `for` `(let i = 0; i < ar.length; i++) {` ` ` `// Remainder of current element` ` ` `let rem = ((ar[i] % k) + k) % k;` ` ` ` ` `// If remainder with current element divides` ` ` `// k into two halves.` ` ` `if` `(2 * rem == k) {` ` ` `// Then there must be even occurrences of` ` ` `// such remainder` ` ` `if` `(hm.get(rem) % 2 == 1)` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// If remainder is 0, then there must be two` ` ` `// elements with 0 remainder` ` ` `else` `if` `(rem == 0) {` ` ` `// Then there must be even occurrences of` ` ` `// such remainder` ` ` `if` `(hm.get(rem) % 2 == 1)` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// Else number of occurrences of remainder` ` ` `// must be equal to number of occurrences of` ` ` `// k - remainder` ` ` `else` `{` ` ` `if` `(hm.get(k - rem) != hm.get(rem))` ` ` `return` `false` `;` ` ` `}` ` ` `}` ` ` `return` `true` `;` ` ` `}` ` ` `// Driver program` ` ` `let arr = [ 92, 75, 65, 48, 45, 35 ];` ` ` `let k = 10;` ` ` ` ` `// Function call` ` ` `let ans = canPairs(arr, k);` ` ` `if` `(ans)` ` ` `document.write(` `"True"` `);` ` ` `else` ` ` `document.write(` `"False"` `);` ` ` `</script>` |

**Output**

True

**Time complexity:** O(n).

**Auxiliary Space:** O(n)

This article is contributed by **Aarti_Rathi **and **Priyanka**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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