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

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

Examples:

```Input: arr[] = {9, 7, 5, 3},
k = 6
Output: True
We can divide 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 = 10
Output: True
We can divide array into (92, 48), (75, 65)
and (45, 35). Sum of all these pairs is a
multiple of 10.

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

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

A Simple Solution is to iterate through every element arr[i]. Find if there is another not yet visited element that has remainder as (k – arr[i]%k). If there is no such element, return false. If a pair is found, then mark both elements as visited. Time complexity of this solution is O(n2 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 in pairs.
2) Traverse input array and count occurrences of
all remainders.
freq[arr[i] % k]++
3) Traverse input array again.
a) Find remainder of 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 remainder is 0, then then there must be
even occurrences.
c) Else, number of occurrences of current
remainder must be equal to number of
occurrences of "k - current remainder".
```

Time complexity of above algorithm is O(n).

Below implementation uses map in C++ STL. The map is typically implemented using Red-Black Tree and takes O(Log n) time for access. Therefore time complexity of below implementation is O(n Log n), but the algorithm can be easily implemented in O(n) time using hash table.

## 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.
map<int, int> freq;

// Count occurrences of all remainders
for (int i = 0; i < n; i++)
freq[arr[i] % 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;

// 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 two
// 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 program to test above function */
int main()
{
int arr[] =  {92, 75, 65, 48, 45, 35};
int k = 10;
int n = sizeof(arr)/sizeof(arr[0]);
canPairs(arr, n, k)? cout << "True": cout << "False";
return 0;
}
```

## Java

```
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;
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;

// 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 to test above functions
public static void main(String[] args)
{
int arr[] = { 92, 75, 65, 48, 45, 35 };
int k = 10;
boolean ans = canPairs(arr, k);
if (ans)
System.out.println("True");
else
System.out.println("False");

}
}

// This code is contributed by Rishabh Mahrsee
```

Output:
`True`

# GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.
3.4 Average Difficulty : 3.4/5.0
Based on 109 vote(s)