# C++ Programming – 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.

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```

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 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++ Program
``````// 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);
canPairs(arr, n, k)? cout << "True": cout << "False";
return 0;
}``````

Output:

`True` #### Wikitechy Editor

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