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(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 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.

**Output:**

True