Count inversions in a sequence generated by appending given array K times

Given an array arr[], the task is to append the given array exactly K - 1 times to its end and print the total number of inversions in the resulting array.

Examples:

Input: arr[]= {2, 1, 3}, K = 3
Output: 12
Explanation:
Appending 2 copies of array arr[] modifies arr[] to {2, 1, 3, 2, 1, 3, 2, 1, 3}
The pairs (arr[i], arr[j]), where i < j and arr[i] > arr[j] are (2, 1), (2, 1), (2, 1), (3, 2), (3, 1), (3, 2), (3, 1), (2, 1), (2, 1), (3, 2), (3, 1), (2, 1)
Therefore, the total number of inversions are 12.

Input: arr[]= {6, 2}, K = 2
Output: 3
Explanation:
Appending 2 copies of array arr[] = {6, 2, 6, 2}
The pairs (arr[i], arr[j]), where i < j and arr[i] > arr[j] are (6, 2), (6, 2), (6, 2)
Therefore, the total number of inversions are 3.

Naive Approach: The simplest approach is to store K copies of the given array in a vector and then, find the count of inversions of the resulting vector. 

Time Complexity: O(N²)
Auxiliary Space: O(K * N)

Efficient Approach: The idea to solve this problem is to first find the total number of inversions in the given array, say inv. Then, count pairs of distinct elements in a single copy, say X. Now, calculate the total number of inversions after appending K copies of the array by the equation: 

(inv*K + ((K*(K-1))/2)*X).

Below is the implementation of the above approach:

12

Time Complexity: O(N²)
Auxiliary Space: O(1)