Count inversions in a permutation of first N natural numbers

Given an array, arr[] of size N denoting a permutation of numbers from 1 to N, the task is to count the number of inversions in the array. 
Note: Two array elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j.

Examples:

Input: arr[] = {2, 3, 1, 5, 4}
Output: 3
Explanation: Given array has 3 inversions: (2, 1), (3, 1), (5, 4).

Input: arr[] = {3, 1, 2}
Output: 2
Explanation: Given array has 2 inversions: (3, 1), (3, 2).

Different methods to solve inversion count has been discussed in the following articles:  

• Naive and Modified Merge Sort
• Using AVL Tree
• Using BIT

Approach: This problem can be solved by using binary search. Follow the steps below to solve the problem:

• Store the numbers in the range [1, N] in increasing order in a vector, V.
• Initialize a variable, ans as 0 to store the number of inversions in the array, arr[].
• Iterate in the range [0, N-1] using the variable i
  • Store the index of occurrence of arr[i] in vector V in a variable index.
  • Add the count of numbers having positions less than the index that are present in the vector, V since they are smaller than the current element and thus form inversions.
  • Remove the element at position index from the vector, V.
• Print the value of ans as the result.

Below is the implementation of the above approach:

3

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)