Approach:
Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2); what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().

How to get the number of inversions in merge()?
In merge process, let i is used for indexing left subarray and j for right subarray. At any step in merge(), if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in leftsubarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j] 
Algorithm:
 The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.
 Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for the first half, and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in leftsubarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].
 Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, the number of inversion in the second half and the number of inversions by merging the two.
 The base case of recursion is when there is only one element in the given half.
 Print the answer

Implementation:
// C program to Count
// Inversions in an array
// using Merge Sort
#include <stdio.h>
#include <stdlib.h>
int
_mergeSort(
int
arr[],
int
temp[],
int
left,
int
right);
int
merge(
int
arr[],
int
temp[],
int
left,
int
mid,
int
right);
/* This function sorts the input array and returns the
number of inversions in the array */
int
mergeSort(
int
arr[],
int
array_size)
{
int
* temp = (
int
*)
malloc
(
sizeof
(
int
) * array_size);
return
_mergeSort(arr, temp, 0,
array_size  1);
}
/* An auxiliary recursive function
that sorts the input
array and returns the number
of inversions in the array.
*/
int
_mergeSort(
int
arr[],
int
temp[],
int
left,
int
right)
{
int
mid, inv_count = 0;
if
(right > left)
{
/* Divide the array into two parts and call
_mergeSortAndCountInv() for each of the parts */
mid = (right + left) / 2;
/* Inversion count will be the sum of inversions in
leftpart, rightpart and number of inversions in
merging */
inv_count += _mergeSort(arr, temp, left, mid);
inv_count += _mergeSort(arr, temp, mid + 1, right);
/*Merge the two parts*/
inv_count += merge(arr, temp, left, mid + 1, right);
}
return
inv_count;
}
/* This funt merges two sorted
arrays and returns inversion
count in the arrays.*/
int
merge(
int
arr[],
int
temp[],
int
left,
int
mid,
int
right)
{
int
i, j, k;
int
inv_count = 0;
i = left;
/* i is index for left subarray*/
j = mid;
/* j is index for right subarray*/
k = left;
/* k is index for resultant merged subarray*/
while
((i <= mid  1) && (j <= right)) {
if
(arr[i] <= arr[j]) {
temp[k++] = arr[i++];
}
else
{
temp[k++] = arr[j++];
/*this is tricky  see above
* explanation/diagram for merge()*/
inv_count = inv_count + (mid  i);
}
}
/* Copy the remaining elements of left subarray
(if there are any) to temp*/
while
(i <= mid  1)
temp[k++] = arr[i++];
/* Copy the remaining elements of right subarray
(if there are any) to temp*/
while
(j <= right)
temp[k++] = arr[j++];
/*Copy back the merged elements to original array*/
for
(i = left; i <= right; i++)
arr[i] = temp[i];
return
inv_count;
}
/* Driver code*/
int
main(
int
argv,
char
** args)
{
int
arr[] = { 1, 20, 6, 4, 5 };
printf
(
" Number of inversions are %d \n"
,
mergeSort(arr, 5));
getchar
();
return
0;
}
Output:
Number of inversions are 5
Complexity Analysis:
 Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level, one full array traversal is needed, and there are log n levels, so the time complexity is O(n log n).
 Space Complexity**:** O(n), Temporary array.