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:
// Java implementation of the approach
import
java.util.Arrays;
public
class
GFG {
// Function to count the number of inversions
// during the merge process
private
static
int
mergeAndCount(
int
[] arr,
int
l,
int
m,
int
r)
{
// Left subarray
int
[] left = Arrays.copyOfRange(arr, l, m +
1
);
// Right subarray
int
[] right = Arrays.copyOfRange(arr, m +
1
, r +
1
);
int
i =
0
, j =
0
, k = l, swaps =
0
;
while
(i < left.length && j < right.length) {
if
(left[i] <= right[j])
arr[k++] = left[i++];
else
{
arr[k++] = right[j++];
swaps += (m +
1
)  (l + i);
}
}
while
(i < left.length)
arr[k++] = left[i++];
while
(j < right.length)
arr[k++] = right[j++];
return
swaps;
}
// Merge sort function
private
static
int
mergeSortAndCount(
int
[] arr,
int
l,
int
r)
{
// Keeps track of the inversion count at a
// particular node of the recursion tree
int
count =
0
;
if
(l < r) {
int
m = (l + r) /
2
;
// Total inversion count = left subarray count
// + right subarray count + merge count
// Left subarray count
count += mergeSortAndCount(arr, l, m);
// Right subarray count
count += mergeSortAndCount(arr, m +
1
, r);
// Merge count
count += mergeAndCount(arr, l, m, r);
}
return
count;
}
// Driver code
public
static
void
main(String[] args)
{
int
[] arr = {
1
,
20
,
6
,
4
,
5
};
System.out.println(
mergeSortAndCount(arr,
0
, arr.length 
1
));
}
}
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.