[Question]: Given an array of N integers, count the inversion of the array (using merge-sort). What is an inversion of an array? Definition: for all i & j < size of array, if i < j then you have to find pair (A[i],A[j]) such that A[j] < A[i].**Example 1:****Input Format**: N = 5, array[] = {1,2,3,4,5}**Result**: 0**Explanation**: we have a sorted array and the sorted array has 0 inversions as for i < j you will never find a pair such that A[j] < A[i]. More clear example: 2 has index 1 and 5 has index 4 now 1 < 5 but 2 < 5 so this is not an inversion.**Example 2:****Input Format**: N = 5, array[] = {5,4,3,2,1}**Result**: 10**Explanation**: we have a reverse sorted array and we will get the maximum inversions as for i < j we will always find a pair such that A[j] < A[i]. Example: 5 has index 0 and 3 has index 2 now (5,3) pair is inversion as 0 < 2 and 5 > 3 which will satisfy out conditions and for reverse sorted array we will get maximum inversions and that is (n)*(n-1) / 2.For above given array there is 4 + 3 + 2 + 1 = 10 inversions.**Example 3:****Input Format**: N = 5, array[] = {5,3,2,1,4}**Result**: 7**Explanation**: There are 7 pairs (5,1), (5,3), (5,2), (5,4),(3,2), (3,1), (2,1) and we have left 2 pairs (2,4) and (1,4) as both are not satisfy our condition.

Time Complexity: O(N*logN), where N = size of the given array.

Reason: We are not changing the merge sort algorithm except by adding a variable to it. So, the time complexity is as same as the merge sort.

Space Complexity: O(N),

```
func numberOfInversions(_ arr: [Int]) -> Int {
var cnt = 0
for i in 0..<arr.count {
for j in i..<arr.count {
if arr[i] > arr[j] {
cnt+=1
}
}
}
return cnt
}
let conversionArrayInput = [5, 4, 3, 2, 1]
let conversionArrayOutput = numberOfInversions(conversionArrayInput)
print("The number of inversions is: ", conversionArrayOutput) // 10
```