Count Inversions in an array

The inversions of an array indicate; how many changes are required to convert the array into its sorted form. When an array is already sorted, it needs 0 inversions, and in another case, the number of inversions will be maximum, if the array is reversed.
To solve this problem, we will follow the Merge sort approach to reduce the time complexity, and make it in Divide and Conquer algorithm.
Input:
A sequence of numbers. (1, 5, 6, 4, 20).
Output:
The number of inversions required to arrange the numbers into ascending order.
Here the number of inversions are 2.
First inversion: (1, 5, 4, 6, 20)
Second inversion: (1, 4, 5, 6, 20)
merge(array, tempArray, left, mid, right)
Input: Two arrays, who have merged, the left, right and the mid indexes.
Output: The merged array in sorted order.
Begin
   i := left, j := mid, k := right
   count := 0
   while i <= mid -1 and j <= right, do
      if array[i] <= array[j], then
         tempArray[k] := array[i]
         increase i and k by 1
      else
         tempArray[k] := array[j]
         increase j and k by 1
         count := count + (mid - i)
   done

   while left part of the array has some extra element, do
      tempArray[k] := array[i]
      increase i and k by 1
   done

   while right part of the array has some extra element, do
      tempArray[k] := array[j]
      increase j and k by 1
   done

   return count
End
mergeSort(array, tempArray, left, right)
Input: Given an array and temporary array, left and right index of the array.
Output − Number of inversions after sorting.
Begin
   count := 0
   if right > left, then
      mid := (right + left)/2
      count := mergeSort(array, tempArray, left, mid)
      count := count + mergeSort(array, tempArray, mid+1, right)
      count := count + merge(array, tempArray, left, mid+1, right)
   return count
End
#include <iostream>
using namespace std;

int merge(intarr[], int temp[], int left, int mid, int right) {
   int i, j, k;
   int count = 0;
   
   i = left;    //i to locate first array location
   j = mid;     //i to locate second array location
   k = left;    //i to locate merged array location

   while ((i <= mid - 1) && (j <= right)) {
      if (arr[i] <= arr[j]) {    //when left item is less than right item
         temp[k++] = arr[i++];
      }else{
         temp[k++] = arr[j++];
         count += (mid - i);    //find how many convertion is performed
      }
   }

    while (i <= mid - 1)    //if first list has remaining item, add them in the list
       temp[k++] = arr[i++];

    while (j <= right)    //if second list has remaining item, add them in the list
       temp[k++] = arr[j++];
   
    for (i=left; i <= right; i++)
       arr[i] = temp[i];    //store temp Array to main array
    return count;
}

intmergeSort(intarr[], int temp[], int left, int right) {
   int mid, count = 0;

   if (right > left) {
      mid = (right + left)/2;    //find mid index of the array
      count  = mergeSort(arr, temp, left, mid);    //merge sort left sub array
      count += mergeSort(arr, temp, mid+1, right);    //merge sort right sub array
         
      count += merge(arr, temp, left, mid+1, right);    //merge two sub arrays
   }
   return count;
}

intarrInversion(intarr[], int n) {
   int temp[n];
   return mergeSort(arr, temp, 0, n - 1);
}

int main() {
   intarr[] = {1, 5, 6, 4, 20};
   int n = 5;
   cout<< "Number of inversions are "<<arrInversion(arr, n);
}
Number of inversions are 2