Count Inversions in an Array?

int MergeSortAndCount(int[] arr, int[] temp, int left, int right)

int invCount = 0;

if (right > left)

int mid = (right + left) / 2;

invCount += MergeSortAndCount(arr, temp, left, mid);

invCount += MergeSortAndCount(arr, temp, mid + 1, right);

invCount += Merge(arr, temp, left, mid + 1, right);

return invCount;

int Merge(int[] arr, int[] temp, int left, int mid, int right)

int i = left, j = mid, k = left;

int invCount = 0;

while (i <= mid - 1 && j <= right)

if (arr[i] <= arr[j])

temp[k++] = arr[i++];

else

temp[k++] = arr[j++];

invCount += (mid - i); // Count inversions

while (i <= mid - 1)

temp[k++] = arr[i++];

Follow on:

while (j <= right)

temp[k++] = arr[j++];

for (int idx = left; idx <= right; idx++)

arr[idx] = temp[idx];

return invCount;

Explanation:

Using a modified merge sort to count pairs where arr[i] > arr[j] for i < j efficiently.