Number of Inversion Couple

 Given an Array, find the number of inversion couples in the Array

 e.g. 2, 4, 3, 1

       4 -> (2,1), (3,1), (4,1), (4,3)

 hint:

Solution 1

 * Iterate the whole array twice

 * Time: O(N^2)

Solution 2

 * Merge Sort

 * Time: O(NlogN)

public class InversionCouple {
    public int mergeSort(int[] A, int start, int end) {
        if (start >= end) return 0;
        int mid = start + (end - start) / 2;
        int left = mergeSort(A, start, mid);
        int right = mergeSort(A, mid+1, end);
        int count = 0;
        int i = start, j = mid + 1;
        int m = mid, n = end;
        int[] tmp = new int[end-start+1];
        int run = 0;
        while (i <= m && j <= end) {
            if (A[i] <= A[j]) {
                tmp[run++] = A[i++];
            } else {
                count += mid - i + 1;
                tmp[run++] = A[j++];
            }
        }
        while (i <= m) {
            tmp[run++] = A[i++];
        }
        while (j <= n) {
            tmp[run++] = A[j++];
        }
        /* copy sorted tmp to original array */
        run = 0;
        while (run < tmp.length) {
            A[start + run] = tmp[run];
            run++;
        }
        tmp = null;
        return count + left + right;
    }
    public static void main(String[] args) {
        int[] A = {1, 7, 2, 9, 6, 4, 5, 3};
        InversionCouple ic = new InversionCouple();
        System.out.println(ic.mergeSort(A, 0, A.length-1));
    }
}