Untitled

package sort;

public class CountInversionInArray {

	public static int merge(int[] arr, int l, int r, int m) {

		int res = 0;

//		int[] arr = {2,4,1,3,5};
		int n1 = m - l + 1;
		int n2 = r - m;

		int left[] = new int[n1];
		int right[] = new int[n2];

		// filling left and right array

		for (int i = 0; i < n1; i++)
			left[i] = arr[i + l];

		for (int i = 0; i < n2; i++)
			right[i] = arr[i + m + 1];

		int i = 0, j = 0, k = l;

		while (i < n1 && j < n2) {

			if (left[i] <= right[j])
				arr[k++] = left[i++];
			else {
				arr[k++] = right[j++];
				res = res + (n1 - i); // most important line
			}
		}

		while (i < n1) {
			arr[k++] = left[i++];
		}

		while (j < n2) {
			arr[k++] = right[j++];
		}

		return res;

	}

	public static int mergeSort(int[] arr, int l, int r) {

		int res = 0;

		if (l < r) {
			int m = (l + r) / 2;

			res = res + mergeSort(arr, l, m);
			res = res + mergeSort(arr, m + 1, r);
			res = res + merge(arr, l, r, m);
		}

		return res;

	}

	public static void main(String[] args) {

		int[] arr = { 2, 4, 1, 3, 5 };

		// inversion: pair with numbers where greater number appear before smaller.
		// eg. (4,1) , (4,3) , (2,1) , o/p: 3

		// int count = 0;

// Naive Solution

//		for(int i=0; i<arr.length-1; i++) {
//		 for(int j=i+1; j<arr.length; j++){
//			
//			 if(arr[i] > arr[j])
//				 count++;
//			 
//			}
//		}
//		
//		System.out.println(count);

// Efficient Solution:

		// use merge sort for efficient solution

		System.out.println(mergeSort(arr, 0, arr.length - 1));

	}

}