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April 15, 2017 15:39
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Inversion Count for an array indicates.java
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| package com.hrishikesh.practices.sort; | |
| import java.util.Arrays; | |
| import static com.hrishikesh.practices.sort.InversionsCountInArray.getCountByMergeSort; | |
| /** | |
| * Problem: | |
| * Inversion Count for an array indicates – | |
| * how far (or close) the array is from being sorted. | |
| * | |
| * @author hrishikesh.mishra | |
| */ | |
| public class InversionsCountInArray { | |
| public static long getCountByBruteForce(int[] a) { | |
| long inversionCount = 0; | |
| for (int i = 0; i < a.length - 1; i++) { | |
| for (int j = i; j < a.length; j++) { | |
| if (a[i] > a[j]) { | |
| inversionCount++; | |
| } | |
| } | |
| } | |
| return inversionCount; | |
| } | |
| public static long getCountByMergeSort(int[] a) { | |
| return mergeSort(a, 0, a.length - 1); | |
| } | |
| public static long mergeSort(int[] a, int left, int right) { | |
| if (right > left) { | |
| int mid = (left + right) / 2; | |
| long inversionCount = mergeSort(a, left, mid); | |
| inversionCount += mergeSort(a, mid + 1, right); | |
| inversionCount += merge(a, left, mid + 1, right, new int[a.length]); | |
| return inversionCount; | |
| } | |
| return 0; | |
| } | |
| private static long merge(int[] a, int left, int mid, int right, int[] temp) { | |
| int i = left; | |
| int j = mid; | |
| int counter = left; | |
| long inversionCount = 0; | |
| while (i < mid && j <= right) { | |
| if (a[i] <= a[j]) { | |
| temp[counter++] = a[i++]; | |
| } else { | |
| temp[counter++] = a[j++]; | |
| inversionCount += (mid - i); | |
| } | |
| } | |
| while (i < mid) { | |
| temp[counter++] = a[i++]; | |
| } | |
| while (j <= right) { | |
| temp[counter++] = a[j++]; | |
| } | |
| for (int c = left; c <= right; c++) { | |
| a[c] = temp[c]; | |
| } | |
| return inversionCount; | |
| } | |
| } | |
| class InversionsCountInArrayTest { | |
| public static void main(String[] args) { | |
| int[] array1 = {1, 1, 1, 2, 2}; | |
| int[] array2 = {2, 1, 3, 1, 2}; | |
| int[] array3 = {1, 1, 1, 2, 2}; | |
| int[] array4 = {2, 1, 3, 1, 2}; | |
| getCountByMergeSort(array2); | |
| System.out.println("Array: " + Arrays.toString(array1)); | |
| System.out.println("Inversion Count by Brute Force: " + InversionsCountInArray.getCountByBruteForce(array1)); | |
| System.out.println("Inversion Count by Brute Force: " + InversionsCountInArray.getCountByBruteForce(array3)); | |
| System.out.println(); | |
| System.out.println("Array: " + Arrays.toString(array2)); | |
| System.out.println("Inversion Count by Brute Force: " + InversionsCountInArray.getCountByBruteForce(array2)); | |
| System.out.println("Inversion Count by Brute Force: " + InversionsCountInArray.getCountByBruteForce(array4)); | |
| } | |
| } |
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