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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 and Output
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)
Algorithm
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
Example
#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);
}Output
Number of inversions are 2
Updated on: 16-Jun-2020
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