Counting Inversions In An Array Using Merge Sort

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  • Microsoft Interview Questions

Problem Statement

Inversion Count for an array indicates – how far (or close) the array is from being sorted. If array is already sorted then inversion count is 0. If array is sorted in reverse order that inversion count is the maximum. So given an array find the no of inversions.

Sample Test Cases

Input: arr[] = {8, 4, 2, 1}
Output: 6

Explanation: Given array has six inversions:
(8,4), (4,2),(8,2), (8,1), (4,1), (2,1).


Input: arr[] = {3, 1, 2}
Output: 2

Explanation: Given array has two inversions:
(3, 1), (3, 2) 

Problem Solution

The idea is similar to merge sort, divide the array into two equal or almost equal halves in each step until the base case is reached.

Suppose the number of inversions in the left half and right half of the array (let be inv1 and inv2), what kinds of inversions are not accounted for in Inv1 + Inv2? The answer is – the inversions that need to be counted during the merge step. Therefore, to get a number of inversions, that needs to be added a number of inversions in the left subarray, right subarray and merge().

Create a function merge that counts the number of inversions when two halves of the array are merged, create two indices i and j, i is the index for first half and j is an index of the second half. if a[i] is greater than a[j], then there are (mid – i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] … a[mid]) will be greater than a[j].

Create a recursive function to divide the array into halves and find the answer by summing the number of inversions is the first half, number of inversion in the second half and the number of inversions by merging the two.

The base case of recursion is when there is only one element in the given half.

Complexity Analysis

Time Complexity: O(n log n), The algorithm used is divide and conquer, So in each level one full array traversal is needed and there are log n levels so the time complexity is O(n log n).

Space Compelxity: O(n), Temporary array.

Code Implementation

#include <bits/stdc++.h>
using namespace std;

int _mergeSort(int arr[], int temp[], int left, int right);
int merge(int arr[], int temp[], int left, int mid, int right);

/* This function sorts the input array and returns the
number of inversions in the array */
int mergeSort(int arr[], int array_size)
{
    int temp[array_size];
    return _mergeSort(arr, temp, 0, array_size - 1);
}

/* An auxiliary recursive function that sorts the input array and
returns the number of inversions in the array. */
int _mergeSort(int arr[], int temp[], int left, int right)
{
    int mid, inv_count = 0;
    if (right > left) {
        /* Divide the array into two parts and
        call _mergeSortAndCountInv()
        for each of the parts */
        mid = (right + left) / 2;

        /* Inversion count will be sum of
        inversions in left-part, right-part
        and number of inversions in merging */
        inv_count += _mergeSort(arr, temp, left, mid);
        inv_count += _mergeSort(arr, temp, mid + 1, right);

        /*Merge the two parts*/
        inv_count += merge(arr, temp, left, mid + 1, right);
    }
    return inv_count;
}

/* This funt merges two sorted arrays
and returns inversion count in the arrays.*/
int merge(int arr[], int temp[], int left,
          int mid, int right)
{
    int i, j, k;
    int inv_count = 0;

    i = left; /* i is index for left subarray*/
    j = mid; /* j is index for right subarray*/
    k = left; /* k is index for resultant merged subarray*/
    while ((i <= mid - 1) && (j <= right)) {
        if (arr[i] <= arr[j]) {
            temp[k++] = arr[i++];
        }
        else {
            temp[k++] = arr[j++];

            /* this is tricky -- see above
            explanation/diagram for merge()*/
            inv_count = inv_count + (mid - i);
        }
    }

    /* Copy the remaining elements of left subarray
(if there are any) to temp*/
    while (i <= mid - 1)
        temp[k++] = arr[i++];

    /* Copy the remaining elements of right subarray
(if there are any) to temp*/
    while (j <= right)
        temp[k++] = arr[j++];

    /*Copy back the merged elements to original array*/
    for (i = left; i <= right; i++)
        arr[i] = temp[i];

    return inv_count;
}

// Driver code
int main()
{
    int arr[] = { 1, 20, 6, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int ans = mergeSort(arr, n);
    cout << " Number of inversions are " << ans;
    return 0;
}

 

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