Count Inversions in an array | Set 1 (Using Merge Sort)

 

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. 
Formally speaking, two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j 
Example: 

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) 

METHOD 1 (Simple)  

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// C++ program to Count Inversions
// in an array
#include <bits/stdc++.h>
using namespace std;
  
int getInvCount(int arr[], int n)
{
    int inv_count = 0;
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if (arr[i] > arr[j])
                inv_count++;
  
    return inv_count;
}
  
// Driver Code
int main()
{
    int arr[] = { 1, 20, 6, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << " Number of inversions are "
         << getInvCount(arr, n);
    return 0;
}
  
// This code is contributed
// by Akanksha Rai
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// C program to Count
// Inversions in an array
#include <stdio.h>
#include <stdlib.h>
int getInvCount(int arr[], int n)
{
    int inv_count = 0;
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if (arr[i] > arr[j])
                inv_count++;
  
    return inv_count;
}
  
/* Driver program to test above functions */
int main()
{
    int arr[] = { 1, 20, 6, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf(" Number of inversions are %d \n", getInvCount(arr, n));
    return 0;
}
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// Java program to count
// inversions in an array
class Test {
    static int arr[] = new int[] { 1, 20, 6, 4, 5 };
  
    static int getInvCount(int n)
    {
        int inv_count = 0;
        for (int i = 0; i < n - 1; i++)
            for (int j = i + 1; j < n; j++)
                if (arr[i] > arr[j])
                    inv_count++;
  
        return inv_count;
    }
  
    // Driver method to test the above function
    public static void main(String[] args)
    {
        System.out.println("Number of inversions are "
                           + getInvCount(arr.length));
    }
}
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# Python3 program to count 
# inversions in an array
  
def getInvCount(arr, n):
  
    inv_count = 0
    for i in range(n):
        for j in range(i + 1, n):
            if (arr[i] > arr[j]):
                inv_count += 1
  
    return inv_count
  
# Driver Code
arr = [1, 20, 6, 4, 5]
n = len(arr)
print("Number of inversions are",
              getInvCount(arr, n))
  
# This code is contributed by Smitha Dinesh Semwal
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// C# program to count inversions
// in an array
using System;
using System.Collections.Generic;
  
class GFG {
  
    static int[] arr = new int[] { 1, 20, 6, 4, 5 };
  
    static int getInvCount(int n)
    {
        int inv_count = 0;
  
        for (int i = 0; i < n - 1; i++)
            for (int j = i + 1; j < n; j++)
                if (arr[i] > arr[j])
                    inv_count++;
  
        return inv_count;
    }
  
    // Driver code
    public static void Main()
    {
        Console.WriteLine("Number of "
                          + "inversions are "
                          + getInvCount(arr.Length));
    }
}
  
// This code is contributed by Sam007
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<?php 
// PHP program to Count Inversions
// in an array
  
function getInvCount(&$arr, $n)
{
    $inv_count = 0;
    for ($i = 0; $i < $n - 1; $i++)
        for ($j = $i + 1; $j < $n; $j++)
            if ($arr[$i] > $arr[$j])
                $inv_count++;
  
    return $inv_count;
}
  
// Driver Code
$arr = array(1, 20, 6, 4, 5 );
$n = sizeof($arr);
echo "Number of inversions are "
           getInvCount($arr, $n);
  
// This code is contributed by ita_c
?>
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Number of inversions are 5

METHOD 2(Enhance Merge Sort) 



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// C++ program to Count
// Inversions in an array
// using Merge Sort
#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;
}
  
// This is code is contributed by rathbhupendra
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// C program to Count
// Inversions in an array
// using Merge Sort
#include <stdio.h>
#include <stdlib.h>
  
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 = (int*)malloc(sizeof(int) * 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 the 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 program to test above functions */
int main(int argv, char** args)
{
    int arr[] = { 1, 20, 6, 4, 5 };
    printf(" Number of inversions are %d \n",
           mergeSort(arr, 5));
    getchar();
    return 0;
}
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// Java implementation of the approach
import java.util.Arrays;
  
public class GFG {
  
    // Function to count the number of inversions
    // during the merge process
    private static int mergeAndCount(int[] arr, 
                                int l, int m, int r)
    {
  
        // Left subarray
        int[] left = Arrays.copyOfRange(arr, l, m + 1);
  
        // Right subarray
        int[] right = Arrays.copyOfRange(arr, m + 1, r + 1);
  
        int i = 0, j = 0, k = l, swaps = 0;
  
        while (i < left.length && j < right.length) 
        {
            if (left[i] <= right[j])
                arr[k++] = left[i++];
            else {
                arr[k++] = right[j++];
                swaps += (m + 1) - (l + i);
            }
        }
        return swaps;
    }
  
    // Merge sort function
    private static int mergeSortAndCount(int[] arr, 
                                        int l, int r)
    {
  
        // Keeps track of the inversion count at a
        // particular node of the recursion tree
        int count = 0;
  
        if (l < r) {
            int m = (l + r) / 2;
  
            // Total inversion count = left subarray count
            // + right subarray count + merge count
  
            // Left subarray count
            count += mergeSortAndCount(arr, l, m);
  
            // Right subarray count
            count += mergeSortAndCount(arr, m + 1, r);
  
            // Merge count
            count += mergeAndCount(arr, l, m, r);
        }
  
        return count;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 1, 20, 6, 4, 5 };
  
        System.out.println(mergeSortAndCount(arr, 0
                                        arr.length - 1));
    }
}
  
// This code is contributed by Pradip Basak
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# Python 3 program to count inversions in an array
  
# Function to Use Inversion Count
def mergeSort(arr, n):
    # A temp_arr is created to store
    # sorted array in merge function
    temp_arr = [0]*n
    return _mergeSort(arr, temp_arr, 0, n-1)
  
# This Function will use MergeSort to count inversions
  
def _mergeSort(arr, temp_arr, left, right):
  
    # A variable inv_count is used to store
    # inversion counts in each recursive call
  
    inv_count = 0
  
    # We will make a recursive call if and only if
    # we have more than one elements
  
    if left < right:
  
        # mid is calculated to divide the array into two subarrays
        # Floor division is must in case of python
  
        mid = (left + right)//2
  
        # It will calculate inversion 
        # counts in the left subarray
  
        inv_count += _mergeSort(arr, temp_arr, 
                                    left, mid)
  
        # It will calculate inversion 
        # counts in right subarray
  
        inv_count += _mergeSort(arr, temp_arr, 
                                  mid + 1, right)
  
        # It will merge two subarrays in 
        # a sorted subarray
  
        inv_count += merge(arr, temp_arr, left, mid, right)
    return inv_count
  
# This function will merge two subarrays 
# in a single sorted subarray
def merge(arr, temp_arr, left, mid, right):
    i = left     # Starting index of left subarray
    j = mid + 1 # Starting index of right subarray
    k = left     # Starting index of to be sorted subarray
    inv_count = 0
  
    # Conditions are checked to make sure that 
    # i and j don't exceed their
    # subarray limits.
  
    while i <= mid and j <= right:
  
        # There will be no inversion if arr[i] <= arr[j]
  
        if arr[i] <= arr[j]:
            temp_arr[k] = arr[i]
            k += 1
            i += 1
        else:
            # Inversion will occur.
            temp_arr[k] = arr[j]
            inv_count += (mid-i + 1)
            k += 1
            j += 1
  
    # Copy the remaining elements of left 
    # subarray into temporary array
    while i <= mid:
        temp_arr[k] = arr[i]
        k += 1
        i += 1
  
    # Copy the remaining elements of right 
    # subarray into temporary array
    while j <= right:
        temp_arr[k] = arr[j]
        k += 1
        j += 1
  
    # Copy the sorted subarray into Original array
    for loop_var in range(left, right + 1):
        arr[loop_var] = temp_arr[loop_var]
          
    return inv_count
  
# Driver Code
# Given array is
arr = [1, 20, 6, 4, 5]
n = len(arr)
result = mergeSort(arr, n)
print("Number of inversions are", result)
  
# This code is contributed by ankush_953
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// C# implementation of counting the
// inversion using merge sort
  
using System;
public class Test {
  
    /* This method sorts the input array and returns the
       number of inversions in the array */
    static int mergeSort(int[] arr, int array_size)
    {
        int[] temp = new int[array_size];
        return _mergeSort(arr, temp, 0, array_size - 1);
    }
  
    /* An auxiliary recursive method that sorts the input
      array and returns the number of inversions in the
      array. */
    static 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 the 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 method merges two sorted arrays and returns
       inversion count in the arrays.*/
    static 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 method to test the above function
    public static void Main()
    {
        int[] arr = new int[] { 1, 20, 6, 4, 5 };
        Console.Write("Number of inversions are "
                      + mergeSort(arr, 5));
    }
}
// This code is contributed by Rajput-Ji
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Number of inversions are 5

Note that the above code modifies (or sorts) the input array. If we want to count only inversions then we need to create a copy of original array and call mergeSort() on copy.

You may like to see. 
Count inversions in an array | Set 2 (Using Self-Balancing BST) 
Counting Inversions using Set in C++ STL 
Count inversions in an array | Set 3 (Using BIT)
References: 
http://www.cs.umd.edu/class/fall2009/cmsc451/lectures/Lec08-inversions.pdf 
http://www.cp.eng.chula.ac.th/~piak/teaching/algo/algo2008/count-inv.htm
Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.

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