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)

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

METHOD 1 (Simple)

Approach :Traverse through the array and for every index find the number of smaller elements on its right side of the array. This can be done using a nested loop. Sum up the counts for all index in the array and print the sum.
Algorithm :
1. Traverse through the array from start to end
2. For every element find the count of elements smaller than the current number upto that index using another loop.
3. Sum up the count of inversion for every index.
4. Print the count of inversions.

Implementation:

C++

// 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 

C

// C program to Count 
// Inversions in an array 
#include <stdio.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; 
} 

Java

// 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)); 
    } 
} 

Python3

# 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 

C#

// 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 

PHP

<?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 
?> 

Output:

Number of inversions are 5

Complexity Analysis:
- Time Complexity: O(n^2), Two nested loops are needed to traverse the array from start to end so the Time complexity is O(n^2)
- Space Compelxity:O(1), No extra space is required.

METHOD 2(Enhance Merge Sort)

Approach:
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().

How to get number of inversions in merge()?
In merge process, let i is used for indexing left sub-array and j for right sub-array. At any step in merge(), 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]

The complete picture:
Algorithm:
1. 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.
2. 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].
3. 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.
4. The base case of recursion is when there is only one element in the given half.
5. Print the answer

Implementation:

C++

// 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 

C

// C program to Count 
// Inversions in an array 
// using Merge Sort 
#include <stdio.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; 
} 

Java

// 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); 
            } 
        } 
  
        // Fill from the rest of the left subarray 
        while (i < left.length) 
            arr[k++] = left[i++]; 
  
        // Fill from the rest of the right subarray 
        while (j < right.length) 
            arr[k++] = right[j++]; 
  
        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 

Python3

# 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 

C#

// 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 

Output:

Number of inversions are 5

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.

Note that 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.

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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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python3

C#

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