Count Inversions of an Array

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Given an integer array arr[] of size n, find the inversion count in the array. Two array elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j.

Note: Inversion Count for an array indicates that how far (or close) the array is from being sorted. If the array is already sorted, then the inversion count is 0, but if the array is sorted in reverse order, the inversion count is maximum. 

Examples: 

Input: arr[] = {4, 3, 2, 1}
Output: 6
Explanation:

Input: arr[] = {1, 2, 3, 4, 5}
Output: 0
Explanation: There is no pair of indexes (i, j) exists in the given array such that arr[i] > arr[j] and i < j

Input: arr[] = {10, 10, 10}
Output: 0

[Naive Approach] Using Two Nested Loops – O(n^2) Time and O(1) Space

Traverse through the array, and for every index, find the number of smaller elements on its right side in the array. This can be done using a nested loop. Sum up the inversion counts for all indices in the array and return the sum.

// C++ program to Count Inversions in an array 
// using nested loop

#include <iostream>
#include <vector>
using namespace std;

// Function to count inversions in the array
int inversionCount(vector<int> &arr) {
    int n = arr.size(); 
    int invCount = 0; 
  
    // Loop through the array
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
          
            // If the current element is greater 
            // than the next, increment the count
            if (arr[i] > arr[j])
                invCount++;
        }
    }
    return invCount; 
}

int main() {
    vector<int> arr = {4, 3, 2, 1};
    cout << inversionCount(arr) << endl; 
    return 0;
}

// C program to Count Inversions in an array 
// using nested loop

#include <stdio.h>

// Function to count inversions in the array
int inversionCount(int arr[], int n) {
    int invCount = 0;

    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            
            // If the current element is greater than the next,
            // increment the count
            if (arr[i] > arr[j])
                invCount++;
        }
    }
    return invCount;
}

int main() {
    int arr[] = {4, 3, 2, 1};
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    printf("%d\n", inversionCount(arr, n));
    return 0;
}

// Java program to Count Inversions in an array 
// using nested loop

import java.util.*;

class GfG {

    // Function to count inversions in the array
    static int inversionCount(int arr[]) {
        int n = arr.length; 
        int invCount = 0;  

        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
              
                // If the current element is greater than the next,
                // increment the count
                if (arr[i] > arr[j])
                    invCount++;
            }
        }
        return invCount; 
    }

    public static void main(String[] args) {
        int arr[] = {4, 3, 2, 1};
        System.out.println(inversionCount(arr));
    }
}

# Python program to Count Inversions in an array 
# using nested loop

# Function to count inversions in the array
def inversionCount(arr):
    n = len(arr) 
    invCount = 0  

    for i in range(n - 1):
        for j in range(i + 1, n):
          
            # If the current element is greater than the next,
            # increment the count
            if arr[i] > arr[j]:
                invCount += 1
    return invCount  

if __name__ == "__main__":
    arr = [4, 3, 2, 1]
    print(inversionCount(arr))

// C# program to Count Inversions in an array 
// using nested loop
using System;

class GfG {
  
    // Function to count inversions in the array
    static int inversionCount(int[] arr) {
        int n = arr.Length; 
        int invCount = 0; 

        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
              
                // If the current element is greater than the next,
                // increment the count
                if (arr[i] > arr[j])
                    invCount++;
            }
        }
        return invCount; 
    }

    static void Main() {
      
        int[] arr = { 4, 3, 2, 1 };
        Console.WriteLine(inversionCount(arr));
    }
}

// JavaScript program to Count Inversions in an array 
// using nested loop

function inversionCount(arr) {
    let n = arr.length; 
    let invCount = 0;  
    
    // Loop through the array
    for (let i = 0; i < n - 1; i++) {
        for (let j = i + 1; j < n; j++) {
        
            // If the current element is greater than the next,
            // increment the count
            if (arr[i] > arr[j]) {
                invCount++;
            }
        }
    }
    return invCount;
}

// driver code
let arr = [4, 3, 2, 1];
console.log(inversionCount(arr));

6

[Expected Approach] Using Merge Step of Merge Sort – O(n*log n) Time and O(n) Space

We can use merge sort to count the inversions in an array. First, we divide the array into two halves: left half and right half. Next, we recursively count the inversions in both halves. While merging the two halves back together, we also count how many elements from the left half array are greater than elements from the right half array, as these represent cross inversions (i.e., element from the left half of the array is greater than an element from the right half during the merging process in the merge sort algorithm). Finally, we sum the inversions from the left half, right half, and the cross inversions to get the total number of inversions in the array. This approach efficiently counts inversions while sorting the array.

Let’s understand the above intuition in more detailed form, as we get to know that we have to perform the merge sort on the given array. Below images represents dividing and merging steps of merge sort.

During each merging step of the merge sort algorithm, we count cross inversions by comparing elements from the left half of the array with those from the right half. If we find an element arr[i] in the left half that is greater than an element arr[j] in the right half, we can conclude that all elements after i in the left half will also be greater than arr[j]. This allows us to count multiple inversions at once. Let’s suppose if there are k elements remaining in the left half after i, then there are k cross inversions for that particular arr[j]. The rest of the merging process continues as usual, where we combine the two halves into a sorted array. This efficient counting method significantly reduces the number of comparisons needed, enhancing the overall performance of the inversion counting algorithm.

Below Illustration represents the cross inversion of a particular step during the merging process in the merge sort algorithm:

1 / 9

// C++ program to Count Inversions in an array using merge sort

#include <iostream>
#include <vector>
using namespace std;

// This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
// and also counts inversions in the whole subarray arr[l..r]
int countAndMerge(vector<int>& arr, int l, int m, int r) {
  
    // Counts in two subarrays
    int n1 = m - l + 1, n2 = r - m;

    // Set up two vectors for left and right halves
    vector<int> left(n1), right(n2);
    for (int i = 0; i < n1; i++)
        left[i] = arr[i + l];
    for (int j = 0; j < n2; j++)
        right[j] = arr[m + 1 + j];

    // Initialize inversion count (or result) and merge two halves
    int res = 0;
    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {

        // No increment in inversion count if left[] has a 
        // smaller or equal element
        if (left[i] <= right[j]) 
            arr[k++] = left[i++];
      
        // If right is smaller, then it is smaller than n1-i 
      	// elements because left[] is sorted
        else {
            arr[k++] = right[j++];
            res += (n1 - i);
        }
    }

    // Merge remaining elements
    while (i < n1)
        arr[k++] = left[i++];
    while (j < n2)
        arr[k++] = right[j++];

    return res;
}

// Function to count inversions in the array
int countInv(vector<int>& arr, int l, int r){
    int res = 0;
    if (l < r) {
        int m = (r + l) / 2;

        // Recursively count inversions in the left and 
        // right halves
        res += countInv(arr, l, m);
        res += countInv(arr, m + 1, r);

        // Count inversions such that greater element is in 
      	// the left half and smaller in the right half
        res += countAndMerge(arr, l, m, r);
    }
    return res;
}

int inversionCount(vector<int> &arr) {
  	int n = arr.size();
  	return countInv(arr, 0, n-1);
}

int main(){
    vector<int> arr = {4, 3, 2, 1};
    
    cout << inversionCount(arr);
    return 0;
}

// C program to Count Inversions in an array using merge sort

#include <stdio.h>

// This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
// and also counts inversions in the whole subarray arr[l..r]
int countAndMerge(int arr[], int l, int m, int r) {
  
    // Counts in two subarrays
    int n1 = m - l + 1, n2 = r - m;

    // Set up two arrays for left and right halves
    int left[n1], right[n2];
    for (int i = 0; i < n1; i++)
        left[i] = arr[i + l];
    for (int j = 0; j < n2; j++)
        right[j] = arr[m + 1 + j];

    // Initialize inversion count (or result)
    // and merge two halves
    int res = 0;
    int i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {

        // No increment in inversion count
        // if left[] has a smaller or equal element
        if (left[i] <= right[j]) 
            arr[k++] = left[i++];
      
        // If right is smaller, then it is smaller than n1-i 
        // elements because left[] is sorted
        else {
            arr[k++] = right[j++];
            res += (n1 - i);
        }
    }

    // Merge remaining elements
    while (i < n1)
        arr[k++] = left[i++];
    while (j < n2)
        arr[k++] = right[j++];

    return res;
}

// Function to count inversions in the array
int countInv(int arr[], int l, int r) {
    int res = 0;
    if (l < r) {
        int m = (r + l) / 2;

        // Recursively count inversions
        // in the left and right halves
        res += countInv(arr, l, m);
        res += countInv(arr, m + 1, r);

        // Count inversions such that greater element is in 
        // the left half and smaller in the right half
        res += countAndMerge(arr, l, m, r);
    }
    return res;
}

int inversionCount(int arr[], int n) {
    return countInv(arr, 0, n - 1);
}

int main() {
    int arr[] = {4, 3, 2, 1};
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("%d", inversionCount(arr, n));
    return 0;
}

// Java program to Count Inversions in an array using merge sort

import java.util.*;

class GfG {

    // This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
    // and also counts inversions in the whole subarray arr[l..r]
    static int countAndMerge(int[] arr, int l, int m, int r) {
      
        // Counts in two subarrays
        int n1 = m - l + 1, n2 = r - m;

        // Set up two arrays for left and right halves
        int[] left = new int[n1];
        int[] right = new int[n2];
        for (int i = 0; i < n1; i++)
            left[i] = arr[i + l];
        for (int j = 0; j < n2; j++)
            right[j] = arr[m + 1 + j];

        // Initialize inversion count (or result)
        // and merge two halves
        int res = 0;
        int i = 0, j = 0, k = l;
        while (i < n1 && j < n2) {

            // No increment in inversion count
            // if left[] has a smaller or equal element
            if (left[i] <= right[j])
                arr[k++] = left[i++];
          
            // If right is smaller, then it is smaller than n1-i 
            // elements because left[] is sorted
            else {
                arr[k++] = right[j++];
                res += (n1 - i);
            }
        }

        // Merge remaining elements
        while (i < n1)
            arr[k++] = left[i++];
        while (j < n2)
            arr[k++] = right[j++];

        return res;
    }

    // Function to count inversions in the array
    static int countInv(int[] arr, int l, int r) {
        int res = 0;
        if (l < r) {
            int m = (r + l) / 2;

            // Recursively count inversions
            // in the left and right halves
            res += countInv(arr, l, m);
            res += countInv(arr, m + 1, r);

            // Count inversions such that greater element is in 
            // the left half and smaller in the right half
            res += countAndMerge(arr, l, m, r);
        }
        return res;
    }

    static int inversionCount(int[] arr) {
        return countInv(arr, 0, arr.length - 1);
    }

    public static void main(String[] args) {
        int[] arr = {4, 3, 2, 1};
        System.out.println(inversionCount(arr));
    }
}

# Python program to Count Inversions in an array using merge sort

# This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
# and also counts inversions in the whole subarray arr[l..r]
def countAndMerge(arr, l, m, r):
  
    # Counts in two subarrays
    n1 = m - l + 1
    n2 = r - m

    # Set up two lists for left and right halves
    left = arr[l:m + 1]
    right = arr[m + 1:r + 1]

    # Initialize inversion count (or result)
    # and merge two halves
    res = 0
    i = 0
    j = 0
    k = l
    while i < n1 and j < n2:

        # No increment in inversion count
        # if left[] has a smaller or equal element
        if left[i] <= right[j]:
            arr[k] = left[i]
            i += 1
        else:
            arr[k] = right[j]
            j += 1
            res += (n1 - i)
        k += 1

    # Merge remaining elements
    while i < n1:
        arr[k] = left[i]
        i += 1
        k += 1
    while j < n2:
        arr[k] = right[j]
        j += 1
        k += 1

    return res

# Function to count inversions in the array
def countInv(arr, l, r):
    res = 0
    if l < r:
        m = (r + l) // 2

        # Recursively count inversions
        # in the left and right halves
        res += countInv(arr, l, m)
        res += countInv(arr, m + 1, r)

        # Count inversions such that greater element is in 
        # the left half and smaller in the right half
        res += countAndMerge(arr, l, m, r)
    return res

def inversionCount(arr):
    return countInv(arr, 0, len(arr) - 1)

if __name__ == "__main__":
    arr = [4, 3, 2, 1]
    print(inversionCount(arr))

// C# program to Count Inversions in an array using merge sort

using System;
using System.Collections.Generic;

class GfG {

    // This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
    // and also counts inversions in the whole subarray arr[l..r]
    static int countAndMerge(int[] arr, int l, int m, int r) {

        // Counts in two subarrays
        int n1 = m - l + 1, n2 = r - m;

        // Set up two arrays for left and right halves
        int[] left = new int[n1];
        int[] right = new int[n2];
        for (int x = 0; x < n1; x++)
            left[x] = arr[x + l];
        for (int x = 0; x < n2; x++)
            right[x] = arr[m + 1 + x];

        // Initialize inversion count (or result)
        // and merge two halves
        int res = 0;
        int i = 0, j = 0, k = l;
        while (i < n1 && j < n2) {

            // No increment in inversion count
            // if left[] has a smaller or equal element
            if (left[i] <= right[j])
                arr[k++] = left[i++];
            else {
                arr[k++] = right[j++];
                res += (n1 - i);
            }
        }

        // Merge remaining elements
        while (i < n1)
            arr[k++] = left[i++];
        while (j < n2)
            arr[k++] = right[j++];

        return res;
    }

    // Function to count inversions in the array
    static int countInv(int[] arr, int l, int r) {
        int res = 0;
        if (l < r) {
            int m = (r + l) / 2;

            // Recursively count inversions
            // in the left and right halves
            res += countInv(arr, l, m);
            res += countInv(arr, m + 1, r);

            // Count inversions such that greater element is in 
            // the left half and smaller in the right half
            res += countAndMerge(arr, l, m, r);
        }
        return res;
    }

    static int inversionCount(int[] arr) {
        return countInv(arr, 0, arr.Length - 1);
    }

    static void Main() {
        int[] arr = {4, 3, 2, 1};
        Console.WriteLine(inversionCount(arr));
    }
}

// JavaScript program to Count Inversions in an array using merge sort

// This function merges two sorted subarrays arr[l..m] and arr[m+1..r] 
// and also counts inversions in the whole subarray arr[l..r]
function countAndMerge(arr, l, m, r) {

    // Counts in two subarrays
    let n1 = m - l + 1, n2 = r - m;

    // Set up two arrays for left and right halves
    let left = arr.slice(l, m + 1);
    let right = arr.slice(m + 1, r + 1);

    // Initialize inversion count (or result)
    // and merge two halves
    let res = 0;
    let i = 0, j = 0, k = l;
    while (i < n1 && j < n2) {

        // No increment in inversion count
        // if left[] has a smaller or equal element
        if (left[i] <= right[j])
            arr[k++] = left[i++];
        else {
            arr[k++] = right[j++];
            res += (n1 - i);
        }
    }

    // Merge remaining elements
    while (i < n1)
        arr[k++] = left[i++];
    while (j < n2)
        arr[k++] = right[j++];

    return res;
}

// Function to count inversions in the array
function countInv(arr, l, r) {
    let res = 0;
    if (l < r) {
        let m = Math.floor((r + l) / 2);

        // Recursively count inversions
        // in the left and right halves
        res += countInv(arr, l, m);
        res += countInv(arr, m + 1, r);

        // Count inversions such that greater element is in 
        // the left half and smaller in the right half
        res += countAndMerge(arr, l, m, r);
    }
    return res;
}

function inversionCount(arr) {
    return countInv(arr, 0, arr.length - 1);
}

// Driver Code
let arr = [4, 3, 2, 1];
console.log(inversionCount(arr));

6

Note: The above code modifies (or sorts) the input array. If we want to count only inversions, we need to create a copy of the original array and perform operation on the copy array.

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)
  

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• Divide and Conquer
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Practice Tags :

• Adobe
• Amazon
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• Microsoft
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• Arrays
• Arrays
• Divide and Conquer
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