Inversion count in Array using Merge Sort

Inversion Count for an array indicates – 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 the maximum. 

Given an array arr[]. The task is to find the inversion count of arr[]. Where two elements arr[i] and arr[j] form an inversion if a[i] > a[j] and i < j.

Examples:

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[] = {1, 20, 6, 4, 5}
Output: 5
Explanation: Given array has five inversions: (20, 6), (20, 4), (20, 5), (6, 4), (6, 5). 

Naive 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 indices in the array and print the sum.

Follow the below steps to Implement the idea:

• Traverse through the array from start to end
• For every element, find the count of elements smaller than the current number up to that index using another loop.
• Sum up the count of inversion for every index.
• Print the count of inversions.

Below is the Implementation of the above approach:

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>
#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;
}

Java

// Java program to count
// inversions in an array

import java.io.*;

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
?>

Javascript

<script>
// Javascript program to count inversions in an array

arr = [1, 20, 6, 4, 5];

function getInvCount(arr){
    let inv_count = 0;
    for(let i=0; i<arr.length-1; i++){
        for(let j=i+1; j<arr.length; j++){
            if(arr[i] > arr[j]) inv_count++;
        }
    }
    return inv_count;
}

// function call
document.write("Number of inversions are "+ getInvCount(arr));

// This code is contributed by Karthik SP
</script>

Output

 Number of inversions are 5

Time Complexity: O(N²), Two nested loops are needed to traverse the array from start to end.
Auxiliary Space: O(1), No extra space is required.

Count Inversions in an array using Merge Sort:

Below is the idea to solve the problem:

Use Merge sort with modification that every time an unsorted pair is found increment count by one and return count at the end.

Illustration:

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 the total number of inversions that needs to be added are the number of inversions in the left subarray, right subarray, and merge().

How to get the 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:

Follow the below steps to Implement the idea:

• 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.
• 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 the 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 in the first half, the number of inversions 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.
• Print the answer.

Below is the Implementation of the above approach:

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 function 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;
    j = mid;
    k = left;
    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>
#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 function 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;
    j = mid;
    k = left;
    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 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);
            }
        }
        while (i < left.length)
            arr[k++] = left[i++];
        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

Javascript

<script>
    // Function to count the number of inversions
    // during the merge process
    function mergeAndCount(arr,l,m,r)
    {

        // Left subarray
        let left = [];
        for(let i = l; i < m + 1; i++)
        {
            left.push(arr[i]);

        }

        // Right subarray
        let right = [];
        for(let i = m + 1; i < r + 1; i++)
        {
            right.push(arr[i]);
        }
        let 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);
            }
        }
        while (i < left.length)
        {
            arr[k++] = left[i++];
        }
        while (j < right.length)
        {
            arr[k++] = right[j++];
        }
        return swaps;
    }

    // Merge sort function
    function mergeSortAndCount(arr, l, r)
    {

        // Keeps track of the inversion count at a
        // particular node of the recursion tree
        let count = 0;
        if (l < r)
        {
            let m = Math.floor((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
    let arr= new Array(1, 20, 6, 4, 5 );
    document.write(mergeSortAndCount(arr, 0, arr.length - 1));

    // This code is contributed by avanitrachhadiya2155
</script>

Output

 Number of inversions are 5

Time Complexity: O(N * log N), The algorithm used is divide and conquer i.e. merge sort whose complexity is O(n log n).
Auxiliary Space: O(N), Temporary array.

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 call mergeSort() on the copy to preserve the original array’s order.

Count Inversions in an array using Heapsort and Bisection:

Follow the below steps to Implement the idea:

• Create a heap with new pair elements,  (element, index). 
• After sorting them, pop out each minimum sequentially and create a new sorted list with the indexes. 
• Calculate the difference between the original index and the index of bisection of the new sorted list.
• Sum up the difference.

Below is the idea to Implement the above approach:

C++

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

int getNumOfInversions(vector<int> A) {
  int N = A.size();
  if (N <= 1) {
    return 0;
  }

  vector<pair<int, int>> sortList;
  int result = 0;

  // Heapsort, O(N*log(N))
  for (int i = 0; i < N; i++) {
    sortList.emplace_back(A[i], i);
    push_heap(sortList.begin(), sortList.end());
  }

  // Create a sorted list of indexes
  vector<int> x;
  while (!sortList.empty()) {
    // O(log(N))
    pair<int, int> v = sortList.front();
    pop_heap(sortList.begin(), sortList.end());
    sortList.pop_back();

    // Find the current minimum's index
    // the index y can represent how many minimums on the left
    int y = lower_bound(x.begin(), x.end(), v.second) - x.begin();

    // i can represent how many elements on the left
    // i - y can find how many bigger nums on the left
    result += v.second - y;

    x.insert(lower_bound(x.begin(), x.end(), v.second), v.second);
  }

  return result;
}

// Driver Code
int main() {
  vector<int> A = {1, 20, 6, 4, 5};
  int result = getNumOfInversions(A);
  cout << "Number of inversions are " << result << endl;
  return 0;
}

Java

import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;

class Main {
  public static int getNumOfInversions(List<Integer> A)
  {
    int N = A.size();
    if (N <= 1) {
      return 0;
    }

    PriorityQueue<int[]> sortList
      = new PriorityQueue<>((a, b) -> a[0] - b[0]);
    int result = 0;

    // Heapsort, O(N*log(N))
    for (int i = 0; i < N; i++) {
      sortList.add(new int[] { A.get(i), i });
    }

    // Create a sorted list of indexes
    List<Integer> x = new ArrayList<>();
    while (!sortList.isEmpty()) {
      // O(log(N))
      int[] v = sortList.poll();

      // Find the current minimum's index
      // the index y can represent how many minimums
      // on the left
      int y = x.size()
        - x.subList(0, x.size()).indexOf(v[1])
        - 1;
      int z = 0;
      if (!x.isEmpty()) {
        z = binarySearch(x, 0, x.size() - 1, v[1]);
        if (z < 0) {
          z = -(z + 1);
        }
      }

      // i can represent how many elements on the left
      // i - y can find how many bigger nums on the
      // left
      result += v[1] - z;

      x.add(v[1]);
      x.sort(null);
    }

    return result;
  }

  // Implementing binary search
  private static int binarySearch(List<Integer> list,
                                  int start, int end,
                                  int key)
  {

    // iterating while the values of start and end are
    // valid
    while (start <= end) {

      // Finding the midpoint
      int mid = start + (end - start) / 2;
      if (list.get(mid) == key) {
        return mid;
      }
      else if (list.get(mid) > key) {
        end = mid - 1;
      }
      else {
        start = mid + 1;
      }
    }
    return -(start + 1);
  }

  // Driver code
  public static void main(String[] args)
  {
    List<Integer> A = List.of(1, 20, 6, 4, 5);
    int result = getNumOfInversions(A);
    System.out.println("Number of inversions are "
                       + result);
  }
}

// This code is contributed by phasing17

Python3

from heapq import heappush, heappop
from bisect import bisect, insort


def getNumOfInversions(A):
    N = len(A)
    if N <= 1:
        return 0

    sortList = []
    result = 0

    # Heapsort, O(N*log(N))
    for i, v in enumerate(A):
        heappush(sortList, (v, i))

    # Create a sorted list of indexes
    x = []
    while sortList:

        # O(log(N))
        v, i = heappop(sortList)

        # Find the current minimum's index
        # the index y can represent how many minimums on the left
        y = bisect(x, i)

        # i can represent how many elements on the left
        # i - y can find how many bigger nums on the left
        result += i - y

        insort(x, i)

    return result

# Driver Code
if __name__ == '__main__':
    A = [1, 20, 6, 4, 5]
    result = getNumOfInversions(A)
    print(f'Number of inversions are {result}')

C#

using System;
using System.Collections.Generic;
using System.Linq;

namespace InversionCount
{
    class MainClass
    {
        public static int GetNumOfInversions(List<int> A)
        {
            int N = A.Count;
            if (N <= 1)
            {
                return 0;
            }

            var sortList = new List<int[]>();
            int result = 0;

            // Heapsort, O(N*log(N))
            for (int i = 0; i < N; i++)
            {
                sortList.Add(new int[] { A[i], i });
            }

            // Create a sorted list of indexes
            List<int> x = new List<int>();
            while (sortList.Any())
            {
                sortList = sortList.OrderBy(a => a[0]).ToList();
                // O(log(N))
                int[] v = sortList[0];
                sortList.RemoveAt(0);

                // Find the current minimum's index
                // the index y can represent how many minimums
                // on the left
                int y = x.Count
                  - x.GetRange(0, x.Count).IndexOf(v[1])
                  - 1;
                int z = 0;
                if (x.Any())
                {
                    z = BinarySearch(x, 0, x.Count - 1, v[1]);
                    if (z < 0)
                    {
                        z = -(z + 1);
                    }
                }

                // i can represent how many elements on the left
                // i - y can find how many bigger nums on the
                // left
                result += v[1] - z;

                x.Add(v[1]);
                x.Sort();
            }

            return result;
        }

        // Implementing binary search
        private static int BinarySearch(List<int> list,
                                        int start, int end,
                                        int key)
        {

            // iterating while the values of start and end are
            // valid
            while (start <= end)
            {

                // Finding the midpoint
                int mid = start + (end - start) / 2;
                if (list[mid] == key)
                {
                    return mid;
                }
                else if (list[mid] > key)
                {
                    end = mid - 1;
                }
                else
                {
                    start = mid + 1;
                }
            }
            return -(start + 1);
        }

        public static void Main(string[] args)
        {
            List<int> A = new List<int> { 1, 20, 6, 4, 5 };
            int result = GetNumOfInversions(A);
            Console.WriteLine("Number of inversions are " + result);
        }
    }
}

// This code is contributed by phasing17

Javascript

// JS program to implement the approach

const GetNumOfInversions = (A) => {
const N = A.length;
if (N <= 1) {
return 0;
}

const sortList = [];
let result = 0;

// Heapsort, O(N*log(N))
for (let i = 0; i < N; i++) {
sortList.push([A[i], i]);
}

// Create a sorted list of indexes
const x = [];
while (sortList.length) {
sortList.sort((a, b) => a[0] - b[0]);

// O(log(N))
const v = sortList[0];
sortList.shift();

// Find the current minimum's index
// the index y can represent how many minimums
// on the left
const y = x.length - x.slice(0, x.length).indexOf(v[1]) - 1;
let z = 0;
if (x.length) {
  z = BinarySearch(x, 0, x.length - 1, v[1]);
  if (z < 0) {
    z = -(z + 1);
  }
}

// i can represent how many elements on the left
// i - y can find how many bigger nums on the
// left
result += v[1] - z;

x.push(v[1]);
x.sort();
}

return result;
}

// Implementing binary search
const BinarySearch = (list, start, end, key) => {

// iterating while the values of start and end are
// valid
while (start <= end) {

// Finding the midpoint
const mid = start + Math.floor((end - start) / 2);
if (list[mid] === key) {
return mid;
} else if (list[mid] > key) {
end = mid - 1;
} else {
start = mid + 1;
}
}
return -(start + 1);
}

// Driver Code
const A = [1, 20, 6, 4, 5];
const result = GetNumOfInversions(A);
console.log(`Number of inversions are ${result}`);

// This code is contributed by phasing17

Output

Number of inversions are 5

Time Complexity: O(N * log N). Both heapsort and bisection can perform sorted insertion in (log n) in each element.
Auxiliary Space: O(N). A heap and a new list are the same length as the original 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)

Please write comments if you find any bug in the above program/algorithm or other ways to solve it.