# Inversion count in Array using Merge Sort

• Difficulty Level : Hard
• Last Updated : 31 Jan, 2023

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 a[]. The task is to find the inversion count of a[]. Where two elements a[i] and a[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).

Recommended Practice

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 ` `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 ` `#include ` `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

 ` ``\$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

 ``

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 â€“ i) inversions. because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] â€¦ 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 â€“ i) inversions because left and right subarrays are sorted, so all the remaining elements in left-subarray (a[i+1], a[i+2] â€¦ 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 ` `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 ` `#include `   `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

 ``

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

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