# Count inversions in an array | Set 3 (Using BIT)

• Difficulty Level : Hard
• Last Updated : 07 Jul, 2021

Inversion Count for an array indicates â€“ how far (or close) the array is from being sorted. If the array is already sorted then the inversion count is 0. If the array is sorted in the reverse order that inversion count is the maximum.
Two elements a[i] and a[j] form an inversion if a[i] > a[j] and i < j. For simplicity, we may assume that all elements are unique.
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).```

## We strongly recommend that you click here and practice it, before moving on to the solution.

We have already discussed below methods to solve inversion count:

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We recommend you to refer Binary Indexed Tree (BIT) before further reading this post.
Solution using BIT of size Î˜(maxElement):

• 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 BIT. Sum up the counts for all indexes in the array and print the sum.
• Background on BIT:
1. BIT basically supports two operations for an array arr[] of size n:
1. Sum of elements till arr[i] in O(Log n) time.
2. Update an array element in O(Log n) time.
2. BIT is implemented using an array and works in form of trees. Note that there are two ways of looking at BIT as a tree.
1. The sum operation where parent of index x is “x – (x & -x)”.
2. The update operation where parent of index x is “x + (x & -x)”.
• Algorithm:
1. Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.
2. Traverse the array from end to start.
3. For every index check how many numbers less than the current element are present in BIT and add it to the result
4. To get the count of smaller elements, getSum() of BIT is used.
5. In his basic idea, BIT is represented as an array of size equal to maximum element plus one. So that elements can be used as an index.
6. After that we add the current element to the BIT[] by doing an update operation that updates the count of the current element from 0 to 1, and therefore updates ancestors of the current element in BIT (See update() in BIT for details).

• Implementation

## C++

 `// C++ program to count inversions using Binary Indexed Tree``#include``using` `namespace` `std;` `// Returns sum of arr[0..index]. This function assumes``// that the array is preprocessed and partial sums of``// array elements are stored in BITree[].``int` `getSum(``int` `BITree[], ``int` `index)``{``    ``int` `sum = 0; ``// Initialize result` `    ``// Traverse ancestors of BITree[index]``    ``while` `(index > 0)``    ``{``        ``// Add current element of BITree to sum``        ``sum += BITree[index];` `        ``// Move index to parent node in getSum View``        ``index -= index & (-index);``    ``}``    ``return` `sum;``}` `// Updates a node in Binary Index Tree (BITree) at given index``// in BITree.  The given value 'val' is added to BITree[i] and``// all of its ancestors in tree.``void` `updateBIT(``int` `BITree[], ``int` `n, ``int` `index, ``int` `val)``{``    ``// Traverse all ancestors and add 'val'``    ``while` `(index <= n)``    ``{``       ``// Add 'val' to current node of BI Tree``       ``BITree[index] += val;` `       ``// Update index to that of parent in update View``       ``index += index & (-index);``    ``}``}` `// Returns inversion count arr[0..n-1]``int` `getInvCount(``int` `arr[], ``int` `n)``{``    ``int` `invcount = 0; ``// Initialize result` `    ``// Find maximum element in arr[]``    ``int` `maxElement = 0;``    ``for` `(``int` `i=0; i=0; i--)``    ``{``        ``// Get count of elements smaller than arr[i]``        ``invcount += getSum(BIT, arr[i]-1);` `        ``// Add current element to BIT``        ``updateBIT(BIT, maxElement, arr[i], 1);``    ``}` `    ``return` `invcount;``}` `// Driver program``int` `main()``{``    ``int` `arr[] = {8, 4, 2, 1};``    ``int` `n = ``sizeof``(arr)/``sizeof``(``int``);``    ``cout << ``"Number of inversions are : "` `<< getInvCount(arr,n);``    ``return` `0;``}`

## Java

 `// Java program to count inversions``// using Binary Indexed Tree` `class` `GFG``{``    ` `// Returns sum of arr[0..index].``// This function assumes that the``// array is preprocessed and partial``// sums of array elements are stored``// in BITree[].``static` `int` `getSum(``int``[] BITree, ``int` `index)``{``    ``int` `sum = ``0``; ``// Initialize result` `    ``// Traverse ancestors of BITree[index]``    ``while` `(index > ``0``)``    ``{``        ``// Add current element of BITree to sum``        ``sum += BITree[index];` `        ``// Move index to parent node in getSum View``        ``index -= index & (-index);``    ``}``    ``return` `sum;``}` `// Updates a node in Binary Index``// Tree (BITree) at given index``// in BITree. The given value 'val'``// is added to BITree[i] and all``// of its ancestors in tree.``static` `void` `updateBIT(``int``[] BITree, ``int` `n,``                        ``int` `index, ``int` `val)``{``    ``// Traverse all ancestors and add 'val'``    ``while` `(index <= n)``    ``{``        ``// Add 'val' to current node of BI Tree``        ``BITree[index] += val;` `        ``// Update index to that of parent in update View``        ``index += index & (-index);``    ``}``}` `// Returns inversion count arr[0..n-1]``static` `int` `getInvCount(``int``[] arr, ``int` `n)``{``    ``int` `invcount = ``0``; ``// Initialize result` `    ``// Find maximum element in arr[]``    ``int` `maxElement = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``if` `(maxElement < arr[i])``            ``maxElement = arr[i];` `    ``// Create a BIT with size equal to``    ``// maxElement+1 (Extra one is used so``    ``// that elements can be directly be``    ``// used as index)``    ``int``[] BIT = ``new` `int``[maxElement + ``1``];``    ``for` `(``int` `i = ``1``; i <= maxElement; i++)``        ``BIT[i] = ``0``;` `    ``// Traverse all elements from right.``    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)``    ``{``        ``// Get count of elements smaller than arr[i]``        ``invcount += getSum(BIT, arr[i] - ``1``);` `        ``// Add current element to BIT``        ``updateBIT(BIT, maxElement, arr[i], ``1``);``    ``}` `    ``return` `invcount;``}` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `[]arr = {``8``, ``4``, ``2``, ``1``};``    ``int` `n = arr.length;``    ``System.out.println(``"Number of inversions are : "` `+``                                ``getInvCount(arr,n));``}``}` `// This code is contributed by mits`

## Python3

 `# Python3 program to count inversions using``# Binary Indexed Tree` `# Returns sum of arr[0..index]. This function``# assumes that the array is preprocessed and``# partial sums of array elements are stored``# in BITree[].``def` `getSum( BITree, index):``    ``sum` `=` `0` `# Initialize result``    ` `    ``# Traverse ancestors of BITree[index]``    ``while` `(index > ``0``):` `        ``# Add current element of BITree to sum``        ``sum` `+``=` `BITree[index]` `        ``# Move index to parent node in getSum View``        ``index ``-``=` `index & (``-``index)` `    ``return` `sum` `# Updates a node in Binary Index Tree (BITree)``# at given index in BITree. The given value``# 'val' is added to BITree[i] and all of its``# ancestors in tree.``def` `updateBIT(BITree, n, index, val):` `    ``# Traverse all ancestors and add 'val'``    ``while` `(index <``=` `n):` `        ``# Add 'val' to current node of BI Tree``        ``BITree[index] ``+``=` `val` `        ``# Update index to that of parent``        ``# in update View``        ``index ``+``=` `index & (``-``index)` `# Returns count of inversions of size three``def` `getInvCount(arr, n):` `    ``invcount ``=` `0` `# Initialize result` `    ``# Find maximum element in arrays``    ``maxElement ``=` `max``(arr)` `    ``# Create a BIT with size equal to``    ``# maxElement+1 (Extra one is used``    ``# so that elements can be directly``    ``# be used as index)``    ``BIT ``=` `[``0``] ``*` `(maxElement ``+` `1``)``    ``for` `i ``in` `range``(``1``, maxElement ``+` `1``):``        ``BIT[i] ``=` `0``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):` `        ``invcount ``+``=` `getSum(BIT, arr[i] ``-` `1``)``        ``updateBIT(BIT, maxElement, arr[i], ``1``)``    ``return` `invcount``    ` `# Driver code``if` `__name__ ``=``=``"__main__"``:``    ``arr ``=` `[``8``, ``4``, ``2``, ``1``]``    ``n ``=` `4``    ``print``(``"Inversion Count : "``,``           ``getInvCount(arr, n))``    ` `# This code is contributed by``# Shubham Singh(SHUBHAMSINGH10)`

## C#

 `// C# program to count inversions``// using Binary Indexed Tree``using` `System;` `class` `GFG``{``    ` `// Returns sum of arr[0..index].``// This function assumes that the``// array is preprocessed and partial``// sums of array elements are stored``// in BITree[].``static` `int` `getSum(``int` `[]BITree, ``int` `index)``{``    ``int` `sum = 0; ``// Initialize result` `    ``// Traverse ancestors of BITree[index]``    ``while` `(index > 0)``    ``{``        ``// Add current element of BITree to sum``        ``sum += BITree[index];` `        ``// Move index to parent node in getSum View``        ``index -= index & (-index);``    ``}``    ``return` `sum;``}` `// Updates a node in Binary Index``// Tree (BITree) at given index``// in BITree. The given value 'val'``// is added to BITree[i] and all``// of its ancestors in tree.``static` `void` `updateBIT(``int` `[]BITree, ``int` `n,``                        ``int` `index, ``int` `val)``{``    ``// Traverse all ancestors and add 'val'``    ``while` `(index <= n)``    ``{``        ``// Add 'val' to current node of BI Tree``        ``BITree[index] += val;` `        ``// Update index to that of parent in update View``        ``index += index & (-index);``    ``}``}` `// Returns inversion count arr[0..n-1]``static` `int` `getInvCount(``int` `[]arr, ``int` `n)``{``    ``int` `invcount = 0; ``// Initialize result` `    ``// Find maximum element in arr[]``    ``int` `maxElement = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``if` `(maxElement < arr[i])``            ``maxElement = arr[i];` `    ``// Create a BIT with size equal to``    ``// maxElement+1 (Extra one is used so``    ``// that elements can be directly be``    ``// used as index)``    ``int``[] BIT = ``new` `int``[maxElement + 1];``    ``for` `(``int` `i = 1; i <= maxElement; i++)``        ``BIT[i] = 0;` `    ``// Traverse all elements from right.``    ``for` `(``int` `i = n - 1; i >= 0; i--)``    ``{``        ``// Get count of elements smaller than arr[i]``        ``invcount += getSum(BIT, arr[i] - 1);` `        ``// Add current element to BIT``        ``updateBIT(BIT, maxElement, arr[i], 1);``    ``}` `    ``return` `invcount;``}` `// Driver code``static` `void` `Main()``{``    ``int` `[]arr = {8, 4, 2, 1};``    ``int` `n = arr.Length;``    ``Console.WriteLine(``"Number of inversions are : "` `+``                                ``getInvCount(arr,n));``}``}` `// This code is contributed by mits`

## PHP

 ` 0)``    ``{``        ``// Add current element of BITree to sum``        ``\$sum` `+= ``\$BITree``[``\$index``];` `        ``// Move index to parent node in getSum View``        ``\$index` `-= ``\$index` `& (-``\$index``);``    ``}``    ``return` `\$sum``;``}` `// Updates a node in Binary Index``// Tree (BITree) at given index``// in BITree. The given value 'val'``// is added to BITree[i] and``// all of its ancestors in tree.``function` `updateBIT(&``\$BITree``, ``\$n``, ``\$index``,``\$val``)``{``    ``// Traverse all ancestors and add 'val'``    ``while` `(``\$index` `<= ``\$n``)``    ``{``        ``// Add 'val' to current node of BI Tree``        ``\$BITree``[``\$index``] += ``\$val``;` `        ``// Update index to that of``        ``// parent in update View``        ``\$index` `+= ``\$index` `& (-``\$index``);``    ``}``}` `// Returns inversion count arr[0..n-1]``function` `getInvCount(``\$arr``, ``\$n``)``{``    ``\$invcount` `= 0; ``// Initialize result` `    ``// Find maximum element in arr[]``    ``\$maxElement` `= 0;``    ``for` `(``\$i``=0; ``\$i``<``\$n``; ``\$i``++)``        ``if` `(``\$maxElement` `< ``\$arr``[``\$i``])``            ``\$maxElement` `= ``\$arr``[``\$i``];` `    ``// Create a BIT with size equal``    ``// to maxElement+1 (Extra one is``    ``// used so that elements can be``    ``// directly be used as index)``    ``\$BIT``=``array_fill``(0,``\$maxElement``+1,0);` `    ``// Traverse all elements from right.``    ``for` `(``\$i``=``\$n``-1; ``\$i``>=0; ``\$i``--)``    ``{``        ``// Get count of elements smaller than arr[i]``        ``\$invcount` `+= getSum(``\$BIT``, ``\$arr``[``\$i``]-1);` `        ``// Add current element to BIT``        ``updateBIT(``\$BIT``, ``\$maxElement``, ``\$arr``[``\$i``], 1);``    ``}` `    ``return` `\$invcount``;``}` `    ``// Driver program``    ``\$arr` `= ``array``(8, 4, 2, 1);``    ``\$n` `= ``count``(``\$arr``);``    ``print``(``"Number of inversions are : "``.getInvCount(``\$arr``,``\$n``));` `// This code is contributed by mits``?>`

## Javascript

 ``
• Output:
`Number of inversions are : 6`
• Complexity Analysis:
• Time Complexity :- The update function and getSum function runs for O(log(maximumelement)). The getSum function has to be run for every element in the array. So overall time complexity is : O(nlog(maximumelement)).
• Auxiliary space : O(maxElement), space required for the BIT is an array of the size of the largest element.

Better solution using BIT of size Î˜(n):

• 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 BIT. Sum up the counts for all indexes in the array and print the sum. The approach remains the same but the problem with the previous approach is that it doesn’t work for negative numbers as the index cannot be negative. The idea is to convert the given array to an array with values from 1 to n and the relative order of smaller and greater elements remains the same.
Example:-
```
arr[] = {7, -90, 100, 1}

It gets  converted to,
arr[] = {3, 1, 4 ,2 }
as -90 < 1 < 7 < 100.

Explanation: Make a BIT array of a number of
elements instead of a maximum element. Changing
element will not have any change in the answer
as the greater elements remain greater and at the
same position. ```
• Algorithm:
1. Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.
2. The previous solution does not work for arrays containing negative elements. So, convert the array into an array containing relative numbering of elements,i.e make a copy of the original array and then sort the copy of the array and replace the elements in the original array with the indices of the same elements in the sorted array.
For example, if the array is {-3, 2, 0} then the array gets converted to {1, 3, 2}
3. Traverse the array from end to start.
4. For every index check how many numbers less than the current element are present in BIT and add it to the result
• Implementation:

## C++

 `// C++ program to count inversions using Binary Indexed Tree``#include``using` `namespace` `std;` `// Returns sum of arr[0..index]. This function assumes``// that the array is preprocessed and partial sums of``// array elements are stored in BITree[].``int` `getSum(``int` `BITree[], ``int` `index)``{``    ``int` `sum = 0; ``// Initialize result` `    ``// Traverse ancestors of BITree[index]``    ``while` `(index > 0)``    ``{``        ``// Add current element of BITree to sum``        ``sum += BITree[index];` `        ``// Move index to parent node in getSum View``        ``index -= index & (-index);``    ``}``    ``return` `sum;``}` `// Updates a node in Binary Index Tree (BITree) at given index``// in BITree.  The given value 'val' is added to BITree[i] and``// all of its ancestors in tree.``void` `updateBIT(``int` `BITree[], ``int` `n, ``int` `index, ``int` `val)``{``    ``// Traverse all ancestors and add 'val'``    ``while` `(index <= n)``    ``{``       ``// Add 'val' to current node of BI Tree``       ``BITree[index] += val;` `       ``// Update index to that of parent in update View``       ``index += index & (-index);``    ``}``}` `// Converts an array to an array with values from 1 to n``// and relative order of smaller and greater elements remains``// same.  For example, {7, -90, 100, 1} is converted to``// {3, 1, 4 ,2 }``void` `convert(``int` `arr[], ``int` `n)``{``    ``// Create a copy of arrp[] in temp and sort the temp array``    ``// in increasing order``    ``int` `temp[n];``    ``for` `(``int` `i=0; i=0; i--)``    ``{``        ``// Get count of elements smaller than arr[i]``        ``invcount += getSum(BIT, arr[i]-1);` `        ``// Add current element to BIT``        ``updateBIT(BIT, n, arr[i], 1);``    ``}` `    ``return` `invcount;``}` `// Driver program``int` `main()``{``    ``int` `arr[] = {8, 4, 2, 1};``    ``int` `n = ``sizeof``(arr)/``sizeof``(``int``);``    ``cout << ``"Number of inversions are : "` `<< getInvCount(arr,n);``    ``return` `0;``}`

## Java

 `// Java program to count inversions``// using Binary Indexed Tree``import` `java.util.*;``class` `GFG{` `// Returns sum of arr[0..index].``// This function assumes that the``// array is preprocessed and partial``// sums of array elements are stored``// in BITree[].``static` `int` `getSum(``int` `BITree[],``                  ``int` `index)``{``  ``// Initialize result``  ``int` `sum = ``0``;` `  ``// Traverse ancestors of``  ``// BITree[index]``  ``while` `(index > ``0``)``  ``{``    ``// Add current element of``    ``// BITree to sum``    ``sum += BITree[index];` `    ``// Move index to parent node``    ``// in getSum View``    ``index -= index & (-index);``  ``}``  ``return` `sum;``}` `// Updates a node in Binary Index Tree``// (BITree) at given index in BITree. ``// The given value 'val' is added to``// BITree[i] and all of its ancestors``// in tree.``static` `void` `updateBIT(``int` `BITree[], ``int` `n,``                      ``int` `index, ``int` `val)``{``  ``// Traverse all ancestors``  ``// and add 'val'``  ``while` `(index <= n)``  ``{``    ``// Add 'val' to current``    ``// node of BI Tree``    ``BITree[index] += val;` `    ``// Update index to that of``    ``// parent in update View``    ``index += index & (-index);``  ``}``}` `// Converts an array to an array``// with values from 1 to n and``// relative order of smaller and``// greater elements remains same. ``// For example, {7, -90, 100, 1}``// is converted to {3, 1, 4 ,2 }``static` `void` `convert(``int` `arr[],``                    ``int` `n)``{``  ``// Create a copy of arrp[] in temp``  ``// and sort the temp array in``  ``// increasing order``  ``int` `[]temp = ``new` `int``[n];``  ` `  ``for` `(``int` `i = ``0``; i < n; i++)``    ``temp[i] = arr[i];``  ``Arrays.sort(temp);` `  ``// Traverse all array elements``  ``for` `(``int` `i = ``0``; i < n; i++)``  ``{``    ``// lower_bound() Returns pointer``    ``// to the first element greater``    ``// than or equal to arr[i]``    ``arr[i] =lower_bound(temp,``0``,``                        ``n, arr[i]) + ``1``;``  ``}``}` `static` `int` `lower_bound(``int``[] a, ``int` `low,``                       ``int` `high, ``int` `element)``{``  ``while``(low < high)``  ``{``    ``int` `middle = low +``                ``(high - low) / ``2``;``    ``if``(element > a[middle])``      ``low = middle + ``1``;``    ``else``      ``high = middle;``  ``}``  ``return` `low;``}` `// Returns inversion count``// arr[0..n-1]``static` `int` `getInvCount(``int` `arr[],``                       ``int` `n)``{``  ``// Initialize result``  ``int` `invcount = ``0``;` `  ``// Convert arr[] to an array``  ``// with values from 1 to n and``  ``// relative order of smaller``  ``// and greater elements remains``  ``// same.  For example, {7, -90,``  ``// 100, 1} is converted to``  ``//  {3, 1, 4 ,2 }``  ``convert(arr, n);` `  ``// Create a BIT with size equal``  ``// to maxElement+1 (Extra one is``  ``// used so that elements can be``  ``// directly be used as index)``  ``int` `[]BIT = ``new` `int``[n + ``1``];``  ` `  ``for` `(``int` `i = ``1``; i <= n; i++)``    ``BIT[i] = ``0``;` `  ``// Traverse all elements``  ``// from right.``  ``for` `(``int` `i = n - ``1``; i >= ``0``; i--)``  ``{``    ``// Get count of elements``    ``// smaller than arr[i]``    ``invcount += getSum(BIT,``                       ``arr[i] - ``1``);` `    ``// Add current element to BIT``    ``updateBIT(BIT, n, arr[i], ``1``);``  ``}` `  ``return` `invcount;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``  ``int` `arr[] = {``8``, ``4``, ``2``, ``1``};``  ``int` `n = arr.length;``  ``System.out.print(``"Number of inversions are : "` `+ ``                    ``getInvCount(arr, n));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program to count inversions using Binary Indexed Tree``from` `bisect ``import` `bisect_left as lower_bound` `# Returns sum of arr[0..index]. This function assumes``# that the array is preprocessed and partial sums of``# array elements are stored in BITree.``def` `getSum(BITree, index):` `    ``sum` `=` `0` `# Initialize result` `    ``# Traverse ancestors of BITree[index]``    ``while` `(index > ``0``):` `        ``# Add current element of BITree to sum``        ``sum` `+``=` `BITree[index]` `        ``# Move index to parent node in getSum View``        ``index ``-``=` `index & (``-``index)` `    ``return` `sum` `# Updates a node in Binary Index Tree (BITree) at given index``# in BITree. The given value 'val' is added to BITree[i] and``# all of its ancestors in tree.``def` `updateBIT(BITree, n, index, val):` `    ``# Traverse all ancestors and add 'val'``    ``while` `(index <``=` `n):` `        ``# Add 'val' to current node of BI Tree``        ``BITree[index] ``+``=` `val` `    ``# Update index to that of parent in update View``    ``index ``+``=` `index & (``-``index)` `# Converts an array to an array with values from 1 to n``# and relative order of smaller and greater elements remains``# same. For example, 7, -90, 100, 1 is converted to``# 3, 1, 4 ,2``def` `convert(arr, n):` `    ``# Create a copy of arrp in temp and sort the temp array``    ``# in increasing order``    ``temp ``=` `[``0``]``*``(n)``    ``for` `i ``in` `range``(n):``        ``temp[i] ``=` `arr[i]``    ``temp ``=` `sorted``(temp)` `    ``# Traverse all array elements``    ``for` `i ``in` `range``(n):` `        ``# lower_bound() Returns pointer to the first element``        ``# greater than or equal to arr[i]``        ``arr[i] ``=` `lower_bound(temp, arr[i]) ``+` `1` `# Returns inversion count arr[0..n-1]``def` `getInvCount(arr, n):` `    ``invcount ``=` `0` `# Initialize result` `    ``# Convert arr to an array with values from 1 to n and``    ``# relative order of smaller and greater elements remains``    ``# same. For example, 7, -90, 100, 1 is converted to``    ``# 3, 1, 4 ,2``    ``convert(arr, n)` `    ``# Create a BIT with size equal to maxElement+1 (Extra``    ``# one is used so that elements can be directly be``    ``# used as index)``    ``BIT ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``# Traverse all elements from right.``    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``):` `        ``# Get count of elements smaller than arr[i]``        ``invcount ``+``=` `getSum(BIT, arr[i] ``-` `1``)` `        ``# Add current element to BIT``        ``updateBIT(BIT, n, arr[i], ``1``)` `    ``return` `invcount` `# Driver program``if` `__name__ ``=``=` `'__main__'``:` `    ``arr ``=` `[``8``, ``4``, ``2``, ``1``]``    ``n ``=` `len``(arr)``    ``print``(``"Number of inversions are : "``,getInvCount(arr, n))` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program to count inversions``// using Binary Indexed Tree``using` `System;``class` `GFG{` `// Returns sum of arr[0..index].``// This function assumes that the``// array is preprocessed and partial``// sums of array elements are stored``// in BITree[].``static` `int` `getSum(``int` `[]BITree,``                  ``int` `index)``{``  ``// Initialize result``  ``int` `sum = 0;` `  ``// Traverse ancestors of``  ``// BITree[index]``  ``while` `(index > 0)``  ``{``    ``// Add current element of``    ``// BITree to sum``    ``sum += BITree[index];` `    ``// Move index to parent node``    ``// in getSum View``    ``index -= index & (-index);``  ``}``  ``return` `sum;``}` `// Updates a node in Binary Index Tree``// (BITree) at given index in BITree. ``// The given value 'val' is added to``// BITree[i] and all of its ancestors``// in tree.``static` `void` `updateBIT(``int` `[]BITree, ``int` `n,``                      ``int` `index, ``int` `val)``{``  ``// Traverse all ancestors``  ``// and add 'val'``  ``while` `(index <= n)``  ``{``    ``// Add 'val' to current``    ``// node of BI Tree``    ``BITree[index] += val;` `    ``// Update index to that of``    ``// parent in update View``    ``index += index & (-index);``  ``}``}` `// Converts an array to an array``// with values from 1 to n and``// relative order of smaller and``// greater elements remains same. ``// For example, {7, -90, 100, 1}``// is converted to {3, 1, 4 ,2 }``static` `void` `convert(``int` `[]arr,``                    ``int` `n)``{``  ``// Create a copy of arrp[] in temp``  ``// and sort the temp array in``  ``// increasing order``  ``int` `[]temp = ``new` `int``[n];``  ` `  ``for` `(``int` `i = 0; i < n; i++)``    ``temp[i] = arr[i];``  ``Array.Sort(temp);` `  ``// Traverse all array elements``  ``for` `(``int` `i = 0; i < n; i++)``  ``{``    ``// lower_bound() Returns pointer``    ``// to the first element greater``    ``// than or equal to arr[i]``    ``arr[i] =lower_bound(temp,0,``                        ``n, arr[i]) + 1;``  ``}``}` `static` `int` `lower_bound(``int``[] a, ``int` `low,``                       ``int` `high, ``int` `element)``{``  ``while``(low < high)``  ``{``    ``int` `middle = low +``                ``(high - low) / 2;``    ``if``(element > a[middle])``      ``low = middle + 1;``    ``else``      ``high = middle;``  ``}``  ``return` `low;``}` `// Returns inversion count``// arr[0..n-1]``static` `int` `getInvCount(``int` `[]arr,``                       ``int` `n)``{``  ``// Initialize result``  ``int` `invcount = 0;` `  ``// Convert []arr to an array``  ``// with values from 1 to n and``  ``// relative order of smaller``  ``// and greater elements remains``  ``// same.  For example, {7, -90,``  ``// 100, 1} is converted to``  ``//  {3, 1, 4 ,2 }``  ``convert(arr, n);` `  ``// Create a BIT with size equal``  ``// to maxElement+1 (Extra one is``  ``// used so that elements can be``  ``// directly be used as index)``  ``int` `[]BIT = ``new` `int``[n + 1];``  ` `  ``for` `(``int` `i = 1; i <= n; i++)``    ``BIT[i] = 0;` `  ``// Traverse all elements``  ``// from right.``  ``for` `(``int` `i = n - 1; i >= 0; i--)``  ``{``    ``// Get count of elements``    ``// smaller than arr[i]``    ``invcount += getSum(BIT,``                       ``arr[i] - 1);` `    ``// Add current element``    ``// to BIT``    ``updateBIT(BIT, n,``              ``arr[i], 1);``  ``}` `  ``return` `invcount;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``  ``int` `[]arr = {8, 4, 2, 1};``  ``int` `n = arr.Length;``  ``Console.Write(``"Number of inversions are : "` `+ ``                 ``getInvCount(arr, n));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
• Output:
`Number of inversions are : 6`
• Complexity Analysis:
• Time Complexity: The update function and getSum function runs for O(log(n)). The getSum function has to be run for every element in the array. So overall time complexity is O(nlog(n)).
• Auxiliary Space: O(n).
Space required for the BIT is an array of the size n.