# Count Inversions of size three in a given array

Given an array arr[] of size n. Three elements arr[i], arr[j] and arr[k] form an inversion of size 3 if a[i] > a[j] >a[k] and i < j < k. Find total number of inversions of size 3.

Example :

```Input:  {8, 4, 2, 1}
Output: 4
The four inversions are (8,4,2), (8,4,1), (4,2,1) and (8,2,1).

Input:  {9, 6, 4, 5, 8}
Output:  2
The two inversions are {9, 6, 4} and {9, 6, 5}
```

We have already discussed inversion count of size two by merge sort, Self Balancing BST and BIT.

Simple approach :- Loop for all possible value of i, j and k and check for the condition a[i] > a[j] > a[k] and i < j < k.

## C++

 `// A Simple C++ O(n^3)  program to count inversions of size 3 ` `#include ` `using` `namespace` `std; ` ` `  `// Returns counts of inversions of size three ` `int` `getInvCount(``int` `arr[],``int` `n) ` `{ ` `    ``int` `invcount = 0;  ``// Initialize result ` ` `  `    ``for` `(``int` `i=0; iarr[j]) ` `            ``{ ` `                ``for` `(``int` `k=j+1; karr[k]) ` `                        ``invcount++; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `invcount; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = {8, 4, 2, 1}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout << ``"Inversion Count : "` `<< getInvCount(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// A simple Java implementation  to count inversion of size 3 ` `class` `Inversion{ ` `     `  `    ``// returns count of inversion of size 3 ` `    ``int` `getInvCount(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `invcount = ``0``; ``// initialize result ` `         `  `        ``for``(``int` `i=``0` `; i< n-``2``; i++) ` `        ``{ ` `            ``for``(``int` `j=i+``1``; j arr[j]) ` `                ``{ ` `                    ``for``(``int` `k=j+``1``; k arr[k]) ` `                            ``invcount++; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `invcount; ` `    ``} ` ` `  `    ``// driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``Inversion inversion = ``new` `Inversion(); ` `        ``int` `arr[] = ``new` `int``[] {``8``, ``4``, ``2``, ``1``}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(``"Inversion count : "` `+  ` `                    ``inversion.getInvCount(arr, n)); ` `    ``} ` `} ` `// This code is contributed by Mayank Jaiswal `

## Python

 `# A simple python O(n^3) program ` `# to count inversions of size 3 ` ` `  `# Returns counts of inversions ` `# of size threee ` `def` `getInvCount(arr): ` `    ``n ``=` `len``(arr) ` `    ``invcount ``=` `0`  `#Initialize result     ` `    ``for` `i ``in` `range``(``0``,n``-``1``): ` `        ``for` `j ``in` `range``(i``+``1` `, n): ` `                ``if` `arr[i] > arr[j]: ` `                    ``for` `k ``in` `range``(j``+``1` `, n): ` `                        ``if` `arr[j] > arr[k]: ` `                            ``invcount ``+``=` `1` `    ``return` `invcount ` ` `  `# Driver program to test above function ` `arr ``=` `[``8` `, ``4``, ``2` `, ``1``] ` `print` `"Inversion Count : %d"` `%``(getInvCount(arr)) ` ` `  `# This code is contributed by Nikhil Kumar Singh(nickzuck_007) `

## C#

 `// A simple C# implementation to ` `// count inversion of size 3 ` `using` `System; ` `class` `GFG { ` `     `  `// returns count of inversion of size 3 ` `static` `int` `getInvCount(``int` `[]arr, ``int` `n) ` `    ``{ ` `         `  `        ``// initialize result ` `        ``int` `invcount = 0;  ` `         `  `         `  `        ``for``(``int` `i = 0 ; i < n - 2; i++) ` `        ``{ ` `            ``for``(``int` `j = i + 1; j < n - 1; j++) ` `            ``{ ` `                ``if``(arr[i] > arr[j]) ` `                ``{ ` `                    ``for``(``int` `k = j + 1; k < n; k++) ` `                    ``{ ` `                        ``if``(arr[j] > arr[k]) ` `                            ``invcount++; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `        ``return` `invcount; ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[]arr = ``new` `int``[] {8, 4, 2, 1}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(``"Inversion count : "` `+  ` `                           ``getInvCount(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` ``\$arr``[``\$j``]) ` `                ``\$small``++; ` ` `  `        ``// Count all greater elements  ` `        ``// on left of arr[i] ` `        ``\$great` `= 0; ` `        ``for``(``\$j` `= ``\$i` `- 1; ``\$j` `>= 0; ``\$j``--) ` `            ``if` `(``\$arr``[``\$i``] < ``\$arr``[``\$j``]) ` `                ``\$great``++; ` ` `  `        ``// Update inversion count by  ` `        ``// adding all inversions ` `        ``// that have arr[i] as  ` `        ``// middle of three elements ` `        ``\$invcount` `+= ``\$great` `* ``\$small``; ` `    ``} ` ` `  `    ``return` `\$invcount``; ` `} ` ` `  `    ``// Driver Code ` `    ``\$arr` `= ``array``(8, 4, 2, 1); ` `    ``\$n` `= sizeof(``\$arr``); ` `    ``echo` `"Inversion Count : "` `        ``, getInvCount(``\$arr``, ``\$n``); ` ` `  `// This code is contributed m_kit ` `?> `

Output:

`Inversion Count : 4 `

Time complexity of this approach is : O(n^3)

Better Approach :
We can reduce the complexity if we consider every element arr[i] as middle element of inversion, find all the numbers greater than a[i] whose index is less than i, find all the numbers which are smaller than a[i] and index is more than i. We multiply the number of elements greater than a[i] to the number of elements smaller than a[i] and add it to the result.
Below is the implementation of the idea.

## C++

 `// A O(n^2) C++  program to count inversions of size 3 ` `#include ` `using` `namespace` `std; ` ` `  `// Returns count of inversions of size 3 ` `int` `getInvCount(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `invcount = 0;  ``// Initialize result ` ` `  `    ``for` `(``int` `i=1; i arr[j]) ` `                ``small++; ` ` `  `        ``// Count all greater elements on left of arr[i] ` `        ``int` `great = 0; ` `        ``for` `(``int` `j=i-1; j>=0; j--) ` `            ``if` `(arr[i] < arr[j]) ` `                ``great++; ` ` `  `        ``// Update inversion count by adding all inversions ` `        ``// that have arr[i] as middle of three elements ` `        ``invcount += great*small; ` `    ``} ` ` `  `    ``return` `invcount; ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = {8, 4, 2, 1}; ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]); ` `    ``cout << ``"Inversion Count : "` `<< getInvCount(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// A O(n^2) Java  program to count inversions of size 3 ` ` `  `class` `Inversion { ` `     `  `    ``// returns count of inversion of size 3 ` `    ``int` `getInvCount(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``int` `invcount = ``0``; ``// initialize result ` `         `  `        ``for` `(``int` `i=``0` `; i< n-``1``; i++) ` `        ``{ ` `            ``// count all smaller elements on right of arr[i] ` `            ``int` `small=``0``; ` `            ``for` `(``int` `j=i+``1``; j arr[j]) ` `                    ``small++; ` `                     `  `            ``// count all greater elements on left of arr[i] ` `            ``int` `great = ``0``; ` `            ``for` `(``int` `j=i-``1``; j>=``0``; j--) ` `                        ``if` `(arr[i] < arr[j]) ` `                            ``great++; ` `                     `  `            ``// update inversion count by adding all inversions ` `            ``// that have arr[i] as middle of three elements ` `            ``invcount += great*small; ` `        ``} ` `        ``return` `invcount; ` `    ``} ` `    ``// driver program to test above function ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``Inversion inversion = ``new` `Inversion(); ` `        ``int` `arr[] = ``new` `int``[] {``8``, ``4``, ``2``, ``1``}; ` `        ``int` `n = arr.length; ` `        ``System.out.print(``"Inversion count : "` `+ ` `                       ``inversion.getInvCount(arr, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by Mayank Jaiswal `

## Python3

 `# A O(n^2) Python3 program to ` `#  count inversions of size 3 ` ` `  `# Returns count of inversions ` `# of size 3 ` `def` `getInvCount(arr, n): ` ` `  `    ``# Initialize result ` `    ``invcount ``=` `0`    ` `  `    ``for` `i ``in` `range``(``1``,n``-``1``): ` `     `  `        ``# Count all smaller elements ` `        ``# on right of arr[i] ` `        ``small ``=` `0` `        ``for` `j ``in` `range``(i``+``1` `,n): ` `            ``if` `(arr[i] > arr[j]): ` `                ``small``+``=``1` ` `  `        ``# Count all greater elements ` `        ``# on left of arr[i] ` `        ``great ``=` `0``; ` `        ``for` `j ``in` `range``(i``-``1``,``-``1``,``-``1``): ` `            ``if` `(arr[i] < arr[j]): ` `                ``great``+``=``1` ` `  `        ``# Update inversion count by ` `        ``# adding all inversions that ` `        ``# have arr[i] as middle of ` `        ``# three elements ` `        ``invcount ``+``=` `great ``*` `small ` `     `  `    ``return` `invcount ` ` `  `# Driver program to test above function ` `arr ``=` `[``8``, ``4``, ``2``, ``1``] ` `n ``=` `len``(arr) ` `print``(``"Inversion Count :"``,getInvCount(arr, n)) ` ` `  `# This code is Contributed by Smitha Dinesh Semwal `

## C#

 `// A O(n^2) Java program to count inversions ` `// of size 3 ` `using` `System; ` ` `  `public` `class` `Inversion { ` `     `  `    ``// returns count of inversion of size 3 ` `    ``static` `int` `getInvCount(``int` `[]arr, ``int` `n) ` `    ``{ ` `        ``int` `invcount = 0; ``// initialize result ` `         `  `        ``for` `(``int` `i = 0 ; i < n-1; i++) ` `        ``{ ` `             `  `            ``// count all smaller elements on  ` `            ``// right of arr[i] ` `            ``int` `small = 0; ` `            ``for` `(``int` `j = i+1; j < n; j++) ` `                ``if` `(arr[i] > arr[j]) ` `                    ``small++; ` `                     `  `            ``// count all greater elements on ` `            ``// left of arr[i] ` `            ``int` `great = 0; ` `            ``for` `(``int` `j = i-1; j >= 0; j--) ` `                        ``if` `(arr[i] < arr[j]) ` `                            ``great++; ` `                     `  `            ``// update inversion count by  ` `            ``// adding all inversions that  ` `            ``// have arr[i] as middle of ` `            ``// three elements ` `            ``invcount += great * small; ` `        ``} ` `         `  `        ``return` `invcount; ` `    ``} ` `     `  `    ``// driver program to test above function ` `    ``public` `static` `void` `Main() ` `    ``{ ` `         `  `        ``int` `[]arr = ``new` `int``[] {8, 4, 2, 1}; ` `        ``int` `n = arr.Length; ` `        ``Console.WriteLine(``"Inversion count : "` `                       ``+ getInvCount(arr, n)); ` `    ``} ` `} ` ` `  `// This code has been contributed by anuj_67. `

## PHP

 ` ``\$arr``[``\$j``]) ` `                ``\$small``++; ` ` `  `        ``// Count all greater elements ` `        ``// on left of arr[i] ` `        ``\$great` `= 0; ` `        ``for` `(``\$j` `= ``\$i` `- 1; ``\$j` `>= 0; ``\$j``--) ` `            ``if` `(``\$arr``[``\$i``] < ``\$arr``[``\$j``]) ` `                ``\$great``++; ` ` `  `        ``// Update inversion count by  ` `        ``// adding all inversions that ` `        ``// have arr[i] as middle of  ` `        ``// three elements ` `        ``\$invcount` `+= ``\$great` `* ``\$small``; ` `    ``} ` ` `  `    ``return` `\$invcount``; ` `} ` ` `  `// Driver Code ` `\$arr` `= ``array` `(8, 4, 2, 1); ` `\$n` `= sizeof(``\$arr``); ` `echo` `"Inversion Count : "` `,  ` `      ``getInvCount(``\$arr``, ``\$n``); ` `     `  `// This code is contributed by m_kit ` `?> `

Output :

`Inversion Count : 4 `

Time Complexity of this approach : O(n^2)

Binary Indexed Tree Approach :
Like inversions of size 2, we can use Binary indexed tree to find inversions of size 3. It is strongly recommended to refer below article first.

Count inversions of size two Using BIT

The idea is similar to above method. We count the number of greater elements and smaller elements for all the elements and then multiply greater[] to smaller[] and add it to the result.

Solution :

1. To find out the number of smaller elements for an index we iterate from n-1 to 0. For every element a[i] we calculate the getSum() function for (a[i]-1) which gives the number of elements till a[i]-1.
2. To find out the number of greater elements for an index we iterate from 0 to n-1. For every element a[i] we calculate the sum of numbers till a[i] (sum smaller or equal to a[i]) by getSum() and subtract it from i (as i is the total number of element till that point) so that we can get number of elements greater than a[i].
3. Like we did for inversions of size 2, here also we convert the input array to an array with values from 1 to n so that the size of BIT remains O(n), and getSum() and update() functions take O(Log n) time. For example, we convert arr[] = {7, -90, 100, 1} to arr[] = {3, 1, 4 ,2 }.

Below is the implementation of above idea.

## C

 `// C++ program to count inversions of size three 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--) ` `    ``{ ` `        ``smaller1[i] = getSum(BIT, arr[i]-1); ` `        ``updateBIT(BIT, n, arr[i], 1); ` `    ``} ` ` `  `    ``// Reset BIT ` `    ``for` `(``int` `i=1; i<=n; i++) ` `        ``BIT[i] = 0; ` ` `  `    ``// Count greater elements ` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to count inversions of size three using  ` `// Binary Indexed Tree ` `import` `java.util.Arrays; ` ` `  `class` `BinaryTree { ` ` `  `   ``// 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[]= ``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] = Arrays.binarySearch(temp, arr[i])  + ``1``; ` `        ``} ` `    ``} ` ` `  `    ``// Returns count of inversions of size three ` `    ``int` `getInvCount(``int` `arr[], ``int` `n) { ` `        `  `        ``// 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 and initialize smaller and greater arrays ` `        ``int` `greater1[]= ``new` `int``[n]; ` `        ``int` `smaller1[]= ``new` `int``[n]; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``greater1[i] = smaller1[i] = ``0``; ` `        ``} ` ` `  `        ``// Create and initialize an array to store Binary ` `        ``// Indexed Tree ` `        ``int` `BIT[]= ``new` `int``[n+``1``]; ` `        ``for` `(``int` `i = ``1``; i <= n; i++) { ` `            ``BIT[i] = ``0``; ` `        ``} ` ` `  `        ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) { ` `            ``smaller1[i] = getSum(BIT, arr[i] - ``1``); ` `            ``updateBIT(BIT, n, arr[i], ``1``); ` `        ``} ` ` `  `        ``// Reset BIT ` `        ``for` `(``int` `i = ``1``; i <= n; i++) { ` `            ``BIT[i] = ``0``; ` `        ``} ` ` `  `        ``// Count greater elements ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``greater1[i] = i - getSum(BIT, arr[i]); ` `            ``updateBIT(BIT, n, arr[i], ``1``); ` `        ``} ` ` `  `        ``// Compute Inversion count using smaller[] and ` `        ``// greater[].  ` `        ``int` `invcount = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++) { ` `            ``invcount += smaller1[i] * greater1[i]; ` `        ``} ` ` `  `        ``return` `invcount; ` `    ``} ` ` `  `    ``// Driver program to test above function ` `    ``public` `static` `void` `main(String args[]) { ` `        ``BinaryTree tree = ``new` `BinaryTree(); ` `        ``int``[] arr = ``new` `int``[]{``8``, ``4``, ``2``, ``1``}; ` `        ``int` `n = arr.length; ` `        ``System.out.print( ``"Inversion Count : "` `+  ` `                           ``tree.getInvCount(arr, n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Mayank Jaiswal `

## C#

 `// C# program to count inversions ` `// of size three 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[]. ` `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 = ``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] = Array.BinarySearch(temp,  ` `                                    ``arr[i]) + 1; ` `    ``} ` `} ` ` `  `// Returns count of inversions  ` `// of size three ` `int` `getInvCount(``int``[] arr, ``int` `n)  ` `{ ` `     `  `    ``// 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 and initialize  ` `    ``// smaller and greater arrays ` `    ``int``[] greater1 = ``new` `int``[n]; ` `    ``int``[] smaller1 = ``new` `int``[n]; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``greater1[i] = smaller1[i] = 0; ` `    ``} ` ` `  `    ``// Create and initialize an  ` `    ``// array to store Binary Indexed Tree ` `    ``int``[] BIT = ``new` `int``[n + 1]; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `    ``{ ` `        ``BIT[i] = 0; ` `    ``} ` ` `  `    ``for` `(``int` `i = n - 1; i >= 0; i--)  ` `    ``{ ` `        ``smaller1[i] = getSum(BIT,  ` `                             ``arr[i] - 1); ` `        ``updateBIT(BIT, n, arr[i], 1); ` `    ``} ` ` `  `    ``// Reset BIT ` `    ``for` `(``int` `i = 1; i <= n; i++)  ` `    ``{ ` `        ``BIT[i] = 0; ` `    ``} ` ` `  `    ``// Count greater elements ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``greater1[i] = i - getSum(BIT, ` `                                 ``arr[i]); ` `        ``updateBIT(BIT, n, arr[i], 1); ` `    ``} ` ` `  `    ``// Compute Inversion count using  ` `    ``// smaller[] and greater[].  ` `    ``int` `invcount = 0; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``invcount += smaller1[i] *  ` `                    ``greater1[i]; ` `    ``} ` ` `  `    ``return` `invcount; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main() ` `{ ` `    ``GFG tree = ``new` `GFG(); ` `    ``int``[] arr = ``new` `int``[]{8, 4, 2, 1}; ` `    ``int` `n = arr.Length; ` `    ``Console.Write( ``"Inversion Count : "` `+  ` `                    ``tree.getInvCount(arr, n)); ` `} ` `} ` ` `  `// This code is contributed  ` `// by ChitraNayal `

## Python3

 `# Python3 program to count inversions of  ` `# size three 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)  ` ` `  `# 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) ` `    ``j ``=` `1` `     `  `    ``# Traverse all array elements  ` `    ``for` `i ``in` `temp:  ` ` `  `        ``# lower_bound() Returns poer to  ` `        ``# the first element greater than ` `        ``# or equal to arr[i]  ` `        ``arr[arr.index(i)] ``=` `j ` `        ``j ``+``=` `1` ` `  `# Returns count of inversions of size three  ` `def` `getInvCount( arr, n): ` ` `  `    ``# 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 and initialize smaller and  ` `    ``# greater arrays  ` `    ``greater1 ``=` `[``0``] ``*` `n ` `    ``smaller1 ``=` `[``0``] ``*` `n  ` `    ``for` `i ``in` `range``(n): ` `        ``greater1[i], smaller1[i] ``=` `0``, ``0` ` `  `    ``# Create and initialize an array to  ` `    ``# store Binary Indexed Tree  ` `    ``BIT ``=` `[``0``] ``*` `(n ``+` `1``)  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``BIT[i] ``=` `0` `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``): ` ` `  `        ``smaller1[i] ``=` `getSum(BIT, arr[i] ``-` `1``)  ` `        ``updateBIT(BIT, n, arr[i], ``1``)  ` ` `  `    ``# Reset BIT  ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):  ` `        ``BIT[i] ``=` `0` ` `  `    ``# Count greater elements  ` `    ``for` `i ``in` `range``(n):  ` ` `  `        ``greater1[i] ``=` `i ``-` `getSum(BIT, arr[i])  ` `        ``updateBIT(BIT, n, arr[i], ``1``)  ` ` `  `    ``# Compute Inversion count using smaller[]  ` `    ``# and greater[].  ` `    ``invcount ``=` `0` `    ``for` `i ``in` `range``(n):  ` `        ``invcount ``+``=` `smaller1[i] ``*` `greater1[i]  ` ` `  `    ``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) `

Output:

`Inversion Count : 4 `

Time Complexity : O(n log n)
Auxiliary Space : O(n)

We can also use Self-Balancing Binary Search Tree to count greater elements on left and smaller on right. Time complexity of this method would also be O(n Log n), But BIT based method is easy to implement.

code