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

27 Apr, 2022
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. 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: 6Explanation: Given array has six inversions: (8,4), (4,2), (8,2), (8,1), (4,1), (2,1).     
Input: arr[] = {3, 1, 2}Output: 2Explanation: Given array has two inversions: (3,1), (3,2).
We have already discussed the below methods to solve inversion count:  
Naive and Modified Merge SortUsing AVL Tree
Naive and Modified Merge Sort
Using AVL Tree
We recommend you to refer Binary Indexed Tree (BIT) before further reading this post. Solution using BIT of size Θ(maxElement):  
Approach: Traverse through the array and for every index find the number of smaller elements on the 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: BIT basically supports two operations for an array arr[] of size n: Sum of elements till arr[i] in O(log n) time.Update an array element in O(log n) time.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. The sum operation where parent of index x is “x – (x & -x)”.The update operation where parent of index x is “x + (x & -x)”.
BIT basically supports two operations for an array arr[] of size n: Sum of elements till arr[i] in O(log n) time.Update an array element in O(log n) time.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. The sum operation where parent of index x is “x – (x & -x)”.The update operation where parent of index x is “x + (x & -x)”.
BIT basically supports two operations for an array arr[] of size n: Sum of elements till arr[i] in O(log n) time.Update an array element in O(log n) time.
Sum of elements till arr[i] in O(log n) time.Update an array element in O(log n) time.
Sum of elements till arr[i] in O(log n) time.
Update an array element in O(log n) time.
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. The sum operation where parent of index x is “x – (x & -x)”.The update operation where parent of index x is “x + (x & -x)”.
The sum operation where parent of index x is “x – (x & -x)”.The update operation where parent of index x is “x + (x & -x)”.
The sum operation where parent of index x is “x – (x & -x)”.
The update operation where parent of index x is “x + (x & -x)”.
Algorithm: Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.Traverse the array from end to start.For every index check how many numbers less than the current element are present in BIT and add it to the resultTo get the count of smaller elements, getSum() of BIT is used.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.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).
Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.Traverse the array from end to start.For every index check how many numbers less than the current element are present in BIT and add it to the resultTo get the count of smaller elements, getSum() of BIT is used.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.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).
Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.
Traverse the array from end to start.
For every index check how many numbers less than the current element are present in BIT and add it to the result
To get the count of smaller elements, getSum() of BIT is used.
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.
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).
Below is the implementation of the above approach:
C++
Java
Python3
C#
PHP
Javascript
// C++ program to count inversions using Binary Indexed Tree#include<bits/stdc++.h>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<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[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 programint main(){    int arr[] = {8, 4, 2, 1};    int n = sizeof(arr)/sizeof(int);    cout << "Number of inversions are : " << getInvCount(arr,n);    return 0;}
// 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 codepublic 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 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 threedef 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(n - 1, -1, -1):         invcount += getSum(BIT, arr[i] - 1)        updateBIT(BIT, maxElement, arr[i], 1)    return invcount     # Driver codeif __name__ =="__main__":    arr = [8, 4, 2, 1]    n = 4    print("Inversion Count : ",           getInvCount(arr, n))     # This code is contributed by# Shubham Singh(SHUBHAMSINGH10)
// C# program to count inversions// using Binary Indexed Treeusing 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 codestatic 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// PHP 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[].function 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 $sum;} // Updates a node in Binary Index// Tree (BITree) at given index// in BITree. The given value 'val'// is added to BITree[i] and// all of its ancestors in tree.function updateBIT(&$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?>
<script>// javascript 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.function getSum(BITree , index){    var 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.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){    var invcount = 0; // Initialize result     // Find maximum element in arr    var 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)    var BIT = Array.from({length: maxElement + 1}, (_, i) => 0);    for (i = 1; i <= maxElement; i++)        BIT[i] = 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 code    var arr = [8, 4, 2, 1];    var n = arr.length;    document.write("Number of inversions are : " +                                getInvCount(arr,n)); // This code is contributed by Amit Katiyar</script>
Number of inversions are : 6
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 Θ(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: 
Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.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}Traverse the array from end to start.For every index check how many numbers less than the current element are present in BIT and add it to the result
Create a BIT, to find the count of the smaller elements in the BIT for a given number and also a variable result = 0.
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}
Traverse the array from end to start.
For every index check how many numbers less than the current element are present in BIT and add it to the result
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ program to count inversions using Binary Indexed Tree#include<bits/stdc++.h>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<n; i++)        temp[i] = arr[i];    sort(temp, temp+n);     // 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, temp+n, arr[i]) - temp + 1;    }} // Returns inversion count arr[0..n-1]int getInvCount(int arr[], int n){    int 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)    int BIT[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 programint main(){    int arr[] = {8, 4, 2, 1};    int n = sizeof(arr)/sizeof(int);    cout << "Number of inversions are : " << getInvCount(arr,n);    return 0;}
// Java program to count inversions// using Binary Indexed Treeimport 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 codepublic 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 program to count inversions using Binary Indexed Treefrom 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 ,2def 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 programif __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# program to count inversions// using Binary Indexed Treeusing 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 codepublic 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
<script>// Javascript 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[].   function getSum(BITree,index){         // Initialize result  let 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.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);  }} // 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 }function convert(arr, n){     // Create a copy of arrp[] in temp  // and sort the temp array in  // increasing order  let temp = new Array(n);      for (let i = 0; i < n; i++)    temp[i] = arr[i];  temp.sort(function(a, b){return a - b;});    // Traverse all array elements  for (let 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;  }} function lower_bound(a,low,high,element){    while(low < high)  {    let middle = low +                Math.floor((high - low) / 2);    if(element > a[middle])      low = middle + 1;    else      high = middle;  }  return low;} // Returns inversion count// arr[0..n-1]function getInvCount(arr,n){    // Initialize result  let 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)  let BIT = new Array(n + 1);      for (let i = 1; i <= n; i++)    BIT[i] = 0;    // Traverse all elements  // from right.  for (let 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 codelet arr=[8, 4, 2, 1];let n = arr.length;document.write("Number of inversions are : " +                    getInvCount(arr, n));     // This code is contributed by unknown2108</script>
Number of inversions are : 6
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.
This article is contributed by Abhiraj Smit. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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Binary Indexed Tree
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