Let us consider the following problem to understand Binary Indexed Tree.

We have an array arr[0 . . . n-1]. We should be able to

**1** Find the sum of first i elements.

**2 ** Change value of a specified element of the array arr[i] = x where 0 <= i <= n-1.

A **simple solution** is to run a loop from 0 to i-1 and calculate sum of elements. To update a value, simply do arr[i] = x. The first operation takes O(n) time and second operation takes O(1) time. Another simple solution is to create another array and store sum from start to i at the i’th index in this array. Sum of a given range can now be calculated in O(1) time, but update operation takes O(n) time now. This works well if the number of query operations are large and very few updates.

**Can we perform both the operations in O(log n) time once given the array? **

One Efficient Solution is to use Segment Tree that does both operations in O(Logn) time.

*Using Binary Indexed Tree, we can do both tasks in O(Logn) time. The advantages of Binary Indexed Tree over Segment are, requires less space and very easy to implement.*.

**Representation**

Binary Indexed Tree is represented as an array. Let the array be BITree[]. Each node of Binary Indexed Tree stores sum of some elements of given array. Size of Binary Indexed Tree is equal to n where n is size of input array. In the below code, we have used size as n+1 for ease of implementation.

**Construction**

We construct the Binary Indexed Tree by first initializing all values in BITree[] as 0. Then we call update() operation for all indexes to store actual sums, update is discussed below.

**Operations**

// Returns sum of arr[0..index] using BITree[0..n]. It assumes that // BITree[] is constructed for given array arr[0..n-1] 1) Initialize sum as 0 and index as index+1. 2) Do following while index is greater than 0. ...a) Add BITree[index] to sum ...b) Go to parent of BITree[index]. Parent can be obtained by removing the last set bit from index, i.e., index = index - (index & (-index)) 3) Return sum.getSum(index): Returns sum of arr[0..index]

The above diagram demonstrates working of getSum(). Following are some important observations.

Node at index 0 is a dummy node.

A node at index y is parent of a node at index x, iff y can be obtained by removing last set bit from binary representation of x.

A child x of a node y stores sum of elements from of y(exclusive y) and of x(inclusive x).

// Note that arr[] is not changed here. It changes // only BI Tree for the already made change in arr[]. 1) Initialize index as index+1. 2) Do following while index is smaller than or equal to n. ...a) Add value to BITree[index] ...b) Go to parent of BITree[index]. Parent can be obtained by removing the last set bit from index, i.e., index = index + (index & (-index))update(index, val): Updates BIT for operation arr[index] += val

The update process needs to make sure that all BITree nodes that have arr[i] as part of the section they cover must be updated. We get all such nodes of BITree by repeatedly adding the decimal number corresponding to the last set bit.

**How does Binary Indexed Tree work?**

The idea is based on the fact that all positive integers can be represented as sum of powers of 2. For example 19 can be represented as 16 + 2 + 1. Every node of BI Tree stores sum of n elements where n is a power of 2. For example, in the above first diagram for getSum(), sum of first 12 elements can be obtained by sum of last 4 elements (from 9 to 12) plus sum of 8 elements (from 1 to 8). The number of set bits in binary representation of a number n is O(Logn). Therefore, we traverse at-most O(Logn) nodes in both getSum() and update() operations. Time complexity of construction is O(nLogn) as it calls update() for all n elements.

**Implementation:**

Following are the implementations of Binary Indexed Tree.

## C++

// C++ code to demonstrate operations of Binary Index Tree #include <iostream> using namespace std; /* n --> No. of elements present in input array. BITree[0..n] --> Array that represents Binary Indexed Tree. arr[0..n-1] --> Input array for whic prefix sum is evaluated. */ // 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; // Iniialize result // index in BITree[] is 1 more than the index in arr[] index = index + 1; // 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) { // index in BITree[] is 1 more than the index in arr[] index = index + 1; // 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); } } // Constructs and returns a Binary Indexed Tree for given // array of size n. int *constructBITree(int arr[], int n) { // Create and initialize BITree[] as 0 int *BITree = new int[n+1]; for (int i=1; i<=n; i++) BITree[i] = 0; // Store the actual values in BITree[] using update() for (int i=0; i<n; i++) updateBIT(BITree, n, i, arr[i]); // Uncomment below lines to see contents of BITree[] //for (int i=1; i<=n; i++) // cout << BITree[i] << " "; return BITree; } // Driver program to test above functions int main() { int freq[] = {2, 1, 1, 3, 2, 3, 4, 5, 6, 7, 8, 9}; int n = sizeof(freq)/sizeof(freq[0]); int *BITree = constructBITree(freq, n); cout << "Sum of elements in arr[0..5] is " << getSum(BITree, 5); // Let use test the update operation freq[3] += 6; updateBIT(BITree, n, 3, 6); //Update BIT for above change in arr[] cout << "\nSum of elements in arr[0..5] after update is " << getSum(BITree, 5); return 0; }

## Java

// Java program to demonstrate lazy // propagation in segment tree import java.util.*; import java.lang.*; import java.io.*; class BinaryIndexedTree { // Max tree size final static int MAX = 1000; static int BITree[] = new int[MAX]; /* n --> No. of elements present in input array. BITree[0..n] --> Array that represents Binary Indexed Tree. arr[0..n-1] --> Input array for whic prefix sum is evaluated. */ // 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 index) { int sum = 0; // Iniialize result // index in BITree[] is 1 more than // the index in arr[] index = index + 1; // 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. public static void updateBIT(int n, int index, int val) { // index in BITree[] is 1 more than // the index in arr[] index = index + 1; // Traverse all ancestors and add 'val' while(index <= n) { // Add 'val' to current node of BIT Tree BITree[index] += val; // Update index to that of parent // in update View index += index & (-index); } } /* Function to construct fenwick tree from given array.*/ void constructBITree(int arr[], int n) { // Initialize BITree[] as 0 for(int i=1; i<=n; i++) BITree[i] = 0; // Store the actual values in BITree[] // using update() for(int i = 0; i < n; i++) updateBIT(n, i, arr[i]); } // Main function public static void main(String args[]) { int freq[] = {2, 1, 1, 3, 2, 3, 4, 5, 6, 7, 8, 9}; int n = freq.length; BinaryIndexedTree tree = new BinaryIndexedTree(); // Build fenwick tree from given array tree.constructBITree(freq, n); System.out.println("Sum of elements in arr[0..5]"+ " is = "+ tree.getSum(5)); // Let use test the update operation freq[3] += 6; // Update BIT for above change in arr[] updateBIT(n, 3, 6); // Find sum after the value is updated System.out.println("Sum of elements in arr[0..5]"+ " after update is = " + tree.getSum(5)); } } // This code is contributed by Ranjan Binwani

## Python

# Python implementation of 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(BITTree,i): s = 0 #initialize result # index in BITree[] is 1 more than the index in arr[] i = i+1 # Traverse ancestors of BITree[index] while i > 0: # Add current element of BITree to sum s += BITTree[i] # Move index to parent node in getSum View i -= i & (-i) return s # 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(BITTree , n , i ,v): # index in BITree[] is 1 more than the index in arr[] i += 1 # Traverse all ancestors and add 'val' while i <= n: # Add 'val' to current node of BI Tree BITTree[i] += v # Update index to that of parent in update View i += i & (-i) # Constructs and returns a Binary Indexed Tree for given # array of size n. def construct(arr, n): # Create and initialize BITree[] as 0 BITTree = [0]*(n+1) # Store the actual values in BITree[] using update() for i in range(n): updatebit(BITTree, n, i, arr[i]) # Uncomment below lines to see contents of BITree[] #for i in range(1,n+1): # print BITTree[i], return BITTree # Driver code to test above methods freq = [2, 1, 1, 3, 2, 3, 4, 5, 6, 7, 8, 9] BITTree = construct(freq,len(freq)) print("Sum of elements in arr[0..5] is " + str(getsum(BITTree,5))) freq[3] += 6 updatebit(BITTree, len(freq), 3, 6) print("Sum of elements in arr[0..5] is " + str(getsum(BITTree,5))) # This code is contributed by Raju Varshney

Output:

Sum of elements in arr[0..5] is 12 Sum of elements in arr[0..5] after update is 18

**Can we extend the Binary Indexed Tree for range Sum in Logn time?**

This is simple to answer. The rangeSum(l, r) can be obtained as getSum(r) – getSum(l-1).

**Applications:**

Used to implement the arithmetic coding algorithm. Development of operations it supports were primarily motivated by use in that case. See this for more details.

**Example Problems:**

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

Two Dimensional Binary Indexed Tree or Fenwick Tree

Counting Triangles in a Rectangular space using BIT

**References:**

http://en.wikipedia.org/wiki/Fenwick_tree

http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binaryIndexedTrees

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

**Practice Tags :**