Binary Indexed Tree : Range Update and Range Queries

Given an array arr[0..n-1]. The following operations need to be performed.

  1. update(l, r, val) : Add ‘val’ to all the elements in the array from [l, r].
  2. getRangeSum(l, r) : Find sum of all elements in array from [l, r].

Initially all the elements in the array are 0. Queries can be in any oder, i.e., there can be many updates before range sum.


Input : n = 5   // {0, 0, 0, 0, 0}
Queries: update : l = 0, r = 4, val = 2
         update : l = 3, r = 4, val = 3 
         getRangeSum : l = 2, r = 4

Output: Sum of elements of range [2, 4] is 12

Explanation : Array after first update becomes
              {2, 2, 2, 2, 2}
              Array after second update becomes
              {2, 2, 2, 5, 5}

In the previous post, we discussed range update and point query solutions using BIT.
rangeUpdate(l, r, val) : We add ‘val’ to element at index ‘l’. We subtract ‘val’ from element at index ‘r+1’.
getElement(index) [or getSum()]: We return sum of elements from 0 to index which can be quickly obtained using BIT.

We can compute rangeSum() using getSum() queries.
rangeSum(l, r) = getSum(r) – getSum(l-1)

A Simple Solution is to use solutions discussed in previous post. Range update query is same. Range sum query can be achieved by doing get query for all elements in range.

An Efficient Solution is to make sure that both queries can be done in O(Log n) time. We get range sum using prefix sums. How to make sure that update is done in a way so that prefix sum can be done quickly? Consider a situation where prefix sum [0, k] (where 0 <= k < n) is needed after range update on range [l, r]. Three cases arises as k can possibly lie in 3 regions.

Case 1: 0 < k < l
The update query won’t affect sum query.

Case 2: l <= k <= r
Consider an example:

Add 2 to range [2, 4], the resultant array would be:
0 0 2 2 2
If k = 3
Sum from [0, k] = 4

How to get this result?
Simply add the val from lth index to kth index. Sum is incremented by “val*(k) – val*(l-1)” after update query.

Case 3: k > r
For this case, we need to add “val” from lth index to rth index. Sum is incremented by “val*r – val*(l-1)” due to update query.

Observations :
Case 1: is simple as sum would remain same as it was before update.

Case 2: Sum was incremented by val*k – val*(l-1). We can find “val”, it is similar to finding the ith element in range update and point query article. So we maintain one BIT for Range Update and Point Queries, this BIT will be helpful in finding the value at kth index. Now val * k is computed, how to handle extra term val*(l-1)?
In order to handle this extra term, we maintain another BIT (BIT2). Update val * (l-1) at lth index, so when getSum query is performed on BIT2 will give result as val*(l-1).

Case 3 : The sum in case 3 was incremented by “val*r – val *(l-1)”, the value of this term can be obtained using BIT2. Instead of adding, we subtract “val*(l-1) – val*r” as we can get this value from BIT2 by adding val*(l-1) as we did in case 2 and subtracting val*r in every update operation.

Update Query 
Update(BITree1, l, val)
Update(BITree1, r+1, -val)
UpdateBIT2(BITree2, l, val*(l-1))
UpdateBIT2(BITree2, r+1, -val*r)

Range Sum 
getSum(BITTree1, k) *k) - getSum(BITTree2, k)

C++ Implementation of above idea

// C++ program to demonstrate Range Update
// and Range Queries using BIT
#include <iostream>
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
    // 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);
// Returns the sum of array from [0, x]
int sum(int x, int BITTree1[], int BITTree2[])
    return (getSum(BITTree1, x) * x) - getSum(BITTree2, x);
void updateRange(int BITTree1[], int BITTree2[], int n,
                 int val, int l, int r)
    // Update Both the Binary Index Trees
    // As discussed in the article
    // Update BIT1
    // Update BIT2
int rangeSum(int l, int r, int BITTree1[], int BITTree2[])
    // Find sum from [0,r] then subtract sum
    // from [0,l-1] in order to find sum from
    // [l,r]
    return sum(r, BITTree1, BITTree2) -
           sum(l-1, BITTree1, BITTree2);
int *constructBITree(int n)
    // Create and initialize BITree[] as 0
    int *BITree = new int[n+1];
    for (int i=1; i<=n; i++)
        BITree[i] = 0;
    return BITree;
// Driver Program to test above function
int main()
    int n = 5;
    // Construct two BIT
    int *BITTree1, *BITTree2;
    // BIT1 to get element at any index
    // in the array
    BITTree1 = constructBITree(n);
    // BIT 2 maintains the extra term
    // which needs to be subtracted
    BITTree2 = constructBITree(n);
    // Add 5 to all the elements from [0,4]
    int l = 0 , r = 4 , val = 5;
    // Add 2 to all the elements from [2,4]
    l = 2 , r = 4 , val = 10;
    // Find sum of all the elements from
    // [1,4]
    l = 1 , r = 4;
    cout << "Sum of elements from [" << l
         << "," << r << "] is ";
    cout << rangeSum(l,r,BITTree1,BITTree2) << " ";
    return 0;


Sum of elements from [1,4] is 50

Time Complexity : O(q*log(n)) where q is number of queries.

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

This article is attributed to GeeksforGeeks.org

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