# Binary Indexed Tree : Range Update and Range Queries

• Difficulty Level : Hard
• Last Updated : 16 Jun, 2021

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 order, i.e., there can be many updates before range sum.
Example:

```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)```

Implementation of above idea

## C++

 `// C++ program to demonstrate Range Update``// and Range Queries using BIT``#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` `    ``// 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``    ``updateBIT(BITTree1,n,l,val);``    ``updateBIT(BITTree1,n,r+1,-val);` `    ``// Update BIT2``    ``updateBIT(BITTree2,n,l,val*(l-1));``    ``updateBIT(BITTree2,n,r+1,-val*r);``}` `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;``    ``updateRange(BITTree1,BITTree2,n,val,l,r);` `    ``// Add 2 to all the elements from [2,4]``    ``l = 2 , r = 4 , val = 10;``    ``updateRange(BITTree1,BITTree2,n,val,l,r);` `    ``// 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) << ``"\n"``;` `    ``return` `0;``}`

## Java

 `// Java program to demonstrate Range Update``// and Range Queries using BIT``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)``{``    ``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.``static` `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]``static` `int` `sum(``int` `x, ``int` `BITTree1[], ``int` `BITTree2[])``{``    ``return` `(getSum(BITTree1, x) * x) - getSum(BITTree2, x);``}`  `static` `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``    ``updateBIT(BITTree1, n, l, val);``    ``updateBIT(BITTree1, n, r + ``1``, -val);` `    ``// Update BIT2``    ``updateBIT(BITTree2, n, l, val * (l - ``1``));``    ``updateBIT(BITTree2, n, r + ``1``, -val * r);``}` `static` `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);``}`  `static` `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``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``5``;` `    ``// Contwo BIT``    ``int` `[]BITTree1;``    ``int` `[]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``;``    ``updateRange(BITTree1, BITTree2, n, val, l, r);` `    ``// Add 2 to all the elements from [2,4]``    ``l = ``2` `; r = ``4` `; val = ``10``;``    ``updateRange(BITTree1, BITTree2, n, val, l, r);` `    ``// Find sum of all the elements from``    ``// [1,4]``    ``l = ``1` `; r = ``4``;``    ``System.out.print(``"Sum of elements from ["` `+ l``        ``+ ``","` `+ r+ ``"] is "``);``    ``System.out.print(rangeSum(l, r, BITTree1, BITTree2)+ ``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python program to demonstrate Range Update``# and Range Queries using BIT` `# 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: ``list``, index: ``int``) ``-``> ``int``:``    ``summ ``=` `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``        ``summ ``+``=` `BITree[index]` `        ``# Move index to parent node in getSum View``        ``index ``-``=` `index & (``-``index)``    ``return` `summ` `# 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: ``list``, n: ``int``, index: ``int``, val: ``int``) ``-``> ``None``:` `    ``# 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``        ``BITTree[index] ``+``=` `val` `        ``# Update index to that of parent in update View``        ``index ``+``=` `index & (``-``index)`  `# Returns the sum of array from [0, x]``def` `summation(x: ``int``, BITTree1: ``list``, BITTree2: ``list``) ``-``> ``int``:``    ``return` `(getSum(BITTree1, x) ``*` `x) ``-` `getSum(BITTree2, x)`  `def` `updateRange(BITTree1: ``list``, BITTree2: ``list``, n: ``int``, val: ``int``, l: ``int``,``                ``r: ``int``) ``-``> ``None``:` `    ``# Update Both the Binary Index Trees``    ``# As discussed in the article` `    ``# Update BIT1``    ``updateBit(BITTree1, n, l, val)``    ``updateBit(BITTree1, n, r ``+` `1``, ``-``val)` `    ``# Update BIT2``    ``updateBit(BITTree2, n, l, val ``*` `(l ``-` `1``))``    ``updateBit(BITTree2, n, r ``+` `1``, ``-``val ``*` `r)` `def` `rangeSum(l: ``int``, r: ``int``, BITTree1: ``list``, BITTree2: ``list``) ``-``> ``int``:` `    ``# Find sum from [0,r] then subtract sum``    ``# from [0,l-1] in order to find sum from``    ``# [l,r]``    ``return` `summation(r, BITTree1, BITTree2) ``-` `summation(``        ``l ``-` `1``, BITTree1, BITTree2)` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `5` `    ``# BIT1 to get element at any index``    ``# in the array``    ``BITTree1 ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``# BIT 2 maintains the extra term``    ``# which needs to be subtracted``    ``BITTree2 ``=` `[``0``] ``*` `(n ``+` `1``)` `    ``# Add 5 to all the elements from [0,4]``    ``l ``=` `0``    ``r ``=` `4``    ``val ``=` `5``    ``updateRange(BITTree1, BITTree2, n, val, l, r)` `    ``# Add 2 to all the elements from [2,4]``    ``l ``=` `2``    ``r ``=` `4``    ``val ``=` `10``    ``updateRange(BITTree1, BITTree2, n, val, l, r)` `    ``# Find sum of all the elements from``    ``# [1,4]``    ``l ``=` `1``    ``r ``=` `4``    ``print``(``"Sum of elements from [%d,%d] is %d"` `%``        ``(l, r, rangeSum(l, r, BITTree1, BITTree2)))` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to demonstrate Range Update``// and Range Queries using BIT``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` `    ``// 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.``static` `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]``static` `int` `sum(``int` `x, ``int` `[]BITTree1,``                      ``int` `[]BITTree2)``{``    ``return` `(getSum(BITTree1, x) * x) -``            ``getSum(BITTree2, x);``}`  `static` `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``    ``updateBIT(BITTree1, n, l, val);``    ``updateBIT(BITTree1, n, r + 1, -val);` `    ``// Update BIT2``    ``updateBIT(BITTree2, n, l, val * (l - 1));``    ``updateBIT(BITTree2, n, r + 1, -val * r);``}` `static` `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);``}` `static` `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 Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 5;` `    ``// Contwo BIT``    ``int` `[]BITTree1;``    ``int` `[]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;``    ``updateRange(BITTree1, BITTree2, n, val, l, r);` `    ``// Add 2 to all the elements from [2,4]``    ``l = 2 ; r = 4 ; val = 10;``    ``updateRange(BITTree1, BITTree2, n, val, l, r);` `    ``// Find sum of all the elements from``    ``// [1,4]``    ``l = 1 ; r = 4;``    ``Console.Write(``"Sum of elements from ["` `+ l +``                             ``","` `+ r + ``"] is "``);``    ``Console.Write(rangeSum(l, r, BITTree1,``                                 ``BITTree2) + ``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`Sum of elements from [1,4] is 50`

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