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 the sum of all elements in the 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}

Naive Approach: To solve the problem follow the below idea:

In the previous post, we discussed range update and point query solutions using BIT. 
rangeUpdate(l, r, val) : We add ‘val’ to the element at index ‘l’. We subtract ‘val’ from the 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 the solutions discussed in the previous post. The range update query is the same. Range sum query can be achieved by doing a get query for all elements in the range. 

Efficient Approach: To solve the problem follow the below idea:

We get range sum using prefix sums. How to make sure that the 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 the range [l, r]. Three cases arise 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 The sum from [0, k] = 4

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

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

Observations:

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

Case 2: Sum was incremented by val*k – val*(l-1). We can find “val”, it is similar to finding the i^th 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 k^th 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 l^th index, so when the getSum query is performed on BIT2 will give the 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)

Follow the below steps to solve the problem:

• Create the two binary index trees using the given function constructBITree()
• To find the Sum in a given range call the function rangeSum() with parameters as the given range and binary indexed trees
  • Call a function sum that will return a sum in the range [0, X]
  • Return sum(R) – sum(L-1)
    • Inside this function call the function getSum(), which will return the sum of the array from [0, X]
    • Return getSum(Tree1, x) * x – getSum(tree2, x)
    • Inside the getSum() function create an integer sum equal to zero and increase the index by 1
    • While the index is greater than zero increase the sum by Tree[index]
    • Decrease index by (index & (-index)) to move the index to the parent node in the tree
    • Return sum
• Print the sum in the given range

Below is the Implementation of the above approach: 

C++

// C++ program to demonstrate Range Update
// and Range Queries using BIT
#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

    // 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 code
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 10 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 10 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

# Python3 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 10 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 10 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

<script>

// JavaScript 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[]
function getSum(BITree,index)
{
    let 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.
function updateBIT(BITree,n,index,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]
function sum(x,BITTree1,BITTree2)
{
    return (getSum(BITTree1, x) * x) - getSum(BITTree2, x);
}

function updateRange(BITTree1,BITTree2,n,val,l,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);
}

function rangeSum(l,r,BITTree1,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);
}

function constructBITree(n)
{
    // Create and initialize BITree[] as 0
    let BITree = new Array(n + 1);
    for (let i = 1; i <= n; i++)
        BITree[i] = 0;

    return BITree;
}

// Driver Program to test above function
let n = 5;

// Contwo BIT
let BITTree1;
let 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]
let l = 0 , r = 4 , val = 5;
updateRange(BITTree1, BITTree2, n, val, l, r);

// Add 10 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;
document.write("Sum of elements from [" + l
                 + "," + r+ "] is ");
document.write(rangeSum(l, r, BITTree1, BITTree2)+ "<br>");

// This code is contributed by rag2127

</script>

Output

Sum of elements from [1,4] is 50

Time Complexity: O(q * log(N)) where q is the number of queries.
Auxiliary Space: O(N)

This article is contributed by Chirag Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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