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Binary Indexed Tree : Range Update and Range Queries

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  • Difficulty Level : Hard
  • Last Updated : 05 Dec, 2022
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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 lth index to kth 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 lth index to rth 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 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 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.

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


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