# Binary Indexed Tree : Range Updates and Point Queries

• Difficulty Level : Medium
• Last Updated : 28 Dec, 2022

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

1. update(l, r, val): Add â€˜valâ€™ to all the elements in the array from [l, r].
2. getElement(i): Find element in the array indexed at â€˜iâ€™.

Initially all the elements in the array are 0. Queries can be in any order, i.e., there can be many updates before point query.

Example:

Input: arr = {0, 0, 0, 0, 0}
Queries: update : l = 0, r = 4, val = 2
getElement : i = 3
update : l = 3, r = 4, val = 3
getElement : i = 3

Output: Element at 3 is 2
Element at 3 is 5

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

Method 1 [update : O(n), getElement() : O(1)]

1. update(l, r, val): Iterate over the subarray from l to r and increase all the elements by val.
2. getElement(i): To get the element at i’th index, simply return arr[i].

The time complexity in the worst case is O(q*n) where q is the number of queries and n is the number of elements.

Method 2 [update: O(1), getElement(): O(n)]

We can avoid updating all elements and can update only 2 indexes of the array!

1. update(l, r, val) : Add â€˜valâ€™ to the lth element and subtract â€˜valâ€™ from the (r+1)th element, do this for all the update queries.
arr[l]   = arr[l] + val
arr[r+1] = arr[r+1] - val
1. getElement(i) : To get ith element in the array find the sum of all integers in the array from 0 to i.(Prefix Sum).

Letâ€™s analyze the update query. Why to add val to lth index? Adding val to lth index means that all the elements after l are increased by val, since we will be computing the prefix sum for every element. Why to subtract val from (r+1)th index? A range update was required from [l,r] but what we have updated is [l, n-1] so we need to remove val from all the elements after r i.e., subtract val from (r+1)th index. Thus the val is added to range [l,r]. Below is the implementation of the above approach.

## C++

 // C++ program to demonstrate Range Update// and Point Queries Without using BIT#include using namespace std; // Updates such that getElement() gets an increased// value when queried from l to r.void update(int arr[], int l, int r, int val){    arr[l] += val;    arr[r+1] -= val;} // Get the element indexed at iint getElement(int arr[], int i){    // To get ith element sum of all the elements    // from 0 to i need to be computed    int res = 0;    for (int j = 0 ; j <= i; j++)        res += arr[j];     return res;} // Driver program to test above functionint main(){    int arr[] = {0, 0, 0, 0, 0};    int n = sizeof(arr) / sizeof(arr[0]);     int l = 2, r = 4, val = 2;    update(arr, l, r, val);     //Find the element at Index 4    int index = 4;    cout << "Element at index " << index << " is " <<         getElement(arr, index) << endl;     l = 0, r = 3, val = 4;    update(arr,l,r,val);     //Find the element at Index 3    index = 3;    cout << "Element at index " << index << " is " <<         getElement(arr, index) << endl;     return 0;}

## Java

 // Java program to demonstrate Range Update// and Point Queries Without using BITclass GfG { // Updates such that getElement() gets an increased// value when queried from l to r.static void update(int arr[], int l, int r, int val){    arr[l] += val;    if(r + 1 < arr.length)    arr[r+1] -= val;} // Get the element indexed at istatic int getElement(int arr[], int i){    // To get ith element sum of all the elements    // from 0 to i need to be computed    int res = 0;    for (int j = 0 ; j <= i; j++)        res += arr[j];     return res;} // Driver program to test above functionpublic static void main(String[] args){    int arr[] = {0, 0, 0, 0, 0};    int n = arr.length;     int l = 2, r = 4, val = 2;    update(arr, l, r, val);     //Find the element at Index 4    int index = 4;    System.out.println("Element at index " + index + " is " +getElement(arr, index));     l = 0;    r = 3;    val = 4;    update(arr,l,r,val);     //Find the element at Index 3    index = 3;    System.out.println("Element at index " + index + " is " +getElement(arr, index)); }}

## Python3

 # Python3 program to demonstrate Range# Update and PoQueries Without using BIT # Updates such that getElement() gets an# increased value when queried from l to r.def update(arr, l, r, val):    arr[l] += val    if r + 1 < len(arr):        arr[r + 1] -= val # Get the element indexed at idef getElement(arr, i):         # To get ith element sum of all the elements    # from 0 to i need to be computed    res = 0    for j in range(i + 1):        res += arr[j]     return res # Driver Codeif __name__ == '__main__':    arr = [0, 0, 0, 0, 0]    n = len(arr)     l = 2    r = 4    val = 2    update(arr, l, r, val)     # Find the element at Index 4    index = 4    print("Element at index", index,          "is", getElement(arr, index))     l = 0    r = 3    val = 4    update(arr, l, r, val)     # Find the element at Index 3    index = 3    print("Element at index", index,          "is", getElement(arr, index)) # This code is contributed by PranchalK

## C#

 // C# program to demonstrate Range Update// and Point Queries Without using BITusing System; class GfG{ // Updates such that getElement()// gets an increased value when// queried from l to r.static void update(int []arr, int l,                    int r, int val){    arr[l] += val;    if(r + 1 < arr.Length)    arr[r + 1] -= val;} // Get the element indexed at istatic int getElement(int []arr, int i){    // To get ith element sum of all the elements    // from 0 to i need to be computed    int res = 0;    for (int j = 0 ; j <= i; j++)        res += arr[j];     return res;} // Driver codepublic static void Main(String[] args){    int []arr = {0, 0, 0, 0, 0};    int n = arr.Length;     int l = 2, r = 4, val = 2;    update(arr, l, r, val);     //Find the element at Index 4    int index = 4;    Console.WriteLine("Element at index " +                        index + " is " +                        getElement(arr, index));     l = 0;    r = 3;    val = 4;    update(arr,l,r,val);     //Find the element at Index 3    index = 3;    Console.WriteLine("Element at index " +                            index + " is " +                            getElement(arr, index));}} // This code is contributed by PrinciRaj1992



## Javascript

 //JavaScript program to demonstrate Range Update// and Point Queries Without using BIT // Updates such that getElement() gets an increased// value when queried from l to r.function update(arr, l, r, val){    arr[l] += val;    arr[r+1] -= val;} // Get the element indexed at ifunction getElement(rr, i){     // To get ith element sum of all the elements    // from 0 to i need to be computed    let res = 0;    for (let j = 0 ; j <= i; j++)        res += arr[j];     return res;} // Driver program to test above function     let arr = [0, 0, 0, 0, 0];    let n = arr.length;     let l = 2, r = 4, val = 2;    update(arr, l, r, val);     // Find the element at Index 4    let index = 4;    console.log("Element at index ",index," is ",getElement(arr, index));     l = 0, r = 3, val = 4;    update(arr,l,r,val);     // Find the element at Index 3    index = 3;    console.log("Element at index ",index," is ",getElement(arr, index)); // This code is contributed by vikkycirus

Output:

Element at index 4 is 2
Element at index 3 is 6

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

Method 3 (Using Binary Indexed Tree)

In method 2, we have seen that the problem can reduced to update and prefix sum queries. We have seen that BIT can be used to do update and prefix sum queries in O(Logn) time. Below is the implementation.

## C++

 // C++ code to demonstrate Range Update and// Point Queries on a Binary Index Tree#include using namespace std; // 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; i0)    {        // Add current element of BITree to sum        sum += BITree[index];         // Move index to parent node in getSum View        index -= index & (-index);    }    return sum;} // Updates such that getElement() gets an increased// value when queried from l to r.void update(int BITree[], int l, int r, int n, int val){    // Increase value at 'l' by 'val'    updateBIT(BITree, n, l, val);     // Decrease value at 'r+1' by 'val'    updateBIT(BITree, n, r+1, -val);} // Driver program to test above functionint main(){    int arr[] = {0, 0, 0, 0, 0};    int n = sizeof(arr)/sizeof(arr[0]);    int *BITree = constructBITree(arr, n);     // Add 2 to all the element from [2,4]    int l = 2, r = 4, val = 2;    update(BITree, l, r, n, val);     // Find the element at Index 4    int index = 4;    cout << "Element at index " << index << " is " <<         getSum(BITree,index) << "\n";     // Add 2 to all the element from [0,3]    l = 0, r = 3, val = 4;    update(BITree, l, r, n, val);     // Find the element at Index 3    index = 3;    cout << "Element at index " << index << " is " <<         getSum(BITree,index) << "\n" ;     return 0;}

## Java

 /* Java code to demonstrate Range Update and* Point Queries on a Binary Index Tree.* This method only works when all array* values are initially 0.*/class GFG{     // Max tree size    final static int MAX = 1000;     static int BITree[] = new int[MAX];     // 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 BITree            BITree[index] += val;             // Update index to that            // of parent in update View            index += index & (-index);        }    }     // Constructs Binary Indexed Tree    // for given array of size n.     public static 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]);         // Uncomment below lines to        // see contents of BITree[]        // for (int i=1; i<=n; i++)        //     cout << BITree[i] << " ";    }     // SERVES THE PURPOSE OF getElement()    // Returns sum of arr[0..index]. This    // function assumes that the array is    // preprocessed and partial sums of    // array elements are stored in BITree[]    public static int getSum(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 the sum        return sum;    }     // Updates such that getElement()    // gets an increased value when    // queried from l to r.    public static void update(int l, int r,                              int n, int val)    {        // Increase value at        // 'l' by 'val'        updateBIT(n, l, val);         // Decrease value at        // 'r+1' by 'val'        updateBIT(n, r + 1, -val);    }      // Driver Code    public static void main(String args[])    {        int arr[] = {0, 0, 0, 0, 0};        int n = arr.length;         constructBITree(arr,n);         // Add 2 to all the        // element from [2,4]        int l = 2, r = 4, val = 2;        update(l, r, n, val);         int index = 4;         System.out.println("Element at index "+                                index + " is "+                                getSum(index));         // Add 2 to all the        // element from [0,3]        l = 0; r = 3; val = 4;        update(l, r, n, val);         // Find the element        // at Index 3        index = 3;        System.out.println("Element at index "+                                index + " is "+                                getSum(index));    }}// This code is contributed// by Puneet Kumar.

## Python3

 # Python3 code to demonstrate Range Update and# PoQueries on a Binary Index Tree # 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(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) # Constructs and returns a Binary Indexed Tree for given# array of size n.def constructBITree(arr, n):         # Create and initialize BITree[] as 0    BITree = [0]*(n+1)     # Store the actual values in BITree[] using update()    for i in range(n):        updateBIT(BITree, n, i, arr[i])     return BITree # SERVES THE PURPOSE OF getElement()# 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, index):    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 such that getElement() gets an increased# value when queried from l to r.def update(BITree, l, r, n, val):         # Increase value at 'l' by 'val'    updateBIT(BITree, n, l, val)     # Decrease value at 'r+1' by 'val'    updateBIT(BITree, n, r+1, -val) # Driver codearr = [0, 0, 0, 0, 0]n = len(arr)BITree = constructBITree(arr, n) # Add 2 to all the element from [2,4]l = 2r = 4val = 2update(BITree, l, r, n, val) # Find the element at Index 4index = 4print("Element at index", index, "is", getSum(BITree, index)) # Add 2 to all the element from [0,3]l = 0r = 3val = 4update(BITree, l, r, n, val) # Find the element at Index 3index = 3print("Element at index", index, "is", getSum(BITree,index)) # This code is contributed by mohit kumar 29

## C#

 using System; /* C# code to demonstrate Range Update and* Point Queries on a Binary Index Tree.* This method only works when all array* values are initially 0.*/public class GFG{     // Max tree size    public const int MAX = 1000;     public static int[] BITree = new int[MAX];     // 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 BITree            BITree[index] += val;             // Update index to that             // of parent in update View            index += index & (-index);        }    }     // Constructs Binary Indexed Tree     // for given array of size n.     public static 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]);        }         // Uncomment below lines to         // see contents of BITree[]        // for (int i=1; i<=n; i++)        //     cout << BITree[i] << " ";    }     // SERVES THE PURPOSE OF getElement()    // Returns sum of arr[0..index]. This     // function assumes that the array is    // preprocessed and partial sums of    // array elements are stored in BITree[]    public static int getSum(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 the sum        return sum;    }     // Updates such that getElement()     // gets an increased value when     // queried from l to r.    public static void update(int l, int r, int n, int val)    {        // Increase value at         // 'l' by 'val'        updateBIT(n, l, val);         // Decrease value at        // 'r+1' by 'val'        updateBIT(n, r + 1, -val);    }      // Driver Code    public static void Main(string[] args)    {        int[] arr = new int[] {0, 0, 0, 0, 0};        int n = arr.Length;         constructBITree(arr,n);         // Add 2 to all the        // element from [2,4]        int l = 2, r = 4, val = 2;        update(l, r, n, val);         int index = 4;         Console.WriteLine("Element at index " + index + " is " + getSum(index));         // Add 2 to all the         // element from [0,3]        l = 0;        r = 3;        val = 4;        update(l, r, n, val);         // Find the element        // at Index 3        index = 3;        Console.WriteLine("Element at index " + index + " is " + getSum(index));    }}    // This code is contributed by Shrikant13

## Javascript

 // 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);  }} // Constructs and returns a Binary Indexed Tree for given// array of size n.function constructBITree(arr, n) {  // Create and initialize BITree[] as 0  let BITree = new Array(n+1).fill(0);   // Store the actual values in BITree[] using update()  for (let i = 0; i < n; i++) {    updateBIT(BITree, n, i, arr[i]);  }   return BITree;} // SERVES THE PURPOSE OF getElement()// 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 such that getElement() gets an increased// value when queried from l to r.function update(BITree, l, r, n, val) {  // Increase value at 'l' by 'val'  updateBIT(BITree, n, l, val);   // Decrease value at 'r+1' by 'val'  updateBIT(BITree, n, r+1, -val);} // Test the functions let arr = [0, 0, 0, 0, 0];let n = arr.length;let BITree = constructBITree(arr, n); // Add 2 to all the element from [2,4]let l = 2, r = 4, val = 2;update(BITree, l, r, n, val); // Find the element at Index 4let index = 4;console.log(`Element at index \${index} is \${getSum(BITree,index)}`); // Add 2 to all the element from [0,3]l = 0, r = 3, val = 4;update(BITree, l, r, n, val); // Find the element at Index 3index = 3;console.log(`Element at index \${index} is \${getSum(BITree,index)}`);

Output:

Element at index 4 is 2
Element at index 3 is 6

Time Complexity : O(q * log n) + O(n * log n) where q is number of queries. Method 1 is efficient when most of the queries are getElement(), method 2 is efficient when most of the queries are updates() and method 3 is preferred when there is mix of both queries. 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 you want to share more information about the topic discussed above.

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