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Binary Indexed Tree : Range Updates and Point Queries
  • Difficulty Level : Medium
  • Last Updated : 25 Feb, 2020
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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. getElement(i) : Find element in the array indexed at ‘i’.

    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 worst case is O(q*n) where q is number of queries and n is 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 ‘val’ to the lth element and subtract ‘val’ 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 

    2. getElement(i) : To get ith element in the array find the sum of all integers in the array from 0 to i.(Prefix Sum).
    Let’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 implementation of above approach.

    C++

    // C++ program to demonstrate Range Update
    // and Point Queries Without using BIT
    #include <bits/stdc++.h>
    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 i
    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 function
    int 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 BIT 
    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 i 
    static 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 function 
    public 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 i 
    def 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 Code
    if __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 BIT 
    using 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 i 
    static 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 code 
    public 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
    PHP



    <?php
    // PHP 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] += $val;
        if($r + 1 < sizeof($arr))
        $arr[$r + 1] -= $val; 
    } 
      
    // Get the element indexed at i 
    function getElement(&$arr, $i) 
    { 
        // To get ith element sum of all the elements 
        // from 0 to i need to be computed 
        $res = 0; 
        for ($j = 0 ; $j <= $i; $j++) 
            $res += $arr[$j]; 
      
        return $res; 
    } 
      
    // Driver Code
    $arr = array(0, 0, 0, 0, 0); 
    $n = sizeof($arr); 
      
    $l = 2; $r = 4; $val = 2; 
    update($arr, $l, $r, $val); 
      
    // Find the element at Index 4 
    $index = 4; 
    echo("Element at index " . $index . 
         " is " . getElement($arr, $index) . "\n"); 
      
    $l = 0;
    $r = 3;
    $val = 4; 
    update($arr, $l, $r, $val); 
      
    // Find the element at Index 3 
    $index = 3; 
    echo("Element at index " . $index . 
         " is " . getElement($arr, $index)); 
      
    // This code is contributed by Code_Mech
    ?>

    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 <bits/stdc++.h>
    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; i<n; i++)
            updateBIT(BITree, n, i, arr[i]);
      
        // Uncomment below lines to see contents of BITree[]
        //for (int i=1; i<=n; i++)
        //      cout << BITree[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[]
    int getSum(int BITree[], int index)
    {
        int sum = 0; // Iniialize 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.
    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 function
    int 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 # Iniialize 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 code
    arr = [0, 0, 0, 0, 0]
    n = len(arr)
    BITree = constructBITree(arr, n)
      
    # Add 2 to all the element from [2,4]
    l = 2
    r = 4
    val = 2
    update(BITree, l, r, n, val)
      
    # Find the element at Index 4
    index = 4
    print("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 3
    index = 3
    print("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
    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 contribute.geeksforgeeks.org or mail your article to contribute@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.
    Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
    My Personal Notes
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