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Segment Tree | Sum of given range

Last Updated : 23 Jul, 2025
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Let us consider the following problem to understand Segment Trees.
We have an array arr[0 . . . n-1]. We should be able to 

  • Find the sum of elements from index l to r where 0 <= l <= r <= n-1
  • Change the value of a specified element of the array to a new value x. We need to do arr[i] = x where 0 <= i <= n-1.
Recommended Practice

Sum of range using Nested Loop :

A simple solution is to run a loop from l to r and calculate the sum of elements in the given range. To update a value, simply do arr[i] = x. The first operation takes O(n) time and the second operation takes O(1) time. 

 

Sum of range using Prefix Sum :

Another solution is to create another array and store the sum from start to i ,at the ith index in this array. The sum of a given range can now be calculated in O(1) time, but update operation takes O(n) time now. This works well if the number of query operations is large and very few updates.

Sum of range using Segment Tree :

The most efficient way is to use a segment tree, we can use a Segment Tree to do both operations in O(log(N)) time.

Representation of Segment trees 

  • Leaf Nodes are the elements of the input array. 
  • Each internal node represents some merging of the leaf nodes. The merging may be different for different problems. For this problem, merging is sum of leaf nodes under a node.
  • An array representation of tree is used to represent Segment Trees. For each node at index i, the left child is at index (2*i+1), right child at (2*i+2) and the parent is at  (⌊(i - 1) / 2⌋).
     

Construction of Segment Tree from the given array:

We start with a segment arr[0 . . . n-1]. and every time we divide the current segment into two (if it has not yet become a segment of length 1), and then call the same procedure on both halves, and for each such segment, we store the sum in the corresponding node. 
All levels of the constructed segment tree will be completely filled except the last level. Also, the tree will be a Full Binary Tree because we always divide segment in two, at every level. Since the constructed tree is always a full binary tree with n leaves, there will be n-1 internal nodes. So the total number of nodes will be 2*n - 1.

What is the height of the segment tree for a given array:

Height of the segment tree will be ⌈log₂N⌉. Since the tree is represented using array and relation between parent and child indexes must be maintained, size of memory allocated for segment tree will be (2 * 2⌈log2n⌉  - 1).

Query for Sum of a given range

Once the tree is constructed, how to get the sum using the constructed segment tree. The following is the algorithm to get the sum of elements.  

int getSum(node, l, r) 
{
   if the range of the node is within l and r
        return value in the node
   else if the range of the node is completely outside l and r
        return 0
   else
    return getSum(node's left child, l, r) + 
           getSum(node's right child, l, r)
}

In the above implementation, there are three cases we need to take into consideration

  • If the range of the current node while traversing the tree is not in the given range then did not add the value of that node in ans
  • If the range of node is partially overlapped with the given range then move either left or right according to the overlapping
  • If the range is completely overlapped by the given range then add it to the ans

Update a value: 

Like tree construction and query operations, the update can also be done recursively. We are given an index which needs to be updated. Let diff be the value to be added. We start from the root of the segment tree and add diff to all nodes which have given index in their range. If a node doesn't have a given index in its range, we don't make any changes to that node. 

The algorithmic steps to implement a segment tree are:

  • Initialize the segment tree with a size equal to 4 * n, where n is the number of elements in the array.
  • In the buildTree function, the base case is when the left and right bounds of the current segment are equal. In this case, the value of the current node in the segment tree is set to the value of the corresponding element in the array.
  • For the rest of the cases, calculate the midpoint of the current segment and recursively call the buildTree function for the left and right subsegments.
  • In the query function, the base case is when the current segment is completely contained within the query range. In this case, the value of the current node in the segment tree is returned.
  • For the rest of the cases, calculate the midpoint of the current segment and recursively call the query function for the left and right subsegments. The minimum (or maximum, or sum, etc.) of the values returned from the left and right subsegments is returned.
  • The query function can be called with the left and right bounds of the desired range to get the desired result.

Note: The implementation details, such as the type of aggregation and the way the midpoint is calculated, can vary based on the specific use case.

Example no1: Below is the implementation of the above approach:

C++
// C++ program to show segment tree operations like construction, query 
// and update 
#include <bits/stdc++.h>
using namespace std;

// A utility function to get the middle index from corner indexes. 
int getMid(int s, int e) { return s + (e -s)/2; } 

/* A recursive function to get the sum of values in the given range 
    of the array. The following are parameters for this function. 

    st --> Pointer to segment tree 
    si --> Index of current node in the segment tree. Initially 
            0 is passed as root is always at index 0 
    ss & se --> Starting and ending indexes of the segment represented 
                by current node, i.e., st[si] 
    qs & qe --> Starting and ending indexes of query range */
int getSumUtil(int *st, int ss, int se, int qs, int qe, int si) 
{ 
    // If segment of this node is a part of given range, then return 
    // the sum of the segment 
    if (qs <= ss && qe >= se) 
        return st[si]; 

    // If segment of this node is outside the given range 
    if (se < qs || ss > qe) 
        return 0; 

    // If a part of this segment overlaps with the given range 
    int mid = getMid(ss, se); 
    return getSumUtil(st, ss, mid, qs, qe, 2*si+1) + 
        getSumUtil(st, mid+1, se, qs, qe, 2*si+2); 
} 

/* A recursive function to update the nodes which have the given 
index in their range. The following are parameters 
    st, si, ss and se are same as getSumUtil() 
    i --> index of the element to be updated. This index is 
            in the input array. 
diff --> Value to be added to all nodes which have i in range */
void updateValueUtil(int *st, int ss, int se, int i, int diff, int si) 
{ 
    // Base Case: If the input index lies outside the range of 
    // this segment 
    if (i < ss || i > se) 
        return; 

    // If the input index is in range of this node, then update 
    // the value of the node and its children 
    st[si] = st[si] + diff; 
    if (se != ss) 
    { 
        int mid = getMid(ss, se); 
        updateValueUtil(st, ss, mid, i, diff, 2*si + 1); 
        updateValueUtil(st, mid+1, se, i, diff, 2*si + 2); 
    } 
} 

// The function to update a value in input array and segment tree. 
// It uses updateValueUtil() to update the value in segment tree 
void updateValue(int arr[], int *st, int n, int i, int new_val) 
{ 
    // Check for erroneous input index 
    if (i < 0 || i > n-1) 
    { 
        cout<<"Invalid Input"; 
        return; 
    } 

    // Get the difference between new value and old value 
    int diff = new_val - arr[i]; 

    // Update the value in array 
    arr[i] = new_val; 

    // Update the values of nodes in segment tree 
    updateValueUtil(st, 0, n-1, i, diff, 0); 
} 

// Return sum of elements in range from index qs (query start) 
// to qe (query end). It mainly uses getSumUtil() 
int getSum(int *st, int n, int qs, int qe) 
{ 
    // Check for erroneous input values 
    if (qs < 0 || qe > n-1 || qs > qe) 
    { 
        cout<<"Invalid Input"; 
        return -1; 
    } 

    return getSumUtil(st, 0, n-1, qs, qe, 0); 
} 

// A recursive function that constructs Segment Tree for array[ss..se]. 
// si is index of current node in segment tree st 
int constructSTUtil(int arr[], int ss, int se, int *st, int si) 
{ 
    // If there is one element in array, store it in current node of 
    // segment tree and return 
    if (ss == se) 
    { 
        st[si] = arr[ss]; 
        return arr[ss]; 
    } 

    // If there are more than one elements, then recur for left and 
    // right subtrees and store the sum of values in this node 
    int mid = getMid(ss, se); 
    st[si] = constructSTUtil(arr, ss, mid, st, si*2+1) + 
            constructSTUtil(arr, mid+1, se, st, si*2+2); 
    return st[si]; 
} 

/* Function to construct segment tree from given array. This function 
allocates memory for segment tree and calls constructSTUtil() to 
fill the allocated memory */
int *constructST(int arr[], int n) 
{ 
    // Allocate memory for the segment tree 

    //Height of segment tree 
    int x = (int)(ceil(log2(n))); 

    //Maximum size of segment tree 
    int max_size = 2*(int)pow(2, x) - 1; 

    // Allocate memory 
    int *st = new int[max_size]; 

    // Fill the allocated memory st 
    constructSTUtil(arr, 0, n-1, st, 0); 

    // Return the constructed segment tree 
    return st; 
} 

// Driver program to test above functions 
int main() 
{ 
    int arr[] = {1, 3, 5, 7, 9, 11}; 
    int n = sizeof(arr)/sizeof(arr[0]); 

    // Build segment tree from given array 
    int *st = constructST(arr, n); 

    // Print sum of values in array from index 1 to 3 
    cout<<"Sum of values in given range = "<<getSum(st, n, 1, 3)<<endl; 

    // Update: set arr[1] = 10 and update corresponding 
    // segment tree nodes 
    updateValue(arr, st, n, 1, 10); 

    // Find sum after the value is updated 
    cout<<"Updated sum of values in given range = "
            <<getSum(st, n, 1, 3)<<endl; 
    return 0; 
} 
//This code is contributed by rathbhupendra
C Java Python3 C# JavaScript

Output
Sum of values in given range = 15
Updated sum of values in given range = 22

Time complexity: O(N*log(N)) 
Auxiliary Space: O(N) 

Example no2:

 Java code to implement a segment tree:

C++
#include <bits/stdc++.h>

using namespace std;

class SegmentTree {
    vector<int> tree;
    int size;

public:
    SegmentTree(vector<int>& array) {
        size = array.size();
        tree.resize(4 * size);
        buildTree(array, 0, 0, size - 1);
    }

private:
    void buildTree(vector<int>& array, int treeIndex, int left, int right) {
        if (left == right) {
            tree[treeIndex] = array[left];
            return;
        }
        int mid = left + (right - left) / 2;
        buildTree(array, 2 * treeIndex + 1, left, mid);
        buildTree(array, 2 * treeIndex + 2, mid + 1, right);
        tree[treeIndex] = min(tree[2 * treeIndex + 1], tree[2 * treeIndex + 2]);
    }

    int query(int treeIndex, int left, int right, int queryLeft, int queryRight) {
        if (queryLeft <= left && right <= queryRight)
            return tree[treeIndex];
        int mid = left + (right - left) / 2;
        int minValue = INT_MAX;
        if (queryLeft <= mid)
            minValue = min(minValue, query(2 * treeIndex + 1, left, mid, queryLeft, queryRight));
        if (queryRight > mid)
            minValue = min(minValue, query(2 * treeIndex + 2, mid + 1, right, queryLeft, queryRight));
        return minValue;
    }

public:
    int query(int left, int right) {
        return query(0, 0, size - 1, left, right);
    }
};

int main() {
    vector<int> array = {1, 3, 2, 5, 4, 6};
    SegmentTree st(array);
    cout << st.query(1, 5) << endl; // 2
    return 0;
}
Java Python3 C# JavaScript

Output
2

Benifits of segment tree usage:

  • Range Queries: One of the main use cases of segment trees is to perform range queries on an array in an efficient manner. The query function in the segment tree can return the minimum, maximum, sum, or any other aggregation of elements within a specified range in the array in O(log n) time.
  • Dynamic Updates: Another advantage of using segment trees is that they support dynamic updates to the array. This means that elements in the array can be changed and the segment tree can be updated accordingly in O(log n) time.
  • Space Optimization: Segment trees are space-optimized compared to other data structures such as the binary indexed tree. The space complexity of a segment tree is O(4n) in the worst case, which is better than the O(2n) space complexity of binary indexed trees.
  • Versatility: Segment trees can be used to solve various range-based problems, such as finding the sum of elements within a range, finding the minimum/maximum element within a range, and finding the number of distinct elements within a range.
  • Easy to Code: Segment trees are relatively easy to code compared to other data structures, and their implementation is straightforward, especially for range minimum/maximum queries.

Segment Tree | Set 2 (Range Minimum Query)


Introduction of Segment Tree
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