Maximum Subarray Sum in a given Range

Given an array of n numbers, the task is to answer the following queries: 

maximumSubarraySum(start, end) : Find the maximum 
subarray sum in the range from array index 'start' 
to 'end'.

Examples:

Input : arr[] = {1, 3, -4, 5, -2}
        Query 1: start = 0, end = 4
        Query 2: start = 0, end = 2
Output : 5
         4
Explanation:
For Query 1, [1, 3, -4, 5] or ( [5] ) 
represent the maximum sum sub arrays 
with sum = 5.

For Query 2, [1, 3] represents the 
maximum sum subarray in the query range
with sum = 4

Segment Trees can be used to solve this problem. Here, we need to keep information regarding various cumulative sums. At every Node we store the following: 

1. Maximum Prefix Sum, 
2. Maximum Suffix Sum, 
3. Total Sum, 
4. Maximum Subarray Sum

A classical Segment Tree with each Node storing the above information should be enough to answer each query. The only focus here is on how the left and the right Nodes of the tree are merged together. Now, we will discuss how each of the information is constructed in each of the segment tree Nodes using the information of its left and right child. 

Constructing the Maximum Prefix Sum using Left and Right child

There can be two cases for maximum prefix sum of a Node: 

• The maximum prefix sum occurs in the left child, 

In this Case,
Maximum Prefix Sum = Maximum Prefix Sum of Left Child

• The maximum prefix sum contains every array element of the left child and the elements contributing to the maximum prefix sum of the right child, 
  

In this Case,
Maximum Prefix Sum = Total Sum of Left Child + 
               Maximum Prefix Sum of Right Child

Constructing the Maximum Suffix Sum using Left and Right child

There can be two cases for maximum suffix sum of a Node: 

• The maximum suffix sum occurs in the right child, 

In this Case,
Maximum Suffix Sum = Maximum Suffix Sum of Right Child

• The maximum suffix sum contains every array element of the Right child and the elements contributing to the maximum suffix sum of the left child, 
  

In this Case,
Maximum Suffix Sum = Total Sum of Right Child + 
               Maximum Suffix Sum of Left Child

Constructing the Maximum Subarray Sum using Left and Right child

There can be three cases for the maximum sub-array sum of a Node: 

• The maximum sub-array sum occurs in the left child, 
  

In this Case,
Maximum Sub-array Sum = Maximum Subarray Sum of Left Child

• The maximum sub-array sum occurs in the right child, 

In this Case,
Maximum Sub-array Sum = Maximum Subarray Sum of Right Child

• The maximum subarray sum, contains array elements of the right child contributing to the maximum prefix sum of the right child, and the array elements of the Left child contributing to the maximum suffix sum of the left child, 
  

In this Case,
Maximum Subarray Sum = Maximum Prefix Sum of Right Child                                           
                                  + 
                       Maximum Suffix Sum of Left Child

Implementation:

CPP

// C++ Program to Implement Maximum Sub-Array Sum in a range
#include <bits/stdc++.h>
using namespace std;

#define inf 0x3f3f

/* Node of the segment tree consisting of:
1. Maximum Prefix Sum,
2. Maximum Suffix Sum,
3. Total Sum,
4. Maximum Sub-Array Sum */
struct Node {
    int maxPrefixSum;
    int maxSuffixSum;
    int totalSum;
    int maxSubarraySum;

    Node()
    {
        maxPrefixSum = maxSuffixSum = maxSubarraySum = -inf;
        totalSum = -inf;
    }
};

// Returns Parent Node after merging its left and right child
Node merge(Node leftChild, Node rightChild)
{
    Node parentNode;
    parentNode.maxPrefixSum = max(leftChild.maxPrefixSum,
                                  leftChild.totalSum +
                                  rightChild.maxPrefixSum);

    parentNode.maxSuffixSum = max(rightChild.maxSuffixSum,
                                  rightChild.totalSum +
                                  leftChild.maxSuffixSum);

    parentNode.totalSum = leftChild.totalSum +
                          rightChild.totalSum;

    parentNode.maxSubarraySum = max({leftChild.maxSubarraySum,
                                     rightChild.maxSubarraySum,
                                     leftChild.maxSuffixSum +
                                     rightChild.maxPrefixSum});

    return parentNode;
}

// Builds the Segment tree recursively
void constructTreeUtil(Node* tree, int arr[], int start,
                                    int end, int index)
{

    /* Leaf Node */
    if (start == end) {

        // single element is covered under this range
        tree[index].totalSum = arr[start];
        tree[index].maxSuffixSum = arr[start];
        tree[index].maxPrefixSum = arr[start];
        tree[index].maxSubarraySum = arr[start];
        return;
    }

    // Recursively Build left and right children
    int mid = (start + end) / 2;
    constructTreeUtil(tree, arr, start, mid, 2 * index);
    constructTreeUtil(tree, arr, mid + 1, end, 2 * index + 1);

    // Merge left and right child into the Parent Node
    tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
}

/* Function to construct segment tree from given array.
   This function allocates memory for segment tree and
   calls constructTreeUtil() to fill the allocated
   memory */
Node* constructTree(int arr[], int n)
{
    // Allocate memory for segment tree
    int x = (int)(ceil(log2(n))); // Height of the tree

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

    // Fill the allocated memory tree
    constructTreeUtil(tree, arr, 0, n - 1, 1);

    // Return the constructed segment tree
    return tree;
}

/* A Recursive function to get the desired
   Maximum Sum Sub-Array,
The following are parameters of the function-

tree     --> Pointer to segment tree
index --> Index of the segment tree Node
ss & se  --> Starting and ending indexes of the
             segment represented by
                 current Node, i.e., tree[index]
qs & qe  --> Starting and ending indexes of query range */
Node queryUtil(Node* tree, int ss, int se, int qs,
                               int qe, int index)
{
    // No overlap
    if (ss > qe || se < qs) {

        // returns a Node for out of bounds condition
        Node nullNode;
        return nullNode;
    }

    // Complete overlap
    if (ss >= qs && se <= qe) {
        return tree[index];
    }

    // Partial Overlap Merge results of Left
    // and Right subtrees
    int mid = (ss + se) / 2;
    Node left = queryUtil(tree, ss, mid, qs, qe,
                                     2 * index);
    Node right = queryUtil(tree, mid + 1, se, qs,
                              qe, 2 * index + 1);

    // merge left and right subtree query results
    Node res = merge(left, right);
    return res;
}

/* Returns the Maximum Subarray Sum between start and end
   It mainly uses queryUtil(). */
int query(Node* tree, int start, int end, int n)
{
    Node res = queryUtil(tree, 0, n - 1, start, end, 1);
    return res.maxSubarraySum;
}

int main()
{
    int arr[] = { 1, 3, -4, 5, -2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Construct Segment Tree
    Node* Tree = constructTree(arr, n);
    int start, end, maxSubarraySum;

    // Answering query 1:
    start = 0;
    end = 4;
    maxSubarraySum = query(Tree, start, end, n);
    cout << "Maximum Sub-Array Sum between "
         << start << " and " << end
         << " = " << maxSubarraySum << "\n";

    // Answering query 2:
    start = 0;
    end = 2;
    maxSubarraySum = query(Tree, start, end, n);
    cout << "Maximum Sub-Array Sum between "
         << start << " and " << end
         << " = " << maxSubarraySum << "\n";

    return 0;
}

Java

// Java Program to Implement Maximum Sub-Array Sum in a
// range

import java.io.*;
import java.util.*;

/* Node of the segment tree consisting of:
1. Maximum Prefix Sum,
2. Maximum Suffix Sum,
3. Total Sum,
4. Maximum Sub-Array Sum */
class Node {
  int maxPrefixSum;
  int maxSuffixSum;
  int totalSum;
  int maxSubarraySum;

  Node()
  {
    maxPrefixSum = maxSuffixSum = maxSubarraySum
      = Integer.MIN_VALUE;
    totalSum = Integer.MIN_VALUE;
  }
}

class GFG {

  static final int inf = 0x3f3f;

  // Returns Parent Node after merging its left and right
  // child
  static Node merge(Node leftChild, Node rightChild)
  {
    Node parentNode = new Node();
    parentNode.maxPrefixSum = Math.max(
      leftChild.maxPrefixSum,
      leftChild.totalSum + rightChild.maxPrefixSum);
    parentNode.maxSuffixSum = Math.max(
      rightChild.maxSuffixSum,
      rightChild.totalSum + leftChild.maxSuffixSum);
    parentNode.totalSum
      = leftChild.totalSum + rightChild.totalSum;
    parentNode.maxSubarraySum
      = Math.max(Math.max(leftChild.maxSubarraySum,
                          rightChild.maxSubarraySum),
                 leftChild.maxSuffixSum
                 + rightChild.maxPrefixSum);
    return parentNode;
  }

  // Builds the Segment tree recursively
  static void constructTreeUtil(Node[] tree, int[] arr,
                                int start, int end,
                                int index)
  {
    /* Leaf Node */
    if (start == end) {
      // single element is covered under this range
      tree[index].totalSum = arr[start];
      tree[index].maxSuffixSum = arr[start];
      tree[index].maxPrefixSum = arr[start];
      tree[index].maxSubarraySum = arr[start];
      return;
    }

    // Recursively Build left and right children
    int mid = (start + end) / 2;
    constructTreeUtil(tree, arr, start, mid, 2 * index);
    constructTreeUtil(tree, arr, mid + 1, end,
                      2 * index + 1);

    // Merge left and right child into the Parent Node
    tree[index]
      = merge(tree[2 * index], tree[2 * index + 1]);
  }

  /* Function to construct segment tree from given array.
     * This function allocates memory for segment tree and
     * calls constructTreeUtil() to fill the allocated
     * memory */
  static Node[] constructTree(int[] arr, int n)
  {
    // Allocate memory for segment tree
    int x = (int)(Math.ceil(
      Math.log(n)
      / Math.log(2))); // Height of the tree

    // Maximum size of segment tree
    int max_size = 2 * (int)Math.pow(2, x) - 1;
    Node[] tree = new Node[max_size];
    for (int i = 0; i < max_size; i++)
      tree[i] = new Node();

    // Fill the allocated memory tree
    constructTreeUtil(tree, arr, 0, n - 1, 1);

    // Return the constructed segment tree
    return tree;
  }

  /* A Recursive function to get the desired
Maximum Sum Sub-Array,
The following are parameters of the function-

tree     --> Pointer to segment tree
index --> Index of the segment tree Node
ss & se  --> Starting and ending indexes of the
         segment represented by
             current Node, i.e., tree[index]
qs & qe  --> Starting and ending indexes of query range */
  static Node queryUtil(Node[] tree, int ss, int se,
                        int qs, int qe, int index)
  {
    // No overlap
    if (ss > qe || se < qs) {

      // returns a Node for out of bounds condition
      Node nullNode = new Node();
      return nullNode;
    }

    // Complete overlap
    if (ss >= qs && se <= qe) {
      return tree[index];
    }

    // Partial Overlap Merge results of Left and Right
    // subtrees
    int mid = (ss + se) / 2;
    Node left
      = queryUtil(tree, ss, mid, qs, qe, 2 * index);
    Node right = queryUtil(tree, mid + 1, se, qs, qe,
                           2 * index + 1);

    // merge left and right subtree query results
    Node res = merge(left, right);
    return res;
  }

  /* Returns the Maximum Subarray Sum between start and
     * end It mainly uses queryUtil(). */
  static int query(Node[] tree, int start, int end, int n)
  {
    Node res = queryUtil(tree, 0, n - 1, start, end, 1);
    return res.maxSubarraySum;
  }

  public static void main(String[] args)
  {
    int[] arr = { 1, 3, -4, 5, -2 };
    int n = arr.length;

    // Construct Segment Tree
    Node[] Tree = constructTree(arr, n);
    int start, end, maxSubarraySum;

    // Answering query 1:
    start = 0;
    end = 4;
    maxSubarraySum = query(Tree, start, end, n);
    System.out.println("Maximum Sub-Array Sum between "
                       + start + " and " + end + " = "
                       + maxSubarraySum);

    // Answering query 2:
    start = 0;
    end = 2;
    maxSubarraySum = query(Tree, start, end, n);
    System.out.println("Maximum Sub-Array Sum between "
                       + start + " and " + end + " = "
                       + maxSubarraySum);
  }
}

// This code is contributed by sankar.

Python3

# Python Program to Implement Maximum Sub-Array Sum in a range

import math

# Node of the segment tree consisting of:
# 1. Maximum Prefix Sum,
# 2. Maximum Suffix Sum,
# 3. Total Sum,
# 4. Maximum Sub-Array Sum
class Node:
    def __init__(self):
        self.maxPrefixSum = float("-inf")
        self.maxSuffixSum = float("-inf")
        self.totalSum = float("-inf")
        self.maxSubarraySum = float("-inf")

# Returns Parent Node after merging its left and right child
def merge(leftChild, rightChild):
    parentNode = Node()
    parentNode.maxPrefixSum = max(leftChild.maxPrefixSum, leftChild.totalSum + rightChild.maxPrefixSum)
    parentNode.maxSuffixSum = max(rightChild.maxSuffixSum, rightChild.totalSum + leftChild.maxSuffixSum)
    parentNode.totalSum = leftChild.totalSum + rightChild.totalSum
    parentNode.maxSubarraySum = max(leftChild.maxSubarraySum, rightChild.maxSubarraySum, leftChild.maxSuffixSum + rightChild.maxPrefixSum)
    return parentNode

# Builds the Segment tree recursively
def constructTreeUtil(tree, arr, start, end, index):
    # Leaf Node
    if start == end:
        # single element is covered under this range
        tree[index] = Node()
        tree[index].totalSum = arr[start]
        tree[index].maxSuffixSum = arr[start]
        tree[index].maxPrefixSum = arr[start]
        tree[index].maxSubarraySum = arr[start]
        return
    # Recursively Build left and right children
    mid = (start + end) // 2
    constructTreeUtil(tree, arr, start, mid, 2 * index)
    constructTreeUtil(tree, arr, mid + 1, end, 2 * index + 1)
    # Merge left and right child into the Parent Node
    tree[index] = merge(tree[2 * index], tree[2 * index + 1])

# Function to construct segment tree from given array.
# This function allocates memory for segment tree and
# calls constructTreeUtil() to fill the allocated
# memory
def constructTree(arr, n):
    # Allocate memory for segment tree
    x = int(math.ceil(math.log2(n))) # Height of the tree
    # Maximum size of segment tree
    max_size = 2 * int(math.pow(2, x)) - 1
    tree = [Node() for i in range(max_size)]
    # Fill the allocated memory tree
    constructTreeUtil(tree, arr, 0, n - 1, 1)
    # Return the constructed segment tree
    return tree

# A Recursive function to get the desired
# Maximum Sum Sub-Array,
# The following are parameters of the function-
# tree     --> Pointer to segment tree
# index --> Index of the segment tree Node
# ss & se  --> Starting and ending indexes of the
# segment represented by
# current Node, i.e., tree[index]
# qs & qe  --> Starting and ending indexes of query range
def queryUtil(tree, ss, se, qs, qe, index):
    # No overlap
    if ss > qe or se < qs:
        # returns a Node for out of bounds condition
        return Node()
    # Complete overlap
    if ss >= qs and se <= qe:
        return tree[index]
    # Partial Overlap
    mid = (ss + se) // 2
    left = queryUtil(tree, ss, mid, qs, qe, 2 * index)
    right = queryUtil(tree, mid + 1, se, qs, qe, 2 * index + 1)
    return merge(left, right)

# Function to get the desired Maximum Sum Sub-Array
def query(tree, n, qs, qe):
    # pass the index of the root node
    return queryUtil(tree, 0, n - 1, qs, qe, 1)

# Driver code
if __name__ == "__main__":
    # test array
    arr = [1, 3, -4, 5, -2]
    n = len(arr)
    # Construct Segment Tree
    Tree = constructTree(arr, n)
    # Answering query 1:
    start = 0
    end = 4
    maxSubarraySum = query(Tree, n, start, end).maxSubarraySum
    print("Maximum Sub-Array Sum between", start, "and", end, "=", maxSubarraySum)

    # Answering query 2:
    start = 0
    end = 2
    maxSubarraySum = query(Tree, n, start, end).maxSubarraySum
    print("Maximum Sub-Array Sum between", start, "and", end, "=", maxSubarraySum)

# This code is contributed by Vikram_Shirsat

C#

// C# Program to Implement Maximum Sub-Array Sum in a range
using System;

/* Node of the segment tree consisting of:
    1. Maximum Prefix Sum,
    2. Maximum Suffix Sum,
    3. Total Sum,
    4. Maximum Sub-Array Sum
*/
public class Node {
    public int maxPrefixSum;
    public int maxSuffixSum;
    public int totalSum;
    public int maxSubarraySum;

    public Node() {
        maxPrefixSum = maxSuffixSum = maxSubarraySum = int.MinValue;
        totalSum = int.MinValue;
    }
}

public class GFG {
    static readonly int inf = 0x3f3f;

    // Returns Parent Node after merging its left and right child
    static Node merge(Node leftChild, Node rightChild) {
        Node parentNode = new Node();
        parentNode.maxPrefixSum = Math.Max(
                                      leftChild.maxPrefixSum,
                                      leftChild.totalSum + rightChild.maxPrefixSum
                                  );
        parentNode.maxSuffixSum = Math.Max(
                                      rightChild.maxSuffixSum,
                                      rightChild.totalSum + leftChild.maxSuffixSum
                                  );
        parentNode.totalSum = leftChild.totalSum + rightChild.totalSum;
        parentNode.maxSubarraySum = Math.Max(
                                        Math.Max(leftChild.maxSubarraySum, rightChild.maxSubarraySum),
                                        leftChild.maxSuffixSum + rightChild.maxPrefixSum
                                    );
        return parentNode;
    }

    // Builds the Segment tree recursively
    static void constructTreeUtil(Node[] tree, int[] arr, int start, int end, int index) {
        /* Leaf Node */
        if (start == end) {
            // single element is covered under this range
            tree[index].totalSum = arr[start];
            tree[index].maxSuffixSum = arr[start];
            tree[index].maxPrefixSum = arr[start];
            tree[index].maxSubarraySum = arr[start];
            return;
        }

        // Recursively Build left and right children
        int mid = (start + end) / 2;
        constructTreeUtil(tree, arr, start, mid, 2 * index);
        constructTreeUtil(tree, arr, mid + 1, end, 2 * index + 1);

        // Merge left and right child into the Parent Node
        tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
    }

    /* Function to construct segment tree from given array.
     * This function allocates memory for segment tree and
     * calls constructTreeUtil() to fill the allocated
     * memory */
    static Node[] constructTree(int[] arr, int n) {
        // Allocate memory for segment tree
        int x = (int)(Math.Ceiling(Math.Log(n) / Math.Log(2)));

        // Maximum size of segment tree
        int max_size = 2 * (int)Math.Pow(2, x) - 1;
        Node[] tree = new Node[max_size];
        for (int i = 0; i < max_size; i++)
            tree[i] = new Node();

        // Fill the allocated memory tree
        constructTreeUtil(tree, arr, 0, n - 1, 1);
        // Return the constructed segment tree
        return tree;
    }

    /* A Recursive function to get the desired
    Maximum Sum Sub-Array,
    The following are parameters of the function-

    tree     --> Pointer to segment tree
    index --> Index of the segment tree Node
    ss & se  --> Starting and ending indexes of the
         segment represented by
             current Node, i.e., tree[index]
    qs & qe  --> Starting and ending indexes of query range */
    static Node queryUtil(Node[] tree, int ss, int se, int qs, int qe, int index) {
        if (ss > qe || se < qs) {
            Node nullNode = new Node();
            return nullNode;
        }
        if (ss >= qs && se <= qe) {
            return tree[index];
        }
        int mid = (ss + se) / 2;
        Node left = queryUtil(tree, ss, mid, qs, qe, 2 * index);
        Node right = queryUtil(tree, mid + 1, se, qs, qe, 2 * index + 1);
        Node res = merge(left, right);
        return res;
    }

    /* Returns the Maximum Subarray Sum between start and
        * end It mainly uses queryUtil(). */
    static int query(Node[] tree, int start, int end, int n) {
        Node res = queryUtil(tree, 0, n - 1, start, end, 1);
        return res.maxSubarraySum;
    }

    public static void Main(string[] args) {
        int[] arr = { 1, 3, -4, 5, -2 };
        int n = arr.Length;
        Node[] Tree = constructTree(arr, n);

        int start, end, maxSubarraySum;

        // Answering query 1:
        start = 0;
        end = 4;
        maxSubarraySum = query(Tree, start, end, n);
        Console.WriteLine("Maximum Sub-Array Sum between " + start + " and " + end + " = " + maxSubarraySum);

        // Answering query 2:
        start = 0;
        end = 2;
        maxSubarraySum = query(Tree, start, end, n);
        Console.WriteLine("Maximum Sub-Array Sum between " + start + " and " + end + " = " + maxSubarraySum);
    }
}

Javascript

// JavaScript Program to Implement Maximum Sub-Array Sum in a range

/* Node of the segment tree consisting of:
1. Maximum Prefix Sum,
2. Maximum Suffix Sum,
3. Total Sum,
4. Maximum Sub-Array Sum */
class Node {
    constructor() {
        this.maxPrefixSum = -Infinity;
        this.maxSuffixSum = -Infinity;
        this.totalSum = -Infinity;
        this.maxSubarraySum = -Infinity;
    }
}

const inf = 0x3f3f;

// Returns Parent Node after merging its left and right child
function merge(leftChild, rightChild) {
    const parentNode = new Node();
    parentNode.maxPrefixSum = Math.max(
        leftChild.maxPrefixSum,
        leftChild.totalSum + rightChild.maxPrefixSum
    );
    parentNode.maxSuffixSum = Math.max(
        rightChild.maxSuffixSum,
        rightChild.totalSum + leftChild.maxSuffixSum
    );
    parentNode.totalSum = leftChild.totalSum + rightChild.totalSum;
    parentNode.maxSubarraySum = Math.max(
        Math.max(
            leftChild.maxSubarraySum,
            rightChild.maxSubarraySum
        ),
        leftChild.maxSuffixSum + rightChild.maxPrefixSum
    );
    return parentNode;
}

// Builds the Segment tree recursively
function constructTreeUtil(tree, arr, start, end, index) {
    /* Leaf Node */
    if (start === end) {
        // single element is covered under this range
        tree[index].totalSum = arr[start];
        tree[index].maxSuffixSum = arr[start];
        tree[index].maxPrefixSum = arr[start];
        tree[index].maxSubarraySum = arr[start];
        return;
    }

    // Recursively Build left and right children
    const mid = Math.floor((start + end) / 2);
    constructTreeUtil(tree, arr, start, mid, 2 * index);
    constructTreeUtil(tree, arr, mid + 1, end, 2 * index + 1);

    // Merge left and right child into the Parent Node
    tree[index] = merge(tree[2 * index], tree[2 * index + 1]);
}

/* Function to construct segment tree from given array.
* This function allocates memory for segment tree and
* calls constructTreeUtil() to fill the allocated
* memory */
function constructTree(arr, n) {
    // Allocate memory for segment tree
    const x = Math.ceil(Math.log2(n)); // Height of the tree

    // Maximum size of segment tree
    const max_size = 2 * Math.pow(2, x) - 1;
    const tree = new Array(max_size);
    for (let i = 0; i < max_size; i++)
        tree[i] = new Node();

    // Fill the allocated memory tree
    constructTreeUtil(tree, arr, 0, n - 1, 1);

    // Return the constructed segment tree
    return tree;
}

/* A Recursive function to get the desired Maximum Sum Sub-Array,
The following are parameters of the function-

tree     --> Pointer to segment tree
index --> Index of the segment tree Node
ss & se  --> Starting and ending indexes of the
         segment represented by
             current Node, i.e., tree[index]
qs & qe  --> Starting and ending indexes of query range */
function queryUtil(tree, ss, se, qs, qe, index) {
    // No overlap
    if (ss > qe || se < qs) {
    // returns a Node for out of bounds condition
    const nullNode = new Node();
    return nullNode;
    }

    // Complete overlap
    if (qs <= ss && qe >= se) {
    return tree[index];
    }

    // Partial Overlap
    const mid = Math.floor((ss + se) / 2);
    const leftChild = queryUtil(tree, ss, mid, qs, qe, 2 * index);
    const rightChild = queryUtil(tree, mid + 1, se, qs, qe, 2 * index + 1);
    return merge(leftChild, rightChild);
}

/* Function to answer the queries */
function query(tree, start, end, n) {
    const res = queryUtil(tree, 0, n - 1, start, end, 1);
    return res.maxSubarraySum;
}

/* Main Function */
const arr = [1, 3, -4, 5, -2];
const n = arr.length;

// Construct Segment Tree
const Tree = constructTree(arr, n);
let start, end, maxSubarraySum;

// Answering query 1:
start = 0;
end = 4;
maxSubarraySum = query(Tree, start, end, n);
console.log(`Maximum Sub-Array Sum between ${start} and ${end} = ${maxSubarraySum} <br>`);

// Answering query 2:
start = 0;
end = 2;
maxSubarraySum = query(Tree, start, end, n);
console.log(`Maximum Sub-Array Sum between ${start} and ${end} = ${maxSubarraySum}`);

// This code is contributed by karthik.

Output:

Maximum Sub-Array Sum between 0 and 4 = 5
Maximum Sub-Array Sum between 0 and 2 = 4

Time Complexity: O(logn) for each query.