Sum of maximum of all subarrays | Divide and Conquer

Last Updated : 11 Jul, 2025

Given an array arr[] of length N, the task is to find the sum of the maximum elements of every possible sub-array of the array.
Examples:

Input : arr[] = {1, 3, 1, 7}
Output : 42
Max of all sub-arrays:
{1} - 1
{1, 3} - 3
{1, 3, 1} - 3
{1, 3, 1, 7} - 7
{3} - 3
{3, 1} - 3
{3, 1, 7} - 7 
{1} - 1
{1, 7} - 7
{7} - 7
1 + 3 + 3 + 7 + 3 + 3 + 7 + 1 + 7 + 7 = 42

Input : arr[] = {1, 1, 1, 1, 1}
Output : 15

We have already discussed an O(N) approach using stack for this problem in this article.
Approach :
In this article, we will learn how to solve this problem using divide and conquer.
Let's assume that element at i^th index is largest of all. For any sub-array that contains index 'i', the element at 'i' will always be maximum in the sub-array.
If element at i^th index is largest, we can safely say, that element ith index will be largest in (i+1)*(N-i) subarrays. So, its total contribution will be arr[i]*(i+1)*(N-i). Now, we will divide the array in two parts, (0, i-1) and (i+1, N-1) and apply the same algorithms to both of them separately.
So our general recurrence relation will be:

maxSumSubarray(arr, l, r) = arr[i]*(r-i+1)*(i-l+1) 
                            + maxSumSubarray(arr, l, i-1)
                            + maxSumSubarray(arr, i+1, r)
where i is index of maximum element in range [l, r].

Now, we need a way to efficiently answer rangeMax() queries. Segment tree will be an efficient way to answer this query. We will need to answer this query at most N times. Thus, the time complexity of our divide and conquer algorithm will O(Nlog(N)).
If we have to answer the problem "Sum of minimum of all subarrays" then we will use the segment tree to answer rangeMin() queries. For this, you can go through the article segment tree range minimum.
Below is the implementation code:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
#define seg_max 51
using namespace std;

// Array to store segment tree.
// In first we will store the maximum
// of a range
// In second, we will store index of
// that range
pair<int, int> seg_tree[seg_max];

// Size of array declared global
// to maintain simplicity in code
int n;

// Function to build segment tree
pair<int, int> buildMaxTree(int l, int r, int i, int arr[])
{
    // Base case
    if (l == r) {
        seg_tree[i] = { arr[l], l };
        return seg_tree[i];
    }

    // Finding the maximum among left and right child
    seg_tree[i] = max(buildMaxTree(l, (l + r) / 2, 2 * i + 1, arr),
                      buildMaxTree((l + r) / 2 + 1, r, 2 * i + 2, arr));

    // Returning the maximum to parent
    return seg_tree[i];
}

// Function to perform range-max query in segment tree
pair<int, int> rangeMax(int l, int r, int arr[],
                        int i = 0, int sl = 0, int sr = n - 1)
{
    // Base cases
    if (sr < l || sl > r)
        return { INT_MIN, -1 };
    if (sl >= l and sr <= r)
        return seg_tree[i];

    // Finding the maximum among left and right child
    return max(rangeMax(l, r, arr, 2 * i + 1, sl, (sl + sr) / 2),
               rangeMax(l, r, arr, 2 * i + 2, (sl + sr) / 2 + 1, sr));
}

// Function to find maximum sum subarray
int maxSumSubarray(int arr[], int l = 0, int r = n - 1)
{
    // base case
    if (l > r)
        return 0;

    // range-max query to determine
    // largest in the range.
    pair<int, int> a = rangeMax(l, r, arr);

    // divide the array in two parts
    return a.first * (r - a.second + 1) * (a.second - l + 1)
           + maxSumSubarray(arr, l, a.second - 1)
           + maxSumSubarray(arr, a.second + 1, r);
}

// Driver Code
int main()
{
    // Input array
    int arr[] = { 1, 3, 1, 7 };

    // Size of array
    n = sizeof(arr) / sizeof(int);

    // Builind the segment-tree
    buildMaxTree(0, n - 1, 0, arr);

    cout << maxSumSubarray(arr);

    return 0;
}

Java

// Java implementation of the above approach
class GFG {
    static class pair {
        int first, second;

        public pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }

    static final int seg_max = 51;
    
    // Array to store segment tree.
    // In first we will store the maximum
    // of a range
    // In second, we will store index of
    // that range
    static pair[] seg_tree = new pair[seg_max];

    // Size of array declared global
    // to maintain simplicity in code
    static int n;

    // Function to build segment tree
    static pair buildMaxTree(int l, int r, int i, int arr[]) 
    {
        // Base case
        if (l == r) {
            seg_tree[i] = new pair(arr[l], l);
            return seg_tree[i];
        }

        // Finding the maximum among left and right child
        seg_tree[i] = max(buildMaxTree(l, (l + r) / 2, 2 * i + 1, arr),
                buildMaxTree((l + r) / 2 + 1, r, 2 * i + 2, arr));

        // Returning the maximum to parent
        return seg_tree[i];
    }

    // Function to perform range-max query in segment tree
    static pair rangeMax(int l, int r, int arr[], 
                        int i, int sl, int sr)
    {
        // Base cases
        if (sr < l || sl > r)
            return new pair(Integer.MIN_VALUE, -1);
        if (sl >= l && sr <= r)
            return seg_tree[i];

        // Finding the maximum among left and right child
        return max(rangeMax(l, r, arr, 2 * i + 1, sl, (sl + sr) / 2),
                rangeMax(l, r, arr, 2 * i + 2, (sl + sr) / 2 + 1, sr));
    }

    static pair max(pair f, pair s) {
        if (f.first > s.first)
            return f;
        else
            return s;
    }

    // Function to find maximum sum subarray
    static int maxSumSubarray(int arr[], int l, int r)
    {
        // base case
        if (l > r)
            return 0;

        // range-max query to determine
        // largest in the range.
        pair a = rangeMax(l, r, arr, 0, 0, n - 1);

        // divide the array in two parts
        return a.first * (r - a.second + 1) * (a.second - l + 1) 
                + maxSumSubarray(arr, l, a.second - 1)
                + maxSumSubarray(arr, a.second + 1, r);
    }

    // Driver Code
    public static void main(String[] args)
    {
        // Input array
        int arr[] = { 1, 3, 1, 7 };

        // Size of array
        n = arr.length;

        // Builind the segment-tree
        buildMaxTree(0, n - 1, 0, arr);

        System.out.print(maxSumSubarray(arr, 0, n - 1));
    }
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the above approach
import sys 

seg_max = 51
      
# Array to store segment tree.
# In first we will store the maximum
# of a range
# In second, we will store index of
# that range
seg_tree = [[] for i in range(seg_max)]

# Function to build segment tree
def buildMaxTree(l, r, i, arr):
    global n, seg_tree, seg_max
    # Base case
    if l == r:
        seg_tree[i] = [arr[l], l]
        return seg_tree[i]

    # Finding the maximum among left and right child
    seg_tree[i] = max(buildMaxTree(l, int((l + r) / 2), 2 * i + 1, arr), buildMaxTree(int((l + r) / 2) + 1, r, 2 * i + 2, arr))

    # Returning the maximum to parent
    return seg_tree[i]

# Function to perform range-max query in segment tree
def rangeMax(l, r, arr, i, sl, sr):
    global n, seg_tree, seg_max
    
    # Base cases
    if sr < l or sl > r:
        return [-sys.maxsize, -1]
    if sl >= l and sr <= r:
        return seg_tree[i]

    # Finding the maximum among left and right child
    return max(rangeMax(l, r, arr, 2 * i + 1, sl, int((sl + sr) / 2)), rangeMax(l, r, arr, 2 * i + 2, int((sl + sr) / 2) + 1, sr))

def Max(f, s):
    if f[0] > s[0]:
        return f
    else:
        return s

# Function to find maximum sum subarray
def maxSumSubarray(arr, l, r):
    # base case
    if l > r:
        return 0

    # range-max query to determine
    # largest in the range.
    a = rangeMax(l, r, arr, 0, 0, n - 1)

    # divide the array in two parts
    return a[0] * (r - a[1] + 1) * (a[1] - l + 1) + maxSumSubarray(arr, l, a[1] - 1) + maxSumSubarray(arr, a[1] + 1, r)

# Input array
arr = [ 1, 3, 1, 7 ]

# Size of array
n = len(arr)

# Builind the segment-tree
buildMaxTree(0, n - 1, 0, arr)

print(maxSumSubarray(arr, 0, n - 1))

# This code is contributed by decode2207.

// C# implementation of the above approach
using System;

class GFG {
    class pair {
        public int first, second;
 
        public pair(int first, int second) {
            this.first = first;
            this.second = second;
        }
    }
 
    static readonly int seg_max = 51;
     
    // Array to store segment tree.
    // In first we will store the maximum
    // of a range
    // In second, we will store index of
    // that range
    static pair[] seg_tree = new pair[seg_max];
 
    // Size of array declared global
    // to maintain simplicity in code
    static int n;
 
    // Function to build segment tree
    static pair buildMaxTree(int l, int r, int i, int []arr) 
    {
        // Base case
        if (l == r) {
            seg_tree[i] = new pair(arr[l], l);
            return seg_tree[i];
        }
 
        // Finding the maximum among left and right child
        seg_tree[i] = max(buildMaxTree(l, (l + r) / 2, 2 * i + 1, arr),
                buildMaxTree((l + r) / 2 + 1, r, 2 * i + 2, arr));
 
        // Returning the maximum to parent
        return seg_tree[i];
    }
 
    // Function to perform range-max query in segment tree
    static pair rangeMax(int l, int r, int []arr, 
                        int i, int sl, int sr)
    {
        // Base cases
        if (sr < l || sl > r)
            return new pair(int.MinValue, -1);
        if (sl >= l && sr <= r)
            return seg_tree[i];
 
        // Finding the maximum among left and right child
        return max(rangeMax(l, r, arr, 2 * i + 1, sl, (sl + sr) / 2),
                rangeMax(l, r, arr, 2 * i + 2, (sl + sr) / 2 + 1, sr));
    }
 
    static pair max(pair f, pair s) {
        if (f.first > s.first)
            return f;
        else
            return s;
    }
 
    // Function to find maximum sum subarray
    static int maxSumSubarray(int []arr, int l, int r)
    {
        // base case
        if (l > r)
            return 0;
 
        // range-max query to determine
        // largest in the range.
        pair a = rangeMax(l, r, arr, 0, 0, n - 1);
 
        // divide the array in two parts
        return a.first * (r - a.second + 1) * (a.second - l + 1) 
                + maxSumSubarray(arr, l, a.second - 1)
                + maxSumSubarray(arr, a.second + 1, r);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        // Input array
        int []arr = { 1, 3, 1, 7 };
 
        // Size of array
        n = arr.Length;
 
        // Builind the segment-tree
        buildMaxTree(0, n - 1, 0, arr);
 
        Console.Write(maxSumSubarray(arr, 0, n - 1));
    }
}

// This code is contributed by PrinciRaj1992

JavaScript

<script>

// JavaScript implementation of the above approach

class pair
{
    constructor(first,second)
    {
        this.first = first;
            this.second = second;
    }
}

let seg_max = 51;

// Array to store segment tree.
    // In first we will store the maximum
    // of a range
    // In second, we will store index of
    // that range
let seg_tree = new Array(seg_max);

// Size of array declared global
    // to maintain simplicity in code
let n;

// Function to build segment tree
function buildMaxTree(l,r,i,arr)
{
     // Base case
        if (l == r) {
            seg_tree[i] = new pair(arr[l], l);
            return seg_tree[i];
        }
  
        // Finding the maximum among left and right child
        seg_tree[i] = max(buildMaxTree(l, Math.floor((l + r) / 2), 
        2 * i + 1, arr), buildMaxTree(Math.floor((l + r) / 2) + 
        1, r, 2 * i + 2, arr));
  
        // Returning the maximum to parent
        return seg_tree[i];
}

// Function to perform range-max query in segment tree
function rangeMax(l,r,arr,i,sl,sr)
{
    // Base cases
        if (sr < l || sl > r)
            return new pair(Number.MIN_VALUE, -1);
        if (sl >= l && sr <= r)
            return seg_tree[i];
  
        // Finding the maximum among left and right child
        return max(rangeMax(l, r, arr, 2 * i + 1, sl, 
                   Math.floor((sl + sr) / 2)),
                rangeMax(l, r, arr, 2 * i + 2, 
                Math.floor((sl + sr) / 2) + 1, sr));
}

function max(f,s)
{
    if (f.first > s.first)
            return f;
        else
            return s;
}

// Function to find maximum sum subarray
function maxSumSubarray(arr,l,r)
{
    // base case
        if (l > r)
            return 0;
  
        // range-max query to determine
        // largest in the range.
        let a = rangeMax(l, r, arr, 0, 0, n - 1);
  
        // divide the array in two parts
        return a.first * (r - a.second + 1) * (a.second - l + 1) 
                + maxSumSubarray(arr, l, a.second - 1)
                + maxSumSubarray(arr, a.second + 1, r);
}

// Driver Code
// Input array
let arr = [ 1, 3, 1, 7 ];

// Size of array
n = arr.length;

// Builind the segment-tree
buildMaxTree(0, n - 1, 0, arr);

document.write(maxSumSubarray(arr, 0, n - 1));

// This code is contributed by rag2127

</script>

Output:

Time complexity : O(Nlog(N))

Analysis of Algorithms

DivyanshuShekhar1

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Article Tags :

Sum of maximum of all subarrays | Divide and Conquer

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