Sum of maximum of all subarrays | Divide and Conquer

Hello Everyone,

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

Approach :
In this article, we will learn how to solve this problem using divide and conquer.
Let’s assume that element at ith 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 ith 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++ 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;

}

Output:

42

Time complexity : O(Nlog(N))