Related Articles

# Maximum sum obtained by dividing Array into several subarrays as per given conditions

• Difficulty Level : Basic
• Last Updated : 06 Jul, 2021

Given an array arr[] of size N, the task is to calculate the maximum sum that can be obtained by dividing the array into several subarrays(possibly one), where each subarray starting at index i and ending at index j (j>=i) contributes arr[j]-arr[i] to the sum.

Examples:

Input: arr[]= {1, 5, 3}, N=3
Output:
4
Explanation: The array can be divided into 2 subarrays:

• {1, 5} -> sum contributed by the subarray = 5-1 = 4
• {3} -> sum contributed by the subarray = 3-3 = 0

Therefore, the answer is 4.(It can be shown that there is no other way of dividing this array in multiple subarrays such that the answer is greater than 4).

Input: arr[] = {6, 2, 1}, N=3
Output:
0

Naive Approach: The naive approach is to consider all possible ways of dividing arr into 1 or more subarrays and calculating the maximum sum obtained for each.

Time Complexity: O(N*2N)
Auxiliary Space: O(1)

Observation: The observations necessary to solve the problem are below:

1. The array should be divided into several(possibly one) subarrays such that each subarray is the longest increasing subarray. For example, if arr[]={3,5,7,9,1}, it is optimal to consider {3,5,7,9} as a subarray as it would contribute 9-3=6 to the sum. Breaking it up further decreases the sum which is not optimal.
2. Every element of a non-increasing subarray should be considered as single element subarrays so that they contribute 0 to the sum. Otherwise, they would be contributing a negative value. For example, if arr[i]>arr[i+1], it is optimal to consider two subarrays of length 1 containing arr[i] and arr[i+1] separately, so that they contribute (arr[i]-arr[i]) +(arr[i+1]-arr[i+1]) =0 to the answer. If they were considered together, they would contribute arr[i+1]-arr[i] which is a negative number, thus decreasing the sum.

Efficient Approach: Follow the steps below to solve the problem:

1. Initialize a variable Sum to 0.
2. Traverse arr from 1 to N-1, using the variable i, and do the following:
1. If arr[i]>arr[i-1], add arr[i]-arr[i-1] to Sum. This works because the sum of differences of adjacent elements in a sorted array is equal to the difference of the elements at extreme ends. Here, only the increasing subarrays are considered as arr[i]>arr[i-1].

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the required answer``int` `maximumSum(``int` `arr[], ``int` `N)``{``    ``// Stores maximum sum``    ``int` `Sum = 0;``    ``for` `(``int` `i = 1; i < N; i++) {` `        ``// Adding the difference of elements at ends of``        ``// increasing subarray to the answer``        ``if` `(arr[i] > arr[i - 1])``            ``Sum += (arr[i] - arr[i - 1]);``    ``}``    ``return` `Sum;``}``// Driver Code``int` `main()``{``    ``// Input``    ``int` `arr[] = { 1, 5, 3 };``    ``int` `N = (``sizeof``(arr) / (``sizeof``(arr)));` `    ``// Functio calling``    ``cout << maximumSum(arr, N);``    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{``    ` `// Function to find the required answer``public` `static` `int` `maximumSum(``int` `arr[], ``int` `N)``{``    ` `    ``// Stores maximum sum``    ``int` `Sum = ``0``;``    ``for``(``int` `i = ``1``; i < N; i++)``    ``{``        ` `        ``// Adding the difference of elements at ends``        ``// of increasing subarray to the answer``        ``if` `(arr[i] > arr[i - ``1``])``            ``Sum += (arr[i] - arr[i - ``1``]);``    ``}``    ``return` `Sum;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Input``    ``int` `arr[] = { ``1``, ``5``, ``3` `};``    ``int` `N = arr.length;` `    ``// Function calling``    ``System.out.println(maximumSum(arr, N));``}``}` `// This code is contributed by Potta Lokesh`

## Python3

 `# Python program for the above approach` `# Function to find the required answer``def` `maximumSum(arr, N):``  ` `    ``# Stores maximum sum``    ``Sum` `=` `0``;``    ``for` `i ``in` `range``(``1``,N):``      ` `        ``# Adding the difference of elements at ends of``        ``# increasing subarray to the answer``        ``if` `(arr[i] > arr[i ``-` `1``]):``            ``Sum` `+``=` `(arr[i] ``-` `arr[i ``-` `1``])``    ` `    ``return` `Sum``;``    ` `# Driver Code` `#Input``arr ``=` `[ ``1``, ``5``, ``3` `];``N ``=` `len``(arr)` `# Functio calling``print``(maximumSum(arr, N));` `# This code is contributed by SoumikMondal`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find the required answer``static` `int` `maximumSum(``int` `[]arr, ``int` `N)``{``    ` `    ``// Stores maximum sum``    ``int` `Sum = 0;``    ``for``(``int` `i = 1; i < N; i++)``    ``{``        ` `        ``// Adding the difference of elements at``        ``// ends of increasing subarray to the answer``        ``if` `(arr[i] > arr[i - 1])``            ``Sum += (arr[i] - arr[i - 1]);``    ``}``    ``return` `Sum;``}` `// Driver Code``public` `static` `void` `Main()``{``    ` `    ``// Input``    ``int` `[]arr = { 1, 5, 3 };``    ``int` `N = arr.Length;` `    ``// Functio calling``    ``Console.Write(maximumSum(arr, N));``}``}` `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``
Output
`4`

Time Complexity: O(N)
Auxiliary Space: O(1)

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.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up