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

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=3Output:

4Explanation: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=3Output:

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*2^{N})**Auxiliary Space: **O(1)

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

- 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. - 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:

- Initialize a variable
**Sum**to 0. - Traverse
**arr**from**1**to**N-1,**using the variable**i**, and do the following:- 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].**

- If

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `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[0])));` ` ` `// 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

`<script>` `// JavaScript program for the above approach` `// Function to find the required answer` `function` `maximumSum(arr, N)` `{` ` ` ` ` `// Stores maximum sum` ` ` `let Sum = 0;` ` ` `for` `(let 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` `// Input` `let arr = [ 1, 5, 3 ];` `let N = arr.length;` `// Function calling` `document.write(maximumSum(arr, N));` `// This code is contributed by Potta Lokesh` `</script>` |

**Output**

4

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

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