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=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*2^N)
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++ 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])));  // Function calling  cout << maximumSum(arr, N);  return 0; } 

// 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 

# 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) # Function calling print(maximumSum(arr, N)); # This code is contributed by SoumikMondal 

// 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;  // Function calling  Console.Write(maximumSum(arr, N)); } } // This code is contributed by SURENDRA_GANGWAR 

<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> 

4

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