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:
- 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].
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 answerint 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 Codeint main(){ // Input int arr[] = { 1, 5, 3 }; int N = (sizeof(arr) / (sizeof(arr[0]))); // Function calling cout << maximumSum(arr, N); return 0;} |
Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to find the required answerpublic 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 Codepublic 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 answerdef 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#Inputarr = [ 1, 5, 3 ];N = len(arr)# Function callingprint(maximumSum(arr, N));# This code is contributed by SoumikMondal |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG{// Function to find the required answerstatic 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 Codepublic 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 |
Javascript
<script>// JavaScript program for the above approach// Function to find the required answerfunction 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// Inputlet arr = [ 1, 5, 3 ];let N = arr.length;// Function callingdocument.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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