Maximize Array sum by replacing middle elements with min of Subarray corners

Last Updated : 04 Dec, 2023

Given an array A[] of length N. Then your task is to output the maximum sum that can be achieved by using the given operation at any number of times (possibly zero):

Select a subarray, whose length is at least 3.
Replace all the middle elements (Except the first and last) with a minimum of elements at both ends.

Examples:

Input: N = 3, A[] = {8, 1, 7}
Output: 22
Explanation: Let us take the subarray A[1, 3] = {8, 1, 7}. Middle element(s) are: {1}. Now minimum of A[1] and A[3] is = min(A[1], A[3]) = min(8, 7) = 7. Then, we replaced all the middle elements equal to min according to the operation. So, updated A[] = {8, 7, 7}. The sum of updated A[] = 22. Which is the maximum possible using a given operation. Therefore, output is 22.

Input: N = 2, A[] = {5, 2}
Output: 7
Explanation: No subarray of length at least 3 can be chosen, Thus can't apply any operation. So the maximum possible sum is 7.

Approach: Implement the idea below to solve the problem

The problem is based on the observations and can be solved using the Prefix and Suffix arrays. It must be noticed that for any index i, we can not make the element A[i] to be anything more than the minimum of the greatest element to its left and the greatest element to its right.

Steps were taken to solve the problem:

Create two arrays let say Prefix[] and Suffix[]
Set prefix[0] = A[0]
Run a loop for i = 1 to i < N and follow the below-mentioned steps under the scope of the loop
- prefix[i] = max(prefix[i - 1], A[i])
Set suffix[N - 1] = A[N - 1]
Run a loop for i = N - 2 to I >=0 and follow the below-mentioned steps under the scope of the loop
- suffix[i] = max(suffix[i + 1], A[i])
Create a variable let say Sum to store the max possible sum.
Run a loop for i = 1 to i < N - 1 and follow below mentioned steps under the scope of loop
- Max1 = Prefix[i]
- Max2 = Suffix[i]
- sum += min(Max1, Max2)
Sum += A[0] + A[N - 1]
Output the value store in Sum.

Code to implement the approach:

C++

//code by FLutterfly #include <iostream> #include <algorithm>  using namespace std;  // Function prototype void Max_sum(int N, int A[]);  int main() {     // Input     int N = 2;     int A[] = {5, 2};      // Function call     Max_sum(N, A);      return 0; }  void Max_sum(int N, int A[]) {     // Prefix and Suffix Arrays     int prefix[N];     int suffix[N];      // Initializing prefix array     prefix[0] = A[0];     for (int i = 1; i < N; i++)     {         prefix[i] = max(prefix[i - 1], A[i]);     }      // Initializing suffix array     suffix[N - 1] = A[N - 1];     for (int i = N - 2; i >= 0; i--)     {         suffix[i] = max(suffix[i + 1], A[i]);     }      // Variable to store max sum     long long sum = 0;      // Calculating sum for all middle elements     for (int i = 1; i < N - 1; i++)     {         int maxi1 = prefix[i], maxi2 = suffix[i];         sum += min(maxi1, maxi2);     }      // Adding last elements of both ends into sum     sum += A[0] + A[N - 1];      // Printing the value of sum     cout << sum << endl; }

Java

// Java code to implement the approach import java.util.*;  class Main {     public static void main(String[] args)     {         // Input         int N = 2;         int A[] = { 5, 2 };          // Function call         Max_sum(N, A);     }     public static void Max_sum(int N, int[] A)     {         // Prefix and Suffix Arrays         int[] prefix = new int[N];         int[] suffix = new int[N];          // Initializing prefix array         prefix[0] = A[0];         for (int i = 1; i < N; i++) {             prefix[i] = Math.max(prefix[i - 1], A[i]);         }          // Initializing suffix array         suffix[N - 1] = A[N - 1];         for (int i = N - 2; i >= 0; i--) {             suffix[i] = Math.max(suffix[i + 1], A[i]);         }          // Variable to store max sum         long sum = 0;          // Calculating sum for all middle elements         for (int i = 1; i < N - 1; i++) {             int maxi1 = prefix[i], maxi2 = suffix[i];             sum += Math.min(maxi1, maxi2);         }          // Adding last elements of both ends into sum         sum += A[0] + A[N - 1];          // Printing the value of sum         System.out.println(sum);     } }

Python

# code by flutterfly def Max_sum(N, A):     # Prefix and Suffix Arrays     prefix = [0] * N     suffix = [0] * N      # Initializing prefix array     prefix[0] = A[0]     for i in range(1, N):         prefix[i] = max(prefix[i - 1], A[i])      # Initializing suffix array     suffix[N - 1] = A[N - 1]     for i in range(N - 2, -1, -1):         suffix[i] = max(suffix[i + 1], A[i])      # Variable to store max sum     sum_val = 0      # Calculating sum for all middle elements     for i in range(1, N - 1):         maxi1, maxi2 = prefix[i], suffix[i]         sum_val += min(maxi1, maxi2)      # Adding last elements of both ends into sum     sum_val += A[0] + A[N - 1]      # Printing the value of sum     print(sum_val)  # Input N = 2 A = [5, 2]  # Function call Max_sum(N, A)

//code by flutterfly using System;  class Program {     static void Main()     {         // Input         int N = 2;         int[] A = { 5, 2 };          // Function call         Max_sum(N, A);     }      static void Max_sum(int N, int[] A)     {         // Prefix and Suffix Arrays         int[] prefix = new int[N];         int[] suffix = new int[N];          // Initializing prefix array         prefix[0] = A[0];         for (int i = 1; i < N; i++)         {             prefix[i] = Math.Max(prefix[i - 1], A[i]);         }          // Initializing suffix array         suffix[N - 1] = A[N - 1];         for (int i = N - 2; i >= 0; i--)         {             suffix[i] = Math.Max(suffix[i + 1], A[i]);         }          // Variable to store max sum         long sum = 0;          // Calculating sum for all middle elements         for (int i = 1; i < N - 1; i++)         {             int maxi1 = prefix[i], maxi2 = suffix[i];             sum += Math.Min(maxi1, maxi2);         }          // Adding last elements of both ends into sum         sum += A[0] + A[N - 1];          // Printing the value of sum         Console.WriteLine(sum);     } }

JavaScript

function Max_sum(N, A) {     // Prefix and Suffix Arrays     let prefix = new Array(N);     let suffix = new Array(N);      // Initializing prefix array     prefix[0] = A[0];     for (let i = 1; i < N; i++) {         prefix[i] = Math.max(prefix[i - 1], A[i]);     }      // Initializing suffix array     suffix[N - 1] = A[N - 1];     for (let i = N - 2; i >= 0; i--) {         suffix[i] = Math.max(suffix[i + 1], A[i]);     }      // Variable to store max sum     let sum = 0;      // Calculating sum for all middle elements     for (let i = 1; i < N - 1; i++) {         let maxi1 = prefix[i], maxi2 = suffix[i];         sum += Math.min(maxi1, maxi2);     }      // Adding last elements of both ends into sum     sum += A[0] + A[N - 1];      // Printing the value of sum     console.log(sum); }  // Input let N = 2; let A = [5, 2];  // Function call Max_sum(N, A);