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Remove array end element to maximize the sum of product

Given an array of N positive integers. We are allowed to remove element from either of the two ends i.e from the left side or right side of the array. Each time we remove an element, score is increased by value of element * (number of element already removed + 1). The task is to find the maximum score that can be obtained by removing all the element.

Examples:

Input : arr[] = { 1, 3, 1, 5, 2 }.
Output : 43
Remove 1 from left side (score = 1*1 = 1)
then remove 2, score = 1 + 2*2 = 5
then remove 3, score = 5 + 3*3 = 14
then remove 1, score = 14 + 1*4 = 18
then remove 5, score = 18 + 5*5 = 43.

Input :  arr[] = { 1, 2 }
Output : 5.



The idea is to use Dynamic Programming. Make a 2D matrix named dp[][] initialised with 0, where dp[i][j] denote the maximum value of score from index from index ito index j of the array. So, our final result will be stored in dp[0][n-1].
Now, value for dp[i][j] will be maximum of arr[i] * (number of element already removed + 1) + dp[i+ 1][j] or arr[j] * (number of element already removed + 1) + dp[i][j – 1].

Below is the the implementation of this approach:

C++

// CPP program to find maximum score we can get
// by removing elements from either end.
#include <bits/stdc++.h>
#define MAX 50
using namespace std;
  
int solve(int dp[][MAX], int a[], int low, int high,
                                          int turn)
{
    // If only one element left.
    if (low == high)
        return a[low] * turn;
  
    // If already calculated, return the value.
    if (dp[low][high] != 0)
        return dp[low][high];
  
    // Computing Maximum value when element at 
    // index i and index j is to be chosed.
    dp[low][high] = max(a[low] * turn + solve(dp, a, 
                            low + 1, high, turn + 1),
                        a[high] * turn + solve(dp, a, 
                           low, high - 1, turn + 1));
  
    return dp[low][high];
}
  
// Driven Program
int main()
{
    int arr[] = { 1, 3, 1, 5, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    int dp[MAX][MAX];
    memset(dp, 0, sizeof(dp));
  
    cout << solve(dp, arr, 0, n - 1, 1) << endl;
    return 0;
}

Java

// Java program to find maximum score we can get
// by removing elements from either end.
  
public class GFG {
  
    static final int MAX = 50;
  
    static int solve(int dp[][], int a[], int low, int high,
            int turn) {
        // If only one element left.
        if (low == high) {
            return a[low] * turn;
        }
  
        // If already calculated, return the value.
        if (dp[low][high] != 0) {
            return dp[low][high];
        }
  
        // Computing Maximum value when element at 
        // index i and index j is to be chosed.
        dp[low][high] = Math.max(a[low] * turn + solve(dp, a,
                low + 1, high, turn + 1),
                a[high] * turn + solve(dp, a,
                        low, high - 1, turn + 1));
  
        return dp[low][high];
    }
  
// Driven Program
    public static void main(String args[]) {
        int arr[] = {1, 3, 1, 5, 2};
        int n = arr.length;
  
        int dp[][] = new int[MAX][MAX];
  
        System.out.println(solve(dp, arr, 0, n - 1, 1));
  
    }
}
  
/*This code is contributed by 29AjayKumar*/

Python 3

# Python 3 program to find maximum
# score we can get by removing
# elements from either end.
MAX = 50
  
def solve(dp, a, low, high, turn):
  
    # If only one element left.
    if (low == high):
        return a[low] * turn
  
    # If already calculated, 
    # return the value.
    if (dp[low][high] != 0):
        return dp[low][high]
  
    # Computing Maximum value when element  
    # at index i and index j is to be chosed.
    dp[low][high] = max(a[low] * turn + solve(dp, a, 
                          low + 1, high, turn + 1),
                        a[high] * turn + solve(dp, a, 
                          low, high - 1, turn + 1));
  
    return dp[low][high]
  
# Driver Code
if __name__ == "__main__":
    arr = [ 1, 3, 1, 5, 2 ]
    n = len(arr)
  
    dp = [[0 for x in range(MAX)]
             for y in range(MAX)]
  
    print(solve(dp, arr, 0, n - 1, 1))
  
# This code is contributed by ChitraNayal

C#

// C# program to find maximum score we can get 
// by removing elements from either end. 
using System; 
  
class GFG 
    static int MAX = 50; 
      
    static int solve(int[,] dp, int[] a, int low,
                            int high, int turn) 
    
        // If only one element left. 
        if (low == high) 
            return a[low] * turn; 
      
        // If already calculated, return the value. 
        if (dp[low, high] != 0) 
            return dp[low, high]; 
      
        // Computing Maximum value when element at 
        // index i and index j is to be chosed. 
        dp[low,high] = Math.Max(a[low] * turn + solve(dp, a, 
                                low + 1, high, turn + 1), 
                            a[high] * turn + solve(dp, a, 
                            low, high - 1, turn + 1)); 
      
        return dp[low, high]; 
    
      
    // Driven code
    static void Main() 
    
        int[] arr = new int[]{ 1, 3, 1, 5, 2 }; 
        int n = arr.Length;
      
        int[,] dp = new int[MAX,MAX]; 
        for(int i = 0; i < MAX; i++)
            for(int j = 0; j < MAX; j++)
                dp[i, j] = 0;
      
        Console.Write(solve(dp, arr, 0, n - 1, 1)); 
    }
}
  
// This code is contributed by DrRoot_

PHP

<?php
// PHP program to find maximum score we can 
// get by removing elements from either end. 
  
$MAX = 50 ;
  
function solve($dp, $a, $low, $high, $turn
    // If only one element left. 
    if ($low == $high
        return $a[$low] * $turn
  
    // If already calculated, return the value. 
    if ($dp[$low][$high] != 0) 
        return $dp[$low][$high]; 
  
    // Computing Maximum value when element at 
    // index i and index j is to be chosed. 
    $dp[$low][$high] = max($a[$low] * $turn
                     solve($dp, $a, $low + 1, 
                           $high, $turn + 1), 
                           $a[$high] * $turn +
                     solve($dp, $a, $low, $high
                                 1, $turn + 1)); 
  
    return $dp[$low][$high]; 
  
// Driver Code
$arr = array(1, 3, 1, 5, 2 ) ; 
$n = count($arr) ;
  
$dp = array() ;
for($i = 0; $i < $MAX ; $i++)
{
    $dp[$i] = array_fill($i, $MAX, 0) ;
}
  
echo solve($dp, $arr, 0, $n - 1, 1); 
  
// This code is contributed by Ryuga
?>


Output:

43

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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