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Remove array end element to maximize the sum of product
• Difficulty Level : Medium
• Last Updated : 29 Apr, 2021

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[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 implementation of this approach:

## C++

 `// CPP program to find maximum score we can get``// by removing elements from either end.``#include ``#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);` `    ``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

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## Javascript

 `   `

Output:

`43`

This article is contributed by Anuj Chauhan. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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