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 C++ implementation of this approach:

`// 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; ` `} ` |

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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

## Recommended Posts:

- Maximize the sum of selected numbers from an array to make it empty
- Number of decimal numbers of length k, that are strict monotone
- Longest alternating (positive and negative) subarray starting at every index
- Minimum number of increment-other operations to make all array elements equal.
- Number of ways to arrange N items under given constraints
- Find element at given index after a number of rotations
- Maximize the binary matrix by filpping submatrix once
- Count all triplets whose sum is equal to a perfect cube
- Maximum sum bitonic subarray
- Unique paths in a Grid with Obstacles
- Convert to Strictly increasing array with minimum changes
- Smallest sum contiguous subarray
- Number of n-digits non-decreasing integers
- Size of array after repeated deletion of LIS
- Longest Increasing Subsequence | DP-3