# LeetCode 1770. Maximum Score from Performing Multiplication Operations

You are given two integer arrays `nums` and `multipliers` of size `n` and `m` respectively, where `n >= m`. The arrays are **1-indexed**.

You begin with a score of `0`. You want to perform **exactly** `m` operations. On the `ith` operation **(1-indexed)**, you will:

* Choose one integer `x` from **either the start or the end** of the array `nums`.
* Add `multipliers[i] * x` to your score.
* Remove `x` from the array `nums`.

Return *the **maximum** score after performing* `m` *operations.*

**Example 1:**

```
Input: nums = [1,2,3], multipliers = [3,2,1]
Output: 14
Explanation: An optimal solution is as follows:
- Choose from the end, [1,2,3], adding 3 * 3 = 9 to the score.
- Choose from the end, [1,2], adding 2 * 2 = 4 to the score.
- Choose from the end, [1], adding 1 * 1 = 1 to the score.
The total score is 9 + 4 + 1 = 14.
```

**Example 2:**

```
Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]
Output: 102
Explanation: An optimal solution is as follows:
- Choose from the start, [-5,-3,-3,-2,7,1], adding -5 * -10 = 50 to the score.
- Choose from the start, [-3,-3,-2,7,1], adding -3 * -5 = 15 to the score.
- Choose from the start, [-3,-2,7,1], adding -3 * 3 = -9 to the score.
- Choose from the end, [-2,7,1], adding 1 * 4 = 4 to the score.
- Choose from the end, [-2,7], adding 7 * 6 = 42 to the score. 
The total score is 50 + 15 - 9 + 4 + 42 = 102.
```

**Constraints:**

* `n == nums.length`
* `m == multipliers.length`
* `1 <= m <= 103`
* `m <= n <= 105`
* `-1000 <= nums[i], multipliers[i] <= 1000`

## Solution:

[English Version in Youtube](https://youtu.be/ggv0pK4BRuI)

[中文版解答Youtube Link](https://youtu.be/SBOp20CzqK0)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1QA411M7Da/)

```
class Solution {
public:
    int maximumScore(vector<int>& nums, vector<int>& multipliers) {
        int n = nums.size();
        int m = multipliers.size();
        vector<vector<int>> dp(m+1, vector<int>(m+1, 0));
        
        int score = INT_MIN;
        for (int k = 1; k <= m; k++) {
            for (int l = 0; l <= k; l++) {
                int pick_left = (l == 0 ? INT_MIN : dp[l-1][k-l] + multipliers[k-1] * nums[l-1]);
                int pick_right = (l == k ? INT_MIN : dp[l][k-l-1] + multipliers[k-1] * nums[n-k+l]);
                dp[l][k-l] = max(pick_left, pick_right);

                if (k == m) {
                    score = max(score, dp[l][k-l]);
                }
            }
        }
        
        return score;
    }
};
```