# 1770. Maximum Score from Performing Multiplication Operations

#### Medium

***

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:**

<pre><code>Input: nums = [1,2,3], multipliers = [3,2,1]
<strong>Output:
</strong> 14
<strong>Explanation:
</strong> 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.
</code></pre>

**Example 2:**

<pre><code>Input: nums = [-5,-3,-3,-2,7,1], multipliers = [-10,-5,3,4,6]
<strong>Output:
</strong> 102
<strong>Explanation: 
</strong>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.
</code></pre>

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**Constraints:**

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

```python
class Solution:
    def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
        n = len(nums)
        m = len(multipliers)
        @lru_cache(2000)
        def dp(left, i):
            if i == m:
                return 0
            l = dp(left+1, i+1) + nums[left]*multipliers[i]
            # l + r == m -> r = m-l
            r = dp(left, i+1) + nums[n - (i-left) - 1]*multipliers[i]
            return max(l,r)
        return dp(0,0)
```