Formatted question description: https://leetcode.ca/all/1770.html

# 1770. Maximum Score from Performing Multiplication Operations

## Level

Medium

## Description

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 `i-th`

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 <= 10^3`

`m <= n <= 10^5`

`-1000 <= nums[i], multipliers[i] <= 1000`

## Solution

Use dynamic programming. Create a 2D array `dp`

of `m + 1`

rows and `n`

columns, where `dp[i][j]`

represents the maximum score using the last `i`

multiplers for the subarray starting from index `j`

. The base case is when `i = 1`

. The final result is `dp[m][0]`

.

To optimize memory, use a 1D array instead, which stores only the last row of the original `dp`

array.

```
class Solution {
public int maximumScore(int[] nums, int[] multipliers) {
int n = nums.length, m = multipliers.length;
int[] dp = new int[n];
int minWindow = n - m + 1;
for (int j = minWindow - 1; j < n; j++) {
int start = j - minWindow + 1;
dp[start] = Math.max(nums[start] * multipliers[m - 1], nums[j] * multipliers[m - 1]);
}
for (int i = 2; i <= m; i++) {
int[] dpNew = new int[n];
int window = n - m + i;
for (int j = window - 1; j < n; j++) {
int start = j - window + 1;
dpNew[start] = Math.max(dp[start + 1] + nums[start] * multipliers[m - i], dp[start] + nums[j] * multipliers[m - i]);
}
dp = dpNew;
}
return dp[0];
}
}
```