1770. Maximum Score from Performing Multiplication Operations

Medium

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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

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)