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[9/16] 1770. Maximum Score from Performing Multiplication Operations

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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

Accepted

61,796

Submissions

177,812

#Google

링크

https://leetcode.com/problems/maximum-score-from-performing-multiplication-operations/ 1752 회 연결

익명
작성일 2022.09.16 07:46

class Solution:
    def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
        @lru_cache(2000)
        def fn(lo, hi, k):
            if k == len(multipliers): return 0
            return max(nums[lo] * multipliers[k] + fn(lo+1, hi, k+1), nums[hi] * multipliers[k] + fn(lo, hi-1, k+1))
        
        return fn(0, len(nums)-1, 0)
        
        # =================================================================================================================================================
        
#         dp = [[0 for _ in range(len(multipliers))] for _ in range(len(multipliers))]
        
#         def backtracking(num_l, mult_idx):
#             if mult_idx == len(multipliers):
#                 return 0
#             if dp[num_l][mult_idx]: return dp[num_l][mult_idx]

#             num_r = len(nums) - 1 - (mult_idx - num_l)
#             dp[num_l][mult_idx] = max(nums[num_l] * multipliers[mult_idx] + backtracking(num_l+1, mult_idx+1), nums[num_r]*multipliers[mult_idx]+backtracking(num_l, mult_idx+1))
            
#             return dp[num_l][mult_idx]
#         return backtracking(0,0)

[code=Python 3]class Solution:
    def maximumScore(self, nums: List[int], multipliers: List[int]) -> int:
        @lru_cache(2000)
        def fn(lo, hi, k):
            if k == len(multipliers): return 0
            return max(nums[lo] * multipliers[k] + fn(lo+1, hi, k+1), nums[hi] * multipliers[k] + fn(lo, hi-1, k+1))
        
        return fn(0, len(nums)-1, 0)
        
        # =================================================================================================================================================
        
#         dp = [[0 for _ in range(len(multipliers))] for _ in range(len(multipliers))]
        
#         def backtracking(num_l, mult_idx):
#             if mult_idx == len(multipliers):
#                 return 0
#             if dp[num_l][mult_idx]: return dp[num_l][mult_idx]

#             num_r = len(nums) - 1 - (mult_idx - num_l)
#             dp[num_l][mult_idx] = max(nums[num_l] * multipliers[mult_idx] + backtracking(num_l+1, mult_idx+1), nums[num_r]*multipliers[mult_idx]+backtracking(num_l, mult_idx+1))
            
#             return dp[num_l][mult_idx]
#         return backtracking(0,0)[/code]