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[9/16] 1770. Maximum Score from Performing Multiplication Operations
본문
Medium
1643395Add to ListShareYou 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 arraynums
. - Add
multipliers[i] * x
to your score. - Remove
x
from the arraynums
.
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
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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)