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
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.
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.
How can we solve this problem?
這題最主要的重點是對於每個multipliers[i]
,它只能挑選最左邊或者最右邊的值。所有,我們需要知道multipliers[i]
拿最左邊的值最後的結果比較大,還是拿最右邊後的結果比較大。為了避免重複計算而超時,所以我們需要使用dp
來幫助我們記錄當前最優解。注:因multipliers最多為m個,所有最多只能從nums拿m個數字