動機

最好幾次的coin change

Problem

You would like to make dessert and are preparing to buy the ingredients. You have n ice cream base flavors and m types of toppings to choose from. You must follow these rules when making your dessert:

  • There must be exactly one ice cream base.
  • You can add one or more types of topping or have no toppings at all.
  • There are at most two of each type of topping.

You are given three inputs:

  • baseCosts, an integer array of length n, where each baseCosts[i] represents the price of the ith ice cream base flavor.
  • toppingCosts, an integer array of length m, where each toppingCosts[i] is the price of one of the ith topping.
  • target, an integer representing your target price for dessert.

You want to make a dessert with a total cost as close to target as possible.

Return the closest possible cost of the dessert to target. If there are multiple, return the lower one.

 

Example 1:

Input: baseCosts = [1,7], toppingCosts = [3,4], target = 10Output: 10Explanation: Consider the following combination (all 0-indexed):- Choose base 1: cost 7- Take 1 of topping 0: cost 1 x 3 = 3- Take 0 of topping 1: cost 0 x 4 = 0Total: 7 + 3 + 0 = 10.

Example 2:

Input: baseCosts = [2,3], toppingCosts = [4,5,100], target = 18Output: 17Explanation: Consider the following combination (all 0-indexed):- Choose base 1: cost 3- Take 1 of topping 0: cost 1 x 4 = 4- Take 2 of topping 1: cost 2 x 5 = 10- Take 0 of topping 2: cost 0 x 100 = 0Total: 3 + 4 + 10 + 0 = 17. You cannot make a dessert with a total cost of 18.

Example 3:

Input: baseCosts = [3,10], toppingCosts = [2,5], target = 9Output: 8Explanation: It is possible to make desserts with cost 8 and 10. Return 8 as it is the lower cost.

Example 4:

Input: baseCosts = [10], toppingCosts = [1], target = 1Output: 10Explanation: Notice that you don't have to have any toppings, but you must have exactly one base.

 

Constraints:

  • n == baseCosts.length
  • m == toppingCosts.length
  • 1 <= n, m <= 10
  • 1 <= baseCosts[i], toppingCosts[i] <= 104
  • 1 <= target <= 104

Sol

class Solution:
    def closestCost(self, bs: List[int], ts: List[int], target: int) -> int:
        @cache
        def coin(cnt,j):
            if j >= len(ts) or cnt >= target:
                return cnt
            else:
                return min([coin(cnt+ts[j]*2,j+1),coin(cnt+ts[j],j+1),coin(cnt,j+1)], key=lambda x: (abs(x-target), x-target))
        return min([coin(x,0) for x in bs], key=lambda x: (abs(x-target), x-target))