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花花酱 LeetCode 1774. Closest Dessert Cost

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 = 10
Output: 10
Explanation: 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 = 0
Total: 7 + 3 + 0 = 10.

Example 2:

Input: baseCosts = [2,3], toppingCosts = [4,5,100], target = 18
Output: 17
Explanation: 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 = 0
Total: 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 = 9
Output: 8
Explanation: 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 = 1
Output: 10
Explanation: 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

Solution: DP / Knapsack

Pre-compute the costs of all possible combinations of toppings.

Time complexity: O(sum(toppings) * 2 * (m + n)) ~ O(10^6)
Space complexity: O(sum(toppings)) ~ O(10^5)

C++

Solution 2: DFS

Combination

Time complexity: O(3^m * n)
Space complexity: O(m)

C++

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