[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

最接近目标价格的甜点成本。

你打算做甜点，现在需要购买配料。目前共有 n 种冰激凌基料和 m 种配料可供选购。而制作甜点需要遵循以下几条规则：

必须选择 一种 冰激凌基料。
可以添加 一种或多种 配料，也可以不添加任何配料。
每种类型的配料 最多两份 。
给你以下三个输入：

baseCosts ，一个长度为 n 的整数数组，其中每个 baseCosts[i] 表示第 i 种冰激凌基料的价格。
toppingCosts，一个长度为 m 的整数数组，其中每个 toppingCosts[i] 表示 一份 第 i 种冰激凌配料的价格。
target ，一个整数，表示你制作甜点的目标价格。
你希望自己做的甜点总成本尽可能接近目标价格 target 。

返回最接近 target 的甜点成本。如果有多种方案，返回 成本相对较低 的一种。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/closest-dessert-cost
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是DFS。这道题是在一定预算范围内找所有的可能性，所以有点类似于backtracking/permutation一类的题。我给出的解法也是类似。如果对permutation一类的题不熟悉可以先做题号比较小的题目。这道题跟一般求permutation类型的题目有点区别的地方就是对于toppings，可以不买，可以买一个，也可以买两个。

时间O(n)

空间O(n)

Java实现

class Solution {
    int res = Integer.MAX_VALUE;

    public int closestCost(int[] baseCosts, int[] toppingCosts, int target) {
        for (int base : baseCosts) {
            helper(base, toppingCosts, 0, target);
        }
        return res;
    }

    private void helper(int curCost, int[] toppingCosts, int index, int target) {
        // 如果总价比res距离target更近
        // 或者总价比target更小，则更新
        if (Math.abs(curCost - target) < Math.abs(res - target)
                || Math.abs(curCost - target) == Math.abs(res - target) && curCost < res) {
            res = curCost;
        }
        // base case
        if (index == toppingCosts.length) {
            return;
        }
        helper(curCost, toppingCosts, index + 1, target); // 不买topping
        helper(curCost + toppingCosts[index], toppingCosts, index + 1, target); // 买一个topping
        helper(curCost + 2 * toppingCosts[index], toppingCosts, index + 1, target); // 买两个topping
    }
}

LeetCode 题目总结