1774. Closest Dessert Cost

Description

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.

 

Constraints:

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

Solutions

Solution 1

Python3JavaC++GoJavaScript

class Solution:
    def closestCost(
        self, baseCosts: List[int], toppingCosts: List[int], target: int
    ) -> int:
        def dfs(i, t):
            if i >= len(toppingCosts):
                arr.append(t)
                return
            dfs(i + 1, t)
            dfs(i + 1, t + toppingCosts[i])

        arr = []
        dfs(0, 0)
        arr.sort()
        d = ans = inf

        # 选择一种冰激淋基料
        for x in baseCosts:
            # 枚举子集和
            for y in arr:
                # 二分查找
                i = bisect_left(arr, target - x - y)
                for j in (i, i - 1):
                    if 0 <= j < len(arr):
                        t = abs(x + y + arr[j] - target)
                        if d > t or (d == t and ans > x + y + arr[j]):
                            d = t
                            ans = x + y + arr[j]
        return ans

class Solution {
    private List<Integer> arr = new ArrayList<>();
    private int[] ts;
    private int inf = 1 << 30;

    public int closestCost(int[] baseCosts, int[] toppingCosts, int target) {
        ts = toppingCosts;
        dfs(0, 0);
        Collections.sort(arr);
        int d = inf, ans = inf;

        // 选择一种冰激淋基料
        for (int x : baseCosts) {
            // 枚举子集和
            for (int y : arr) {
                // 二分查找
                int i = search(target - x - y);
                for (int j : new int[] {i, i - 1}) {
                    if (j >= 0 && j < arr.size()) {
                        int t = Math.abs(x + y + arr.get(j) - target);
                        if (d > t || (d == t && ans > x + y + arr.get(j))) {
                            d = t;
                            ans = x + y + arr.get(j);
                        }
                    }
                }
            }
        }
        return ans;
    }

    private int search(int x) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr.get(mid) >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private void dfs(int i, int t) {
        if (i >= ts.length) {
            arr.add(t);
            return;
        }
        dfs(i + 1, t);
        dfs(i + 1, t + ts[i]);
    }
}

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