1774. Closest Dessert Cost

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 = 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] <= 10^4

• 1 <= target <= 10^4

Solution (Java)

class Solution {
    private int finalValue = Integer.MAX_VALUE;

    public int closestCost(int[] baseCosts, int[] toppingCosts, int target) {
        for (int baseCost : baseCosts) {
            closestCost(baseCost, toppingCosts, target, 0);
        }
        return finalValue;
    }

    private void closestCost(int curCost, int[] toppingCosts, int target, int index) {
        if (index >= toppingCosts.length || curCost >= target) {
            if (Math.abs(target - curCost) < Math.abs(target - finalValue)) {
                finalValue = curCost;
            } else if (Math.abs(target - curCost) == Math.abs(target - finalValue)
                    && target < finalValue) {
                finalValue = curCost;
            }
            return;
        }
        closestCost(curCost, toppingCosts, target, index + 1);
        closestCost(curCost + toppingCosts[index], toppingCosts, target, index + 1);
        closestCost(curCost + toppingCosts[index] * 2, toppingCosts, target, index + 1);
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).