g1701_1800.s1774_closest_dessert_cost.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated
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package g1701_1800.s1774_closest_dessert_cost;

// #Medium #Array #Dynamic_Programming #Backtracking
// #2022_04_27_Time_5_ms_(82.32%)_Space_42.3_MB_(20.34%)

/**
 * 1774 - Closest Dessert Cost\.
 *
 * Medium
 *
 * 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 i^th ice cream base flavor.
 * *   `toppingCosts`, an integer array of length `m`, where each `toppingCosts[i]` is the price of **one** of the i^th 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
**/
@SuppressWarnings("java:S1871")
public 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);
    }
}