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: Backtracking

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

 

JavaScript

/**
 * @param {number[]} baseCosts
 * @param {number[]} toppingCosts
 * @param {number} target
 * @return {number}
 */
const closestCost = function (baseCosts, toppingCosts, target) {
  let result = Number.MAX_SAFE_INTEGER;
  const toppingSize = toppingCosts.length;
  const addTopping = (cost, index) => {
    if (Math.abs(target - cost) < Math.abs(target - result)) result = cost;
    if (Math.abs(target - cost) === Math.abs(target - result) && cost < result) {
      result = cost;
    }
    if (index >= toppingSize) return;
    const toppingCost = toppingCosts[index];

    addTopping(cost, index + 1);
    addTopping(cost + toppingCost, index + 1);
    addTopping(cost + toppingCost * 2, index + 1);
  };

  for (const cost of baseCosts) {
    addTopping(cost, 0);
  }
  return result;
};