Closest Dessert Cost

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RecursionDynamic Programming

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the values in the `baseCosts` and `toppingCosts` arrays, and for `target`?
Can the `baseCosts` or `toppingCosts` arrays be empty?
Is it possible for the `target` to be negative or zero?
If multiple dessert costs are equally close to the `target`, should I return the smaller or larger cost?
For the `toppingCosts` array, can I choose to use a topping 0, 1, or 2 times?

Brute Force Solution

Approach

The goal is to find the dessert cost closest to a target value by trying all possible combinations of base costs and topping costs. We'll explore every possible way to combine the ingredients, considering that we can have zero, one, or two of each topping. We will compare the total cost to the target and find the one that's closest.

Here's how the algorithm would work step-by-step:

First, pick one base ingredient cost.
Then, for each topping, consider three options: don't include it, include it once, or include it twice.
For every possible combination of toppings, add the topping costs to the base cost to get a total cost.
Compare this total cost to the target cost.
Keep track of the total cost that's closest to the target.
Repeat the topping choices for all base costs
After trying all base ingredients and all topping combinations, return the total cost that was closest to the target.

Code Implementation

def closestDessertCost(baseCosts, toppingCosts, target):
    closest_cost = float('inf')

    # Iterate through each base cost
    for base_cost in baseCosts:
        
        # Implement recursion to generate topping combinations
        def find_all_costs(topping_index, current_cost):
            nonlocal closest_cost

            # Update closest cost if current cost is closer to target
            if abs(current_cost - target) < abs(closest_cost - target):
                closest_cost = current_cost
            elif abs(current_cost - target) == abs(closest_cost - target) and current_cost < closest_cost:
                closest_cost = current_cost

            # Base case: all toppings have been considered
            if topping_index == len(toppingCosts):
                return

            # Explore options: 0, 1, or 2 of the current topping
            find_all_costs(topping_index + 1, current_cost) # Option 1: 0 toppings

            find_all_costs(topping_index + 1, current_cost + toppingCosts[topping_index]) # Option 2: 1 topping
            
            find_all_costs(topping_index + 1, current_cost + 2 * toppingCosts[topping_index]) # Option 3: 2 toppings

        # Initiate recursive calls to find best combination for each base
        find_all_costs(0, base_cost)

    return closest_cost

Big(O) Analysis

Time Complexity

O(3^m * n) – Let n be the number of base costs and m be the number of topping costs. For each base cost, we explore all possible combinations of toppings. Since we can choose each topping zero, one, or two times, there are 3 options for each topping. Therefore, for m toppings, there are 3^m possible combinations. We iterate through n base costs, and for each base cost, we generate 3^m topping combinations. Thus the time complexity is O(3^m * n).

Space Complexity

O(1) – The algorithm explores combinations of toppings using a recursive approach or iterative equivalent. The plain English explanation doesn't describe the creation of any auxiliary data structures whose size depends on the input size (number of base costs or topping costs). The space used is dominated by a constant number of variables to keep track of the closest cost and potentially the current cost being explored, resulting in constant auxiliary space.

Optimal Solution

Approach

The goal is to find the dessert cost closest to a target value by combining base costs and topping costs. We can efficiently explore possible costs using a technique that avoids checking every single combination, focusing on getting closer to the target with each decision.

Here's how the algorithm would work step-by-step:

Start with the base cost as your current dessert cost.
For each topping, we have three options: include it zero times, once, or twice. We will explore all three possibilities for each topping.
As we add each topping option, calculate the new total dessert cost.
Keep track of the dessert cost that is closest to the target value so far.
When deciding about adding a topping, if including that topping option makes the cost go further away from the target, stop exploring that path and consider the next topping. If we have already gone beyond the target value, adding more toppings of the same kind will only make the total dessert cost larger than the target, so we don't need to continue down that path.
After considering all toppings, return the closest dessert cost found.

Code Implementation

def closest_dessert_cost(base_costs, topping_costs, target):    closest_cost = min(base_costs, key=lambda x: abs(x - target))
    def calculate_cost(current_cost, topping_index):
        nonlocal closest_cost
        if abs(current_cost - target) < abs(closest_cost - target):
            closest_cost = current_cost
        elif abs(current_cost - target) == abs(closest_cost - target):
            closest_cost = min(closest_cost, current_cost)

        if topping_index == len(topping_costs) or current_cost > target + abs(closest_cost-target):
            return

        # Explore the possibility of adding 0, 1, or 2 of the current topping.
        for num_toppings in range(3):
            new_cost = current_cost + num_toppings * topping_costs[topping_index]

            # Recursively calculate with the next topping
            calculate_cost(new_cost, topping_index + 1)

    # Iterate through each base cost to begin calculation.
    for base_cost in base_costs:
        calculate_cost(base_cost, 0)

    return closest_cost

Big(O) Analysis

Time Complexity

O(3^m) – The time complexity is determined by the topping costs array, where m represents the number of different toppings. For each topping, we explore three possibilities: adding it zero times, once, or twice. Since we are considering all combinations of these topping options, the number of paths we explore grows exponentially with the number of toppings. Therefore, the time complexity is O(3^m), where m is the length of the toppingCosts array, as it represents the maximum number of combinations we might evaluate to get close to our target.

Space Complexity

O(T) – The algorithm's space complexity primarily stems from the recursion depth of exploring topping options. For each topping, there are three choices: 0, 1, or 2 times. In the worst case, where the algorithm explores all possible combinations of toppings before exceeding the target and backtracking, the recursion depth will be proportional to the number of topping types, T. Thus, the space used by the recursion stack is O(T), where T represents the number of different types of toppings. Constant extra space is used for variables tracking current cost and closest cost, so it can be ignored.

Edge Cases

Case	How to Handle
Empty baseCosts or toppingCosts arrays	If baseCosts is empty, return -1 or throw an exception as no dessert can be made; if toppingCosts is empty, the closest cost is the minimum baseCost.
baseCosts or toppingCosts contain negative numbers	Return an error or throw an exception as costs cannot be negative, assuming the prompt specifies otherwise.
toppingCosts is very large	The recursive approach should be optimized to avoid stack overflow, possibly using memoization or an iterative approach.
target is smaller than all baseCosts	The solution must correctly identify the smallest baseCost as the closest.
target is larger than the sum of all baseCosts and toppings (even if doubled)	The solution must correctly identify the maximum possible cost as the closest.
Integer overflow during sum calculation	Use long integers for intermediate sums or perform modular arithmetic if the problem specifies a modulo.
Multiple dessert costs are equally close to target	The solution should return the dessert cost that is smaller when the absolute differences from the target are equal, according to the prompt requirements.
baseCosts or toppingCosts contains extremely large numbers	The recursive approach might lead to overflow or stack overflow; an iterative dynamic programming approach can alleviate this.

Empty baseCosts or toppingCosts arrays

How to Handle:

If baseCosts is empty, return -1 or throw an exception as no dessert can be made; if toppingCosts is empty, the closest cost is the minimum baseCost.

baseCosts or toppingCosts contain negative numbers

How to Handle:

Return an error or throw an exception as costs cannot be negative, assuming the prompt specifies otherwise.

toppingCosts is very large

How to Handle:

The recursive approach should be optimized to avoid stack overflow, possibly using memoization or an iterative approach.

target is smaller than all baseCosts

How to Handle:

The solution must correctly identify the smallest baseCost as the closest.

target is larger than the sum of all baseCosts and toppings (even if doubled)

How to Handle:

The solution must correctly identify the maximum possible cost as the closest.

Integer overflow during sum calculation

How to Handle:

Use long integers for intermediate sums or perform modular arithmetic if the problem specifies a modulo.

Multiple dessert costs are equally close to target

How to Handle:

The solution should return the dessert cost that is smaller when the absolute differences from the target are equal, according to the prompt requirements.

baseCosts or toppingCosts contains extremely large numbers

How to Handle:

The recursive approach might lead to overflow or stack overflow; an iterative dynamic programming approach can alleviate this.