How do you solve Closest Dessert Cost on LeetCode?

Closest Dessert Cost (LeetCode #1774) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Understanding how to combine items with constraints is crucial.

What is the brute force solution for Closest Dessert Cost?

The brute force approach for Closest Dessert Cost is Brute Force, with O(n²) time complexity and O(1) space complexity. We will try every possible combination of base and toppings to find the closest cost to the target. This means calculating the total cost for each base flavor combined with all possible topping combinations. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Closest Dessert Cost?

Closest Dessert Cost uses the following concepts: Array, Dynamic Programming, Backtracking. The recommended patterns to study are: Dynamic Programming, Combinatorial Optimization.

What are the interview tips for Closest Dessert Cost?

Always clarify the constraints and requirements before jumping into coding. Think about edge cases, such as no toppings or very high target values. Practice explaining your thought process while coding to simulate the interview environment.

What are common mistakes when solving Closest Dessert Cost?

Not considering all combinations of toppings correctly. Overlooking the need to minimize costs when multiple options are equally close to the target.

#1774

Closest Dessert Cost

Medium

ArrayDynamic ProgrammingBacktrackingDynamic ProgrammingCombinatorial Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

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Intuition

Time O(n * m)Space O(n)

We can use a dynamic programming approach to efficiently calculate all possible costs with toppings. By iterating through the toppings and updating possible costs, we can find the closest cost to the target without generating all combinations explicitly.

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Algorithm

3 steps

1Step 1: Initialize a set of possible costs with just the base costs.
2Step 2: For each topping, update the possible costs by considering adding 0, 1, or 2 of that topping to each existing cost.
3Step 3: After processing all toppings, find the closest cost to the target from the set of possible costs.

solution.py11 lines

1# Full working Python code
2
3def closestCost(baseCosts, toppingCosts, target):
4    possibleCosts = set(baseCosts)
5    for topping in toppingCosts:
6        newCosts = set()
7        for cost in possibleCosts:
8            newCosts.add(cost + topping)
9            newCosts.add(cost + 2 * topping)
10        possibleCosts.update(newCosts)
11    return min(possibleCosts, key=lambda x: (abs(x - target), x))

ℹ

Complexity note: The time complexity is O(n * m) where n is the number of base costs and m is the number of toppings. This is because we iterate through each topping for each base cost.

1Understanding how to combine items with constraints is crucial.
2Dynamic programming can often simplify problems that seem combinatorial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.