How do you solve Soup Servings on LeetCode?

Soup Servings (LeetCode #808) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The probability stabilizes for large n, allowing us to return a constant value.

What is the time complexity of Soup Servings?

The optimal solution for Soup Servings runs in O(n) time and uses O(n) space. The time complexity is O(n) due to the memoization of results for each state, which reduces redundant calculations.

What is the brute force solution for Soup Servings?

The brute force approach for Soup Servings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates every possible sequence of operations until one of the soups runs out. By calculating the probabilities for each sequence, we can determine the overall probability of soup A running out first. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Soup Servings?

Soup Servings uses the following concepts: Math, Dynamic Programming, Probability and Statistics. The recommended patterns to study are: Dynamic Programming, Memoization.

What are the interview tips for Soup Servings?

Always discuss your thought process and approach before diving into coding. Consider edge cases, especially when dealing with probabilities. Explain the significance of your chosen data structures and algorithms.

What are common mistakes when solving Soup Servings?

Forgetting to handle the base cases correctly in recursion. Not using memoization, leading to excessive recursion depth and time.

#808

Soup Servings

Medium

MathDynamic ProgrammingProbability and StatisticsDynamic ProgrammingMemoization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

For larger values of n, the probabilities stabilize and can be approximated. We can use dynamic programming to store results for smaller values and avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: If n > 5000, return 1.0 as the probability stabilizes.
2Step 2: Use a memoization technique to store results of dp(a, b) for each state.
3Step 3: Calculate probabilities recursively as in the brute force approach but with memoization.

solution.py18 lines

1# Full working Python code
2from functools import lru_cache
3
4def soupServings(n: int) -> float:
5    if n > 5000:
6        return 1.0
7
8    @lru_cache(None)
9    def dp(a, b):
10        if a <= 0 and b <= 0:
11            return 0.5
12        if a <= 0:
13            return 1.0
14        if b <= 0:
15            return 0.0
16        return 0.25 * (dp(a - 100, b) + dp(a - 75, b - 25) + dp(a - 50, b - 50) + dp(a - 25, b - 75))
17
18    return dp(n, n)

ℹ

Complexity note: The time complexity is O(n) due to the memoization of results for each state, which reduces redundant calculations.

1The probability stabilizes for large n, allowing us to return a constant value.
2Using memoization significantly reduces the number of calculations needed.

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