Soup Servings — LeetCode 808 Python Solution

The problem

You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns: pour 100 mL from type A and 0 mL from type B pour 75 mL from type A and 25 mL from type B pour 50 mL from type A and 50 mL from type B pour 25 mL from type A and 75 mL from type B Note: There is no operation that pours 0 mL from A and 100 mL from B.

Example

Input

n = 50

Output

0.62500

Explanation

If we perform either of the first two serving operations, soup A will become empty first.

Python solution

class Solution:
    def soupServings(self, n: int) -> float:
        @cache
        def dfs(i: int, j: int) -> float:
            if i <= 0 and j <= 0:
                return 0.5
            if i <= 0:
                return 1
            if j <= 0:
                return 0
            return 0.25 * (
                dfs(i - 4, j)
                + dfs(i - 3, j - 1)
                + dfs(i - 2, j - 2)
                + dfs(i - 1, j - 3)
            )

        return 1 if n > 4800 else dfs((n + 24) // 25, (n + 24) // 25)

Complexity

Pattern: Dynamic Programming

Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 808. Soup Servings is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.

The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.

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