Soup Servings Python Solution with Tests

lcpy gen -n 808   # by problem number
lcpy gen -s soup_servings   # by problem name

Input: n = 50
Output: 0.62500
Explanation:
If we perform either of the first two serving operations, soup A will become empty first.
If we perform the third operation, A and B will become empty at the same time.
If we perform the fourth operation, B will become empty first.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.

Input: n = 100
Output: 0.71875
Explanation:
If we perform the first serving operation, soup A will become empty first.
If we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4.
If we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3.
If we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.

from functools import cache


class Solution:
    # Time: O(n^2) after scaling n down to units of 25 mL, capped by the
    # shortcut that treats n >= 5000 as an immediate A loss
    # Space: O(n^2) for the memo table
    def soup_servings(self, n: int) -> float:
        units = (n + 24) // 25
        if units >= 200:
            return 1.0

        @cache
        def dp(soup_a: int, soup_b: int) -> float:
            if soup_a <= 0 and soup_b <= 0:
                return 0.5
            if soup_a <= 0:
                return 1.0
            if soup_b <= 0:
                return 0.0
            return (
                dp(soup_a - 4, soup_b)
                + dp(soup_a - 3, soup_b - 1)
                + dp(soup_a - 2, soup_b - 2)
                + dp(soup_a - 1, soup_b - 3)
            ) / 4

        return dp(units, units)