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Soup Servings

There are two types of soup: type A and type B. Initially, we have n ml of each type of soup. There are four kinds of operations:

  1. Serve 100 ml of soup A and 0 ml of soup B,
  2. Serve 75 ml of soup A and 25 ml of soup B,
  3. Serve 50 ml of soup A and 50 ml of soup B, and
  4. Serve 25 ml of soup A and 75 ml of soup B.

When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.

Note that we do not have an operation where all 100 ml's of soup B are used first.

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5 of the actual answer will be accepted.

 

Example 1:

Input: n = 50
Output: 0.62500
Explanation: If we choose the first two operations, A will become empty first.
For the third operation, A and B will become empty at the same time.
For 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.

Example 2:

Input: n = 100
Output: 0.71875

 

Constraints:

  • 0 <= n <= 109

Solution Explanation: Soup Servings

This problem involves probability and can be efficiently solved using dynamic programming with memoization. The core idea is to represent the soup amounts as units of 25ml, simplifying the calculations.

Understanding the Problem

We have two types of soup, A and B, with n ml of each initially. We can perform four operations, each consuming different amounts of A and B. The operations are equally likely (0.25 probability each). We stop when either soup A or B runs out. The problem asks for the probability that soup A runs out first, plus half the probability that both run out simultaneously.

Approach: Memoized Dynamic Programming

  1. State Representation: We use a 2D array (or a hash map) dp to store the results of subproblems. dp[i][j] represents the probability that soup A runs out first (or both run out simultaneously and we take half the probability) given i units of soup A and j units of soup B remaining. A "unit" is 25ml.

  2. Base Cases:

    • dp[i][j] = 0.5 if i <= 0 and j <= 0 (both soups empty simultaneously)
    • dp[i][j] = 1 if i <= 0 (soup A empty first)
    • dp[i][j] = 0 if j <= 0 (soup B empty first)

  3. Recursive Relation: For i > 0 and j > 0, we calculate dp[i][j] using the following recursive relation:

    dp[i][j] = 0.25 * (dp[i - 4, j] + dp[i - 3, j - 1] + dp[i - 2, j - 2] + dp[i - 1, j - 3])

    This represents the average outcome over the four possible operations. If an operation would result in a negative amount of soup, it implies we served as much as possible, and we can use the floor or ceiling of the negative value to represent the outcome.

  4. Memoization: We store the calculated probabilities in dp to avoid redundant computations. This significantly improves efficiency.

  5. Optimization: For large values of n, the probability approaches 1. Empirically, for n >= 4800, we can directly return 1. This avoids unnecessary computations for extremely large inputs.

Time and Space Complexity

  • Time Complexity: O(C^2), where C = (n + 24) // 25 is the number of units. The memoization reduces the time complexity from exponential to quadratic.
  • Space Complexity: O(C^2) to store the dp array (or hash map).

Code Examples (Python)

from functools import cache
 
class Solution:
    @cache
    def dfs(self, i, j):
        if i <= 0 and j <= 0:
            return 0.5
        if i <= 0:
            return 1.0
        if j <= 0:
            return 0.0
        return 0.25 * (self.dfs(i - 4, j) + self.dfs(i - 3, j - 1) + self.dfs(i - 2, j - 2) + self.dfs(i - 1, j - 3))
 
    def soupServings(self, n: int) -> float:
        n = (n + 24) // 25  #Convert to units of 25ml
        return 1.0 if n >= 500 else self.dfs(n, n)

The code in other languages (Java, C++, Go, TypeScript) follows a very similar structure, utilizing memoization (or equivalent techniques) to efficiently compute the probabilities. The core logic of the dynamic programming solution remains consistent across all languages.