808. Soup Servings

The Soup Servings problem is a classic Dynamic Programming or Recursion with Memoization challenge. It sounds like a simple probability question, but it has a very important trick: because the probability of one soup running out first becomes overwhelmingly high as the volume increases, you can solve it for large inputs using a simple constant.

This problem evaluates your ability to handle probabilities in recursive states and your observation skills for edge cases.

Problem

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

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

Each operation has a 0.25 probability. If the remaining volume of soup is less than the amount needed for an operation, we use what is left. We stop serving when we no longer have enough for both types (i.e., at least one type becomes 0).

Your task:

Return the probability that Soup A will be empty first, plus half the probability that A and B become empty at the same time.

Rules and constraints:

• 0 <= n <= 10^9 (Note the very large upper bound!)
• Answers within 10^-5 of the actual answer will be accepted.

Solution

The key to this problem is the “The Threshold Observation.” Since Type A is served in larger average amounts than Type B (Average A: 62.5ml, Average B: 37.5ml), as n becomes very large, the probability of A running out first approaches 1. In fact, for any n > 5000, the answer is so close to 1 that it falls within the allowed error margin.

Key Idea: State Compression

We can divide n by 25 to simplify the operations:

1. Serve (4, 0)
2. Serve (3, 1)
3. Serve (2, 2)
4. Serve (1, 3)

Algorithm Outline

1. If n > 5000, return 1.0 immediately.
2. Normalize n: n = (n + 24) // 25 (This converts the volume into “units” of 25ml).
3. Use a Memoization table (or a dictionary) to store the results of solve(a, b):
   • Base Case 1: If a <= 0 and b <= 0, return 0.5 (Both empty at the same time).
   • Base Case 2: If a <= 0, return 1.0 (A is empty first).
   • Base Case 3: If b <= 0, return 0.0 (B is empty first).
4. Recursive Step:
   result = 0.25 * (solve(a-4, b) + solve(a-3, b-1) + solve(a-2, b-2) + solve(a-1, b-3))
5. Store and return the result.

Sample Implementation (Python)

Complexity

• Time: O(1) (Technically O(5000^2 / 25), but since we cap n at 5000, it is a constant time operation).
• Space: O(1) (The memoization table size is capped by the threshold).