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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 serving operations:

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

We need to calculate the probability that soup type A will be empty first, plus half the probability that both types of soup will be empty at the same time. Given N, return this probability. If N is very large, you can assume the probability approaches 1.

Clarifying Questions

  1. What should the range of N be?
    • N is a non-negative integer. The problem typically considers large values of N to see how the probability asymptotes to 1.
  2. Is there any constraint on the precision of the output?
    • The expected answer should be a floating-point number with reasonable precision for probability values (e.g., up to six decimal places).
  3. Should we consider the memoization technique or a recursive approach?
    • Given the structure and the need for repeated calculations, memoization or dynamic programming would be beneficial for higher efficiency.

Strategy

  1. Base Case:
    • If N is larger than a certain threshold (one could choose 4800 based on the nature of the problem), return 1 directly.
  2. Recursive Relation with Memoization:
    • Use recursion to break down the problem. Calculate the probabilities using the given serving operations and utilize memoization to avoid redundant computations.
  3. Probability Calculation:
    • For each operation, calculate the resultant state and recursively compute the probability of that state. Sum up these probabilities accordingly:
      • Probability of only A getting empty first.
      • Probability of both A and B getting empty at the same time.
  4. Memoization:
    • Use a dictionary to store the results of subproblems to save computation time.

Code

def soupServings(N: int) -> float:
    if N > 4800:
        return 1.0

    # Scaling down N by a factor of 25 to simplify the problem
    N = (N + 24) // 25
    
    memo = {}

    def dp(a, b):
        if a <= 0 and b <= 0:
            return 0.5
        if a <= 0:
            return 1.0
        if b <= 0:
            return 0.0
        if (a, b) in memo:
            return memo[(a, b)]

        memo[(a, b)] = 0.25 * (dp(a - 4, b) + dp(a - 3, b - 1) +
                               dp(a - 2, b - 2) + dp(a - 1, b - 3))
        return memo[(a, b)]

    return dp(N, N)

# Example Usage:
N = 50
print(soupServings(N))  # Expected Output based on the given value of N

Time Complexity

The time complexity is approximately O(N^2), where N is the scaled-down value of the initial volume of soup. However, the memoization approach ensures that each state is computed only once, making the solution efficient even for relatively large values of N.

Explanation

  1. Scaling:
    • We scale N down by rounding it to the nearest multiple of 25 to reduce the state space.
  2. Memoization:
    • The dictionary memo helps in storing the probability values for each state (a, b) to avoid redundant recursive calls.
  3. Recursive Calculation:
    • For each possible serving operation, the resultant state is computed, and its probability is summed up to derive the final probability value for the given state (a, b).

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