Soup Servings

class Solution {
public:
    double calculateSoupProbability(int N) {
        int M = ceil(N / 25.0);
        unordered_map<int, unordered_map<int, double>> memo;
    
        function<double(int, int)> solve = [&](int x, int y) -> double {
            if (x <= 0 && y <= 0) {
                return 0.5;
            }
            if (x <= 0) {
                return 1;
            }
            if (y <= 0) {
                return 0;
            }
            if (memo[x].count(y)) {
                return memo[x][y];
            }
            return memo[x][y] = (solve(x - 4, y) + solve(x - 3, y - 1) +
                                 solve(x - 2, y - 2) + solve(x - 1, y - 3)) / 4;
        };
    
        for (int iter = 1; iter <= M; iter++) {
            if (solve(iter, iter) > 1 - 1e-5) {
                return 1;
            }
        }
        return solve(M, M);
    }
};

The "Soup Servings" problem involves calculating the probability of one soup finishing before the other when serving them in certain quantities. The solution is implemented in C++.

Here’s a step-by-step breakdown:

1. The function calculateSoupProbability takes an integer N, which represents the amount of initial soup in milliliters, normalized by dividing by 25.
2. An unordered map named memo is used for memoization to store the computed probabilities for different states (x, y), where x and y denote the remaining amounts of the two types of soups.
3. A lambda function solve defines the recursion for calculating the probability that soup A finishes before soup B based on their current amounts. The logic includes:
   • Checking if both soups are finished or if one finishes before the other.
   • Utilizing memoization to save and retrieve precomputed results for specific soup amounts.
   • Recursively calculating using different decrements to the soup amounts and averaging the results.
4. The loop iterates over possible states and uses the solve function to calculate if the probability reaches a certain threshold (1 - 1e-5), representing near certainty.
5. Finally, solve(M, M) is returned to provide the probability for the initial full amounts of both soups.

This solution utilizes dynamic programming with memoization to efficiently calculate the desired probability and minimize recomputation.

class Solution {
    public double calculateSoupServing(int servings) {
        int adjustedServings = (int)Math.ceil(servings / 25.0);
        Map<Integer, Map<Integer, Double>> memoization = new HashMap<>();
    
        for (int index = 1; index <= adjustedServings; index++) {
            if (findProbability(index, index, memoization) > 1 - 1e-5) {
                return 1.0;
            }
        }
        return findProbability(adjustedServings, adjustedServings, memoization);
    }
    
    private double findProbability(int A, int B, Map<Integer, Map<Integer, Double>> memo) {
        if (A <= 0 && B <= 0) {
            return 0.5;
        }
        if (A <= 0) {
            return 1.0;
        }
        if (B <= 0) {
            return 0.0;
        }
        memo.computeIfAbsent(A, x -> new HashMap<>());
        if (memo.get(A).containsKey(B)) {
            return memo.get(A).get(B);
        }
        double probability = (findProbability(A - 4, B, memo) + findProbability(A - 3, B - 1, memo) +
                findProbability(A - 2, B - 2, memo) + findProbability(A - 1, B - 3, memo)) / 4.0;
        memo.get(A).put(B, probability);
        return probability;
    }
}

The Java solution provided addresses the task of calculating the probability of one type of soup becoming empty before or at the same time as the other soup by modeling the probabilities recursively and uses memoization to optimize the calculations.

• Initially, the total number of servings is adjusted by dividing by 25 and rounding up, as each serving decrement in the problem is effectively in 25-unit decrements.
• A HashMap named memoization is created to store the previously computed probabilities, which reduces the time complexity by avoiding repetitive calculations.
• The method calculateSoupServing determines the probability by checking if the soup will finish nearly simultaneously under the condition of recursively reducing servings in 4 possible ways: 4 units from A only, 3 units from A and 1 from B, 2 units from each, or 1 unit from A and 3 from B.
• A secondary helper function findProbability is defined where:
  • It handles the base cases:
    • Both A and B are 0, returning 0.5 as it represents a simultaneous finish.
    • A is 0, returns 1.0, indicating soup A finished first.
    • B is 0, returns 0.0, indicating soup B finished first.
  • For other cases, it recursively calls itself adjusting the servings based on the four decrement options, averages their probabilities, and memoizes the result.
  • It checks the existence of the calculation in the memoization map before performing calculations to avoid redundancy.
• This approach offers an efficient probability calculation where the complexities of direct combinatorial approaches are circumvented, therefore suitable for higher serving numbers.
• The probability for each situation is weighted equally, ensuring a fair representation of each scenario's impact on the outcome of the soup servings.

class Solution:
    def probabilityOfSoup(self, quantity: int) -> float:
        scaled_qty = ceil(quantity / 25)
        memoization = collections.defaultdict(dict)
            
        def helper(x: int, y: int) -> float:
            if x <= 0 and y <= 0:
                return 0.5
            if x <= 0:
                return 1.0
            if y <= 0:
                return 0.0
            if x in memoization and y in memoization[x]:
                return memoization[x][y]
    
            memoization[x][y] = (
                helper(x - 4, y)
                + helper(x - 3, y - 1)
                + helper(x - 2, y - 2)
                + helper(x - 1, y - 3)
            ) / 4.0
            return memoization[x][y]
    
        for idx in range(1, scaled_qty + 1):
            if helper(idx, idx) > 1 - 1e-5:
                return 1.0
        return helper(scaled_qty, scaled_qty)

Solution Summary

The "Soup Servings" problem focuses on calculating the probability of finishing one type of soup before the other given servings are taken in different combinations. You solve this problem using dynamic programming and recursion in Python.

• Start by scaling down the quantity by dividing by 25 and using ceil to round up, this reduces the problem size and handles larger inputs efficiently.
• Utilize a memoization technique for optimizing the recursive calls. This is implemented using a defaultdict from the collections module where each state of the soup quantities (x, y) serves as a key.
• Define a helper function for recursion which:
  • Returns 0.5 if both soup quantities are zero simultaneously (indicating equal chance as both finished together).
  • Returns 1.0 if the first soup finishes (quantity x <= 0) and the other is still available.
  • Returns 0.0 if the second soup finishes first.
  • Stores and retrieves computed results from memoization to avoid redundant calculations.
• In the helper function, recursive calls calculate probabilities based on current quantities:
  • Reduce the quantity of the first soup by 4, 3, 2, and 1 respectively and adjust the second soup correspondingly.
  • Average the four resulting probabilities to get the current state's probability.
• Iterate through each possible quantity using a for loop to ascertain at what point (if any) the probability of the first soup finishing first reaches or exceeds 1 - 1e-5 (essentially 1).
• Return total probability if no such point is found within the loop.

This approach ensures that the computation is efficient and precise for both small and large quantities using scaling, memoization, and careful probability calculations.