808. Soup Servings

MediumMathDynamic ProgrammingProbability and Statistics
Leetcode Link

Problem Description

The problem states that there are two types of soup, type A and type B, each initially with n ml. There are four types of serving operations with equal probability where we serve different amounts of soup A and B:

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

The key idea is that we serve soup until we can't serve both types of soup due to insufficiency. We wish to find the probability that soup A runs out before soup B, including half the probability of both soups running out simultaneously. The problem challenges us to calculate this probability given the constraints of quantity and serving operations.

Intuition

The solution utilizes recursive depth-first search (DFS) to explore all possible serving sequences until one or both soups are depleted. Since serving operations are equally likely, the probability at each step is divided by 4 (since there are four operations).

The intuition for reaching this solution is understanding that the problem can be modeled as a probability tree where each node represents the state of the soups, and each operation leads to a new state. The probability of reaching the end where soup A is empty is the sum of probabilities along the paths leading to this state.

Moreover, to optimize the computation, memoization (caching) is used to store the results of subproblems. Since the function dfs(i, j) will be called multiple times for the same values of i and j, caching results saves us from recomputing them.

An edge case optimization is introduced for large n. If n is larger than 4800, the probability of soup A running out first is approximately 1. This is derived from the law of large numbers: as the number of operations increases, the probability of both soups running out at exactly the same time becomes negligible, so we can assume soup A will run out first almost certainly.

The scaling factor (n + 24) // 25 is applied to both i and j to reduce computations. Since all serving sizes are multiples of 25, we can scale down the problem and still preserve the probabilities. Adding 24 and then using integer division by 25 ensures that we round up and avoid underestimating the amount of soup we start with.

In summary, the recursive solution splits the problem into smaller subproblems, computes the probability of each possible event, and ultimately builds up the solution using probabilities from the base cases back up to the original amounts of soup A and B.

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Solution Approach

The implementation of the solution employs several techniques to compute the desired probability.

Recursive Depth-First Search (DFS)

A recursive DFS approach is used to traverse the state space of soup A and soup B quantities until one or both are depleted. Each call to the function represents a choice of serving operation that leads to a new state. The base cases for this recursion are:

  • When both i (A) and j (B) are less than or equal to 0, the probability that A runs out first is 0.5 since both soups run out simultaneously.
  • If only i (A) is less than or equal to 0, the probability A runs out first is 1.
  • If only j (B) is less than or equal to 0, the probability A runs out first is 0 since B has already run out.

Memoization

To avoid recalculating the probabilities for states we have already seen, memoization is utilized by decorating the DFS function with @cache. Memoization is an optimization technique where the return values of a function are stored so that the next time the function is called with the same arguments, the previously calculated result is returned.

Probability Calculation

In each recursive call, we have four options, each with equal probability. Since these are independent and mutually exclusive events, the total probability is the sum of the individual probabilities. The recursive formula that computes the probability of soup A running out first is:

1dfs(i - 4, j) * 0.25 +
2dfs(i - 3, j - 1) * 0.25 +
3dfs(i - 2, j - 2) * 0.25 +
4dfs(i - 1, j - 3) * 0.25

Where i and j represent the current milliliters of soup A and soup B, respectively, adjusted by a scaling factor.

Scaling Factor

To decrease runtime and memory usage, scaling is applied by dividing the initial quantity of soup by 25 (since we deal with multiples of 25 ml) before passing it to the DFS function. The ceiling function (n + 24) // 25 ensures that we do not underestimate the starting quantity.

Edge Case Optimization

For a large n (greater than 4800 in this case), we can return 1 immediately since the probability of A running out first is virtually guaranteed, bypassing the need for further computation.

Using these techniques, the soupServings function computation is efficient enough to calculate the probability for reasonable quantities of soup and does so only using the information about serving rules and initial quantities.

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The three-steps of Depth First Search are:

  1. Identify states;
  2. Draw the state-space tree;
  3. DFS on the state-space tree.

Example Walkthrough

Let's illustrate the solution approach with a small, easy-to-follow example. Assume we initially have n = 50 ml of soups A and B, so we do not yet employ the edge case optimization for a large n.

To find the probability of soup A running out before soup B, we will use recursive depth-first search (DFS), memoization, and apply the scaling factor for optimization.

Step 1: Apply Scaling Factor

Since n = 50 and the servings are multiples of 25, we apply the scaling factor. Here's the calculation:

1(n + 24) // 25 = (50 + 24) // 25 = 74 // 25 = 2

This means we will run our recursive depth-first search with the scaled values, treating 2 as our starting value instead of 50.

Step 2: Begin Recursive DFS with Memoization

We start our DFS with soupServings(2, 2) representing the scaled quantities of soups A and B.

During the DFS, we consider all possible serving combinations at each stage and calculate their probabilities.

Step 3: Explore Each Serving Operation

For each recursive call, there are four possible serving operations (scaled down for our example):

  1. Serve 4 parts of A and 0 parts of B (Equivalent to 100 ml A, 0 ml B)
  2. Serve 3 parts of A and 1 part of B (Equivalent to 75 ml A, 25 ml B)
  3. Serve 2 parts of A and 2 parts of B (Equivalent to 50 ml A, 50 ml B)
  4. Serve 1 part of A and 3 parts of B (Equivalent to 25 ml A, 75 ml B)

If we call dfs(2, 2), we would explore dfs(1, 2), dfs(0, 1), dfs(0, 0), and dfs(1, 0).

Step 4: Base Cases and Probability Calculation

Our base cases kick in when one or both of the soup quantities becomes 0 or less:

  • dfs(0, 1) implies that A runs out first, with a probability of 1.
  • dfs(1, 0) does not contribute since B runs out first, with a probability of 0.
  • dfs(0, 0) represents both running out simultaneously, with a probability of 0.5.

Adding these up, we have 1 * 0.25 + 0 * 0.25 + 0.5 * 0.25 from these three events. However, we also need to consider dfs(1, 2), which will recursively calculate further probabilities.

Step 5: Continue Recursion Until Base Cases are Reached

Since dfs(1, 2) does not hit a base case, it explores:

  • dfs(0, 2) (A runs out, giving probability 1),
  • dfs(0, 1) (A runs out, giving probability 1),
  • dfs(0, 0) (Simultaneous run out, giving probability 0.5),
  • dfs(1, 0) (B runs out, giving probability 0).

Adding these gives us another 2 * 0.25 + 0.5 * 0.25.

Step 6: Final Probability

By summing up the probabilities calculated in previous recursive calls and adjusting for the equal chance of each serving event, we arrive at the final probability of A running out before B.

In our example, a complete calculation with proper values would involve more recursive steps, but the general process is the same: break down into smaller subproblems, cache results for efficiency, and build up the final answer using the base cases and equation provided.

For n = 50, our rough estimation would be dfs(2, 2) = 0.25 + 0.25 + 0.125 + 0.25 + 0.25 + 0.125 = 1.25. Since we can't have a probability greater than 1, we know that the final answer must be adjusted for the actual recursive calls, but it is evident that the probability of A running out before B will be high in this example.

Keep in mind that in an actual program, the memoization would prevent redundant calculations for the same recursive calls, optimizing the entire process considerably.

Solution Implementation

1from functools import lru_cache
2
3class Solution:
4    def soupServings(self, quantity: int) -> float:
5        # Using Least Recently Used (LRU) cache as a decorator to memoize the intermediate results
6        @lru_cache(None)
7        def recursive_serve(a_quantity: int, b_quantity: int) -> float:
8            # Base case: both soups have been served completely or below zero
9            if a_quantity <= 0 and b_quantity <= 0:
10                return 0.5  # Equal chance of serving A or B completely last
11
12            # Base case: only A has been completely served
13            if a_quantity <= 0:
14                return 1  # A has been served completely
15
16            # Base case: only B has been completely served
17            if b_quantity <= 0:
18                return 0  # B has been served completely
19
20            # Recursively calculate the probability of serving A completely first
21            # following the four possible operations stated in the problem
22            return 0.25 * (
23                recursive_serve(a_quantity - 4, b_quantity) +
24                recursive_serve(a_quantity - 3, b_quantity - 1) +
25                recursive_serve(a_quantity - 2, b_quantity - 2) +
26                recursive_serve(a_quantity - 1, b_quantity - 3)
27            )
28
29        # An optimization for large n, as the probability tends to 1 when n becomes very large (> 4800)
30        if quantity > 4800:
31            return 1.0
32
33        # Normalize the soup quantity and cache the calculations to optimize further calls
34        # that have reduced soup quantities after servings
35        normalized_quantity = (quantity + 24) // 25
36        return recursive_serve(normalized_quantity, normalized_quantity)
37
38# Example Usage:
39# sol = Solution()
40# result = sol.soupServings(100)
41# print(result)  # prints the probability of serving soup A completely first
42
1class Solution {
2    // Cache for storing results of subproblems
3    private double[][] memo = new double[200][200];
4
5    // Entry method which adjusts servings (if necessary) and starts the depth-first search
6    public double soupServings(int n) {
7        // If n is larger than 4800, we assume probability is practically 1
8        return n > 4800 ? 1.0 : dfs((n + 24) / 25, (n + 24) / 25);
9    }
10
11    // Depth-first search to compute probability
12    private double dfs(int a, int b) {
13        // If both servings are not positive, it's a tie, return 0.5
14        if (a <= 0 && b <= 0) {
15            return 0.5;
16        }
17        // If only A is depleted, A finished serving first, return 1.0
18        if (a <= 0) {
19            return 1.0;
20        }
21        // If only B is depleted, B finished serving first, return 0
22        if (b <= 0) {
23            return 0;
24        }
25        // If result is already computed, return it from the memo cache
26        if (memo[a][b] > 0) {
27            return memo[a][b];
28        }
29        // Recursively compute the probability considering all 4 possible operation
30        double probability = 0.25 * (
31            dfs(a - 4, b) + 
32            dfs(a - 3, b - 1) + 
33            dfs(a - 2, b - 2) + 
34            dfs(a - 1, b - 3)
35        );
36        // Store the computed probability in the memo cache
37        memo[a][b] = probability;
38        // Return the computed probability
39        return probability;
40    }
41}
42
1class Solution {
2public:
3    double soupServings(int n) {
4        // Cache probability results to avoid redundant calculations
5        double probabilityCache[200][200] = {0.0};
6      
7        // Recursive function to calculate probabilities using memorization
8        function<double(int, int)> calculateProbabilities = [&](int soupA, int soupB) -> double {
9            // Base cases when either or both soup quantities become non-positive
10            if (soupA <= 0 && soupB <= 0) return 0.5; 
11            if (soupA <= 0) return 1;
12            if (soupB <= 0) return 0;
13            // Return the cached result if it has already been computed
14            if (probabilityCache[soupA][soupB] > 0) return probabilityCache[soupA][soupB];
15          
16            // Recursive case: calculate the probability and cache it
17            double probability = 0.25 * (calculateProbabilities(soupA - 4, soupB) + 
18                                         calculateProbabilities(soupA - 3, soupB - 1) +
19                                         calculateProbabilities(soupA - 2, soupB - 2) + 
20                                         calculateProbabilities(soupA - 1, soupB - 3));
21            probabilityCache[soupA][soupB] = probability;
22            return probability;
23        };
24      
25        // According to problem description, for large 'n' values, probability converges to 1
26        // The value 4800 is determined empirically; for inputs larger than this, return 1
27        if (n > 4800) {
28            return 1;
29        }
30      
31        // For smaller values of 'n', use the recursive function to calculate the probability,
32        // normalizing 'n' so that the maximum index does not exceed 200.
33        return calculateProbabilities((n + 24) / 25, (n + 24) / 25);
34    }
35};
36
1// Calculates the probability that Soup A will be empty before Soup B, or simultaneously empty.
2// For very large servings (n >= 4800), the probability is approximated as 1.
3function soupServings(n: number): number {
4  // Memoization array to store probabilities that are already calculated.
5  // Initialized to -1 to represent uncalculated probabilities.
6  const memo = new Array(200).fill(0).map(() => new Array(200).fill(-1));
7
8  // Helper function for depth-first search to calculate probability recursively.
9  // i: remaining units of Soup A, j: remaining units of Soup B
10  const dfs = (i: number, j: number): number => {
11    // Base cases when one or both soups are empty
12    if (i <= 0 && j <= 0) {
13      return 0.5; // Probability when both soups are empty simultaneously
14    }
15    if (i <= 0) {
16      return 1; // Probability when Soup A is empty and Soup B is not
17    }
18    if (j <= 0) {
19      return 0; // Probability when Soup B is empty and Soup A is not
20    }
21    // If the probability is already calculated, return it
22    if (memo[i][j] !== -1) {
23      return memo[i][j];
24    }
25    // Calculate probability recursively for all four operations
26    memo[i][j] =
27      0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
28    return memo[i][j];
29  };
30
31  // Special case for large servings where probability is considered to be 1
32  if (n >= 4800) {
33    return 1;
34  }
35  // Scale down the servings for computational efficiency and make recursive call
36  return dfs(Math.ceil(n / 25), Math.ceil(n / 25));
37}
38

Time and Space Complexity

The given Python code defines a function soupServings that calculates the probability of soup A being empty before soup B when starting with n milliliters of each soup. The function uses dynamic programming with memoization to prevent recalculations of the same state.

Time Complexity

The time complexity of the soupServings function primarily depends on the number of recursive calls made by the dfs helper function and the amount of computation in each call.

For every function call to dfs(i, j), four more calls are made recursively with quantities (i-4, j), (i-3, j-1), (i-2, j-2), (i-1, j-3). However, due to memoization (caching), once a particular state is computed, it is stored and reused.

The problem is solved for n/25 instead of n to reduce the number of states since the problem's granularity allows for this optimization without changing the result significantly (rounding up to the nearest integer after dividing by 25). Thus the possible unique calls to dfs are reduced to the order of N^2, where N = (n + 24)//25.

Due to memoization, each state (i, j) will be computed once, and since we have approximately N^2/2 states (since we only care about i <= j due to the symmetry of the problem), the total number of computations is O(N^2).

The time complexity is therefore O((n/25)^2), which simplifies to O(n^2) when not considering the constant factor of 1/25.

Special Case: If n > 4800, the function returns 1 immediately, and the time complexity is O(1) since the probability is almost 1 that soup A will finish first.

Space Complexity

The space complexity is related to the storage of computed states for memoization. Since there are approximately N^2/2 unique states to cache, the space complexity is O(N^2).

Additionally, dfs uses a stack for recursion. In the worst case, where n is the smallest and we don't hit the base cases early, the maximum depth of the recursion will be O(N). However, this contributes less than the memoization cache to the space complexity, so it's dominated by O(N^2) after considering memoization.

In conclusion, the space complexity is O((n/25)^2), which simplifies to O(n^2) for the main computation. For n > 4800, the space complexity is O(1) as no additional space is used beyond the input and constant space for the output.

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