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We can use a dynamic programming approach with memoization to store the probability results for known quantities of soup A and B. We'll treat the problem states as a grid where each cell holds the probability of that state leading to soup A becoming empty first. The key to optimizing this solution is to memoize intermediate results to avoid redundant calculations by storing probabilities in a 2D array or dictionary.
Time Complexity: O(n^2) due to memoization.
Space Complexity: O(n^2) for the memoization table.
1public class SoupServings {
2 public double soupServings(int N) {
3 if (N > 4800) return 1.0;
4 return calculate((N + 24) / 25, (N + 24) / 25, new HashMap<>());
5 }
6 private double calculate(int a, int b, Map<String, Double> memo) {
7 if (a <= 0 && b <= 0) return 0.5;
8 if (a <= 0) return 1.0;
9 if (b <= 0) return 0.0;
10 String key = a + "," + b;
11 if (memo.containsKey(key)) return memo.get(key);
12 double result = 0.25 * (calculate(a - 4, b, memo) +
13 calculate(a - 3, b - 1, memo) +
14 calculate(a - 2, b - 2, memo) +
15 calculate(a - 1, b - 3, memo));
16 memo.put(key, result);
17 return result;
18 }
19}This Java solution uses a recursive function with memoization stored in a HashMap. Similar base cases are checked and results stored with a concatenated string key to identify states. Reduction by integer division by 25 ensures manageable state space.
An alternative is solving the problem iteratively using a DP table to store results for each state. We build up from small values to larger values of soup A and B, filling out the DP table based on possible outcomes until we reach the desired amounts.
Time Complexity: O(n^2) as we fill up each cell of an N-by-N table.
Space Complexity: O(n^2) for maintaining the DP table.
1function soupServings(N) {
2 if (
We use a 2D array dp to compute probabilities iteratively. Each cell in the DP table is filled based on the results of possible soup servings. The getDP helper function handles edge cases like negative indices to ensure correctness.