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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.
1def soupServings(N):
2 if N > 4800:
3 return 1
4
5 memo = {}
6
7 def dp(a, b):
8 if a <= 0 and b <= 0:
9 return 0.5
10 if a <= 0:
11 return 1
12 if b <= 0:
13 return 0
14 if (a, b) in memo:
15 return memo[(a, b)]
16
17 memo[(a, b)] = 0.25 * (dp(a - 100, b) + dp(a - 75, b - 25) + dp(a - 50, b - 50) + dp(a - 25, b - 75))
18 return memo[(a, b)]
19
20 N = (N + 24) // 25
21 return dp(N, N)In this Python solution, we use a recursive function with memoization. If the amount of soup A or B is less than or equal to zero, we return the base probabilities. The recursive step calculates the probability for each serving option and stores the result in a memo dictionary to avoid recalculating. For large n, greater than 4800, the probability approaches 1, so we return 1 directly.
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.