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 operations:
100 ml of soup A and 0 ml of soup B,75 ml of soup A and 25 ml of soup B,50 ml of soup A and 50 ml of soup B, and25 ml of soup A and 75 ml of soup B.When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: n = 50 Output: 0.62500 Explanation: If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.
Example 2:
Input: n = 100 Output: 0.71875
Constraints:
0 <= n <= 109We 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.
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.
Java
Time Complexity: O(n^2) due to memoization.
Space Complexity: O(n^2) for the memoization table.
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.
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.
C++
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.
| Approach | Complexity |
|---|---|
| DP with Memoization | Time Complexity: O(n^2) due to memoization. |
| Iterative DP Solution | Time Complexity: O(n^2) as we fill up each cell of an N-by-N table. |
Soup Servings | INTUITIVE | Recursion | Memoization | GOOGLE | Leetcode-808 • codestorywithMIK • 6,078 views views
Watch 9 more video solutions →Practice Soup Servings with our built-in code editor and test cases.
Practice on FleetCodePractice this problem
Open in Editor