Soup Servings
Problem Description
You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns:
- pour 100 mL from type A and 0 mL from type B
- pour 75 mL from type A and 25 mL from type B
- pour 50 mL from type A and 50 mL from type B
- pour 25 mL from type A and 75 mL from type B
Note:
- There is no operation that pours 0 mL from A and 100 mL from B.
- The amounts from A and B are poured simultaneously during the turn.
- If an operation asks you to pour more than you have left of a soup, pour all that remains of that soup.
The process stops immediately after any turn in which one of the soups is used up.
Return the probability that A is used up before B, plus half the probability that both soups are used up in the same turn. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: n = 50
Output: 0.62500
Explanation:
If we perform either of the first two serving operations, soup A will become empty first.
If we perform the third operation, A and B will become empty at the same time.
If we perform 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
Explanation:
If we perform the first serving operation, soup A will become empty first.
If we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4.
If we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3.
If we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.
Constraints:
- 0 <= n <= 109
Solve it on LeetCode
Approaches
1. Recursive Approach with Memoization
Intuition:
This problem involves serving units of soup A and soup B. The operations have an equal probability of reducing different volumes of soups. Our goal is to find the probability that soup A will be empty first. We will use a recursive function to simulate the serving process, and optimize it with memoization to avoid redundant calculations.
Approach:
- Use a recursive function
serve(A, B)to model the serving process. - For each call, consider the probability of the next four outcomes.
- The base cases are when A or B is empty.
- Use memoization to store the results of
serve(A, B)to prevent recalculating.
Code:
- Time Complexity:
O(N^2)- Each state (A,B) is computed at most once. - Space Complexity:
O(N^2)- Due to memoization storage.
2. Dynamic Programming
Intuition:
To improve upon the recursive approach, we can instead use dynamic programming. By iteratively building solutions for smaller subproblems, we can avoid recursion altogether.
Approach:
- Use a 2-D table
dp[i][j]wheredp[i][j]is the probability that when A = i and B = j, soup A will run out first. - Fill this table by considering the outcomes of serving operations.
- Start from base cases and build up to (N, N).
Code:
- Time Complexity: O(N^2)
- Space Complexity: O(N^2)
3. Optimized Mathematical Approach
Intuition:
When N becomes very large, the probability approaches 1 due to constraints of problem dynamics. We can mathematically approximate the result to improve efficiency.
Approach:
- For a sufficiently large N (found empirically to be around 4800), the probability that soup A will run out first is virtually 1.
- For smaller N, use the DP solution as described.
Code:
- Time Complexity:
O(1)for large N, elseO(N^2) - Space Complexity:
O(1)for large N, elseO(N^2)