This problem requires calculating the probability that soup A runs out before soup B using a discrete state transition approach. Each serving reduces soup amounts randomly, and dynamic programming caches intermediate probabilities for efficiency. The solution balances precise probability computation with optimized recursion or iterative DP to handle large n values.
Problem Statement
You have two soups, A and B, each starting with n milliliters. On each turn, one of four serving operations is randomly selected with equal probability. Each operation reduces the amounts of A and B by fixed quantities. The goal is to determine the probability that soup A will become empty before soup B, with ties counting as half probability.
The process stops immediately once either soup reaches zero. Because the quantities decrease in discrete steps and operations are random, naive simulation is inefficient for large n. You must use state transition dynamic programming to efficiently track all possible states and their probabilities, handling overlaps and memoization carefully.
Examples
Example 1
Input: n = 50
Output: 0.62500
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
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
Solution Approach
Model as discrete state transitions
Define DP states as dp[a][b], representing the probability that soup A empties first when A has a mL and B has b mL. Each serving operation leads to a new state with adjusted soup quantities, updating the probability as the sum of all outcomes multiplied by 0.25.
Apply memoization or iterative DP
Use recursion with memoization to avoid recomputing overlapping subproblems, or implement a bottom-up iterative DP table. Store intermediate probabilities for all states until both soups are exhausted or one reaches zero, ensuring accurate accumulation of probabilities for large n.
Optimize for large n using threshold approximation
For very large n, the probability approaches 1. Use a cutoff where further recursion yields negligible probability change. This approximation maintains O(1) practical complexity while preserving correctness within an acceptable error margin.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(1) |
| Space | O(1) |
The DP approach effectively treats the problem as bounded by the number of discrete 25 mL units, allowing O(1) time and space in practical execution using approximation and memoization, despite the theoretical unbounded n.
What Interviewers Usually Probe
- Watch for floating-point accumulation errors in probability sums.
- Notice if the candidate considers naive simulation versus DP state modeling.
- Check understanding of stopping conditions and tie probabilities.
Common Pitfalls or Variants
Common pitfalls
- Ignoring half probability when both soups empty simultaneously.
- Failing to memoize overlapping DP states leading to exponential recursion.
- Incorrectly handling state transitions when soup quantities drop below zero.
Follow-up variants
- Change serving sizes or probabilities to test flexible DP modeling.
- Ask for probability that soup B empties first instead of A.
- Introduce more soups or more complex serving patterns to extend state space.
How GhostInterview Helps
- Automatically tracks DP states and updates probabilities accurately for each turn.
- Highlights threshold approximations for large n to save computation time.
- Explains step-by-step why each serving choice affects the total probability of soup A emptying first.
Topic Pages
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- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main idea behind solving Soup Servings?
Use state transition dynamic programming to model the probability that soup A empties before B, caching intermediate results to optimize recursion.
How do I handle large n values efficiently?
Approximate probabilities when n is large by using thresholds where further recursion contributes negligibly, ensuring practical O(1) performance.
Why is memoization necessary in this problem?
Without memoization, the recursion recomputes many overlapping states, leading to exponential runtime and potential stack overflow.
How are ties between soups handled?
If both soups become empty simultaneously, count it as half the probability for soup A emptying first, following the problem definition.
Can I use iterative DP instead of recursion for Soup Servings?
Yes, iterative DP builds the probability table bottom-up and avoids recursion depth issues while still applying the state transition logic.
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