There are buckets buckets of liquid, where exactly one of the buckets is poisonous. To figure out which one is poisonous, you feed some number of (poor) pigs the liquid to see whether they will die or not. Unfortunately, you only have minutesToTest minutes to determine which bucket is poisonous.
You can feed the pigs according to these steps:
minutesToDie minutes. You may not feed any other pigs during this time.minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive.Given buckets, minutesToDie, and minutesToTest, return the minimum number of pigs needed to figure out which bucket is poisonous within the allotted time.
Example 1:
Input: buckets = 4, minutesToDie = 15, minutesToTest = 15 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig buckets 1 and 2, and feed the second pig buckets 2 and 3. At time 15, there are 4 possible outcomes: - If only the first pig dies, then bucket 1 must be poisonous. - If only the second pig dies, then bucket 3 must be poisonous. - If both pigs die, then bucket 2 must be poisonous. - If neither pig dies, then bucket 4 must be poisonous.
Example 2:
Input: buckets = 4, minutesToDie = 15, minutesToTest = 30 Output: 2 Explanation: We can determine the poisonous bucket as follows: At time 0, feed the first pig bucket 1, and feed the second pig bucket 2. At time 15, there are 2 possible outcomes: - If either pig dies, then the poisonous bucket is the one it was fed. - If neither pig dies, then feed the first pig bucket 3, and feed the second pig bucket 4. At time 30, one of the two pigs must die, and the poisonous bucket is the one it was fed.
Constraints:
1 <= buckets <= 10001 <= minutesToDie <= minutesToTest <= 100Given the problem's constraints and the binary outcomes (either a pig dies or it doesn't), think of the problem in terms of binary representations. A pig can sample several buckets at once, die or survive based on which buckets contain poison. This gives a binary outcome for each test. By using logarithmic calculations with the number of possible outcomes, you can determine the number of states (base 2) that each pig can check within the available time.
This solution leverages the mathematical relationship between the problem variables. The number of tests each pig can take during the allocated time is minutesToTest // minutesToDie. Since each pig can be in one of several states (dead/alive), each state helps narrow down the possibilities. Using logarithms, you derive the minimum number of pigs required to cover all buckets.
Time Complexity: O(1). The use of logarithmic calculations on constants results in constant time complexity.
Space Complexity: O(1). The storage used does not scale with input size.
In this approach, consider how many different states each pig can "check" through its outcome. Since a pig can be either alive or dead at the end of testing, each pig effectively provides binary outcomes, which can be thought of as a base. The total number of scenarios (outcomes) that can be tested is the product of the number of states each pig can obtain.
This implementation iteratively increases the number of pigs until the number of scenarios they can test matches or exceeds the number of buckets. Each pig increases the power in which states (base) are multiplied, and hence increases the number of possible outcomes they can cover.
Time Complexity: O(log(buckets)). The loop iteratively increases until states raised to pigs is greater or equal to buckets, which is logarithmic relative to the base determined by the states.
Space Complexity: O(1). The memory usage is constant.
To solve this problem, we need to understand how many states or outcomes a single pig can have. Given the time constraint, a pig can have two states after testing a bucket: it dies, or it survives. If
we allow for tests over a given time duration, a single pig can represent multiple states by modifying the test configuration over time. In essence, with x pigs and y tests, the states can be represented as (y + 1)^(x), where (y + 1) accounts for the initial baseline test and each subsequent test outcome.
Therefore, for 'n' buckets, the equation becomes (minutesToTest / minutesToDie + 1) ^ pigs >= buckets. We solve for pigs using logarithms, over a bounded integer domain.
In C, we use the math library to calculate the logarithms needed for the calculation of pigs. The function calculates 'tests' as the number of complete test cycles within the given time window. It then applies the mathematical formula as described to compute and return the number of pigs needed.
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Time Complexity: O(1) - Direct calculation using logarithms.
Space Complexity: O(1) - Only constant space is used.
Another perspective involves using a binary search. This method articulates the problem in terms of balancing the number of pigs against the possible combinations of outcomes they represent per test. Use a binary search over the pig count, harnessing calculations to determine whether given a mid-point pig count, all necessary conditions for success are achievable. This method, while initially more complex, offers a systematic exploration of feasible pig counts.
This solution defines a function to determine if a specific number of pigs meets the criteria. By systematically narrowing down using binary search, it minimizes computational steps, offering an optimized solution pathway.
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Java
Python
C#
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Time Complexity: O(log n) - Efficient binary search of potential pig values.
Space Complexity: O(1) - Constant space usage.
| Approach | Complexity |
|---|---|
| Binary Strategy | Time Complexity: O(1). The use of logarithmic calculations on constants results in constant time complexity. |
| State Expansion Strategy | Time Complexity: O(log(buckets)). The loop iteratively increases until states raised to pigs is greater or equal to buckets, which is logarithmic relative to the base determined by the states. |
| Mathematical Representation Using States | Time Complexity: O(1) - Direct calculation using logarithms. |
| Binary Search for Minimum Pigs | Time Complexity: O(log n) - Efficient binary search of potential pig values. |
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