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Given 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.
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.
1import math
2
3def poor_pigs(buckets: int, minutesToDie: int, minutesToTest: int) -> int:
4 states = minutesToTest // minutesToDie + 1
5 return math.ceil(math.log(buckets) / math.log(states))
6This 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.
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.
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.
1import math
2
3def poor_pigs(buckets: int, minutesToDieTo 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.
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.
Time Complexity: O(log n) - Efficient binary search of potential pig values.
Space Complexity: O(1) - Constant space usage.
1
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.
In Python, the math library provides easy access to logarithmic and rounding operations. The solution proceeds by calculating the number of tests possible and solving the core equation for pigs.
This Python solution parallels others in logic, emphasizing a streamlined binary search over possible pig values. It checks each midpoint and narrows the bounds efficiently.