Poor Pigs - Leetcode Solution

Problem Statement:

There are buckets of water placed in a line, numbered from left to right from 0 to 1000000. Initially, all the buckets are full of water. A poor pig is to be fed a poison which will kill it after T minutes, but before that, the poor pig can drink from any water-filled bucket that is within 15 minutes of its location at a rate of 1 bucket per minute. What is the minimum number of pigs you need to test to find the "poison" bucket within T minutes?

Solution:

This is a interesting problem which requires us to think and come up with an optimal solution to minimize the number of pigs required to find the poisoned bucket. To understand and solve the problem, we need to make a few observations about the problem.

Observation 1:

Given the time T, we can calculate the maximum number of buckets that can be tested by a single pig. Since the pig can drink from any water-filled bucket that is within 15 minutes of its location and can drink 1 bucket of water per minute, a pig can test (T/15 + 1) buckets.

Observation 2:

If we have only one pig, then we can only test a fixed number of buckets within T minutes. Let's assume a pig can test x buckets within T minutes, then we have to make sure that x is not less than the number of buckets that could contain the poison. If we have n buckets, then we can write the equation as follows: (x + 1) ^ n >= Buckets with poison (log (x+1)) * n >= log(Buckets with poison) n >= log(Buckets with poison) / log(x+1)

Observation 3:

From the above equation, we can deduce that the number of pigs required to test all the buckets within T minutes is mainly dependent on two factors - the time T and the number of buckets n. We can use the equation derived in the previous observation to calculate the minimum number of pigs required to test all the buckets within T minutes.

Now we can go ahead to define an algorithm to solve the problem as follows:

Algorithm:

1. Calculate the number of pigs required to test all the buckets within T minutes, given the number of buckets n.
2. Return the answer.

To implement step 1, we can use the following equation derived in observation 3:

pigs = ceil(log(buckets) / log(minutesToTest / minutesToDie + 1));

Let's see the implementation of the solution in python:

class Solution:
    def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
        pigs = 0
        while (pow(minutesToTest // minutesToDie + 1, pigs) < buckets):
            pigs += 1
        return pigs

Time Complexity:

The time complexity of the above algorithm is O(log(n)).

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