Introduction

In this blog post, we will explore how to solve the LeetCode problem "458" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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: Choose some live pigs to feed. For each pig, choose which buckets to feed it. The pig will consume all the chosen buckets simultaneously and will take no time. Each pig can feed from any number of buckets, and each bucket can be fed from by any number of pigs. Wait for minutesToDie minutes. You may not feed any other pigs during this time. After minutesToDie minutes have passed, any pigs that have been fed the poisonous bucket will die, and all others will survive. Repeat this process until you run out of time. 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 <= 1000 1 <= minutesToDie <= minutesToTest <= 100

Explanation

Here's the solution to the "Poor Pigs" problem:

• Key Idea: The core idea is to use each pig to represent a dimension of information. The number of states a pig can be in (dead or alive at different time intervals) determines the base of our number system. We want to find the smallest number of dimensions (pigs) such that the total number of possible states is greater than or equal to the number of buckets.

• Calculating States: Calculate the number of trials possible during the given time: trials = minutesToTest / minutesToDie + 1. This value determines the number of different outcomes for each pig (alive at different times, or dead after a specific time).

• Finding Minimum Pigs: Determine the minimum number of pigs needed such that trials^pigs >= buckets. This is equivalent to finding the smallest pigs such that log(buckets) / log(trials) <= pigs.

• Complexity: O(1) runtime, O(1) storage.

Code

    import math

def poorPigs(buckets: int, minutesToDie: int, minutesToTest: int) -> int:
    """
    Determines the minimum number of pigs needed to find the poisonous bucket.

    Args:
        buckets: The number of buckets.
        minutesToDie: The time it takes for a pig to die after drinking poison.
        minutesToTest: The total time available for testing.

    Returns:
        The minimum number of pigs needed.
    """
    if buckets == 1:
        return 0

    trials = minutesToTest // minutesToDie + 1
    return math.ceil(math.log(buckets) / math.log(trials))