Leetcode 458. Poor Pigs

This problem belongs to the category of puzzles which involve figuring out an arrangement to identify a unique case. Here, we have buckets filled with either water or poison and pigs which we can use to sample the buckets. The main point in this problem is that we can sample as many buckets as we want at the same time and if the pig drinks poison, it will die within a specific period of time.

Say if the time to die is m minutes, and we have p minutes in total. Then if we think about it, we can do around p/m tests in total. And if each trial, we allow x pigs to drink. This means that the number of buckets we can test in total is (x^(p/m)). Since the problem asks for the minimum number of pigs to identify the poison bucket, we'll aim to make (x^(p/m)) >= total number of buckets.

The above equation can be simplified as x >= ceil(log(n) / log(p/m+1)) where n is the total number of buckets and log represents logarithm function, ceil represents the smallest integer not less than a given value. This is based on the property of logarithm which says log(a^b) = b*log(a).

Walkthrough example: Suppose we have 1000 buckets, 1 hour and the poison effect after 15 minutes. So this gives us 60/15 = 4 trials we can make. If suppose we have 5 pigs, we can try 5^4 = 625 buckets. But we need to test 1000 buckets, so need more pigs. On trying for 6 pigs, we get 6^4 = 1296 which is more than 1000. Thus minimum pigs required are 6.

Here is an illustration explaining how each step of the algorithm works

Using the code,


3import math
4class Solution:
5  def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
6    return math.ceil(math.log(buckets) / math.log(minutesToTest / minutesToDie + 1))


3import java.lang.Math;
4class Solution {
5  public int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
6      return (int)Math.ceil(Math.log(buckets) / Math.log(minutesToTest / minutesToDie + 1));
7  }


3class Solution {
4  poorPigs(buckets, minutesToDie, minutesToTest) {
5    return Math.ceil(Math.log(buckets) / Math.log(minutesToTest / minutesToDie + 1));
6  }


3#include <cmath>
4class Solution {
6    int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
7        return std::ceil(std::log(buckets) / std::log(minutesToTest / minutesToDie + 1));
8    }


3using System;
4public class Solution {
5    public int PoorPigs(int buckets, int minutesToDie, int minutesToTest) {
6        return (int)Math.Ceiling((Math.Log(buckets) / Math.Log(minutesToTest / minutesToDie + 1)));
7    }

The primary idea behind the algorithm is to make use of the pigs as binary placeholders of sort to identify the deadly drink. The pigs essentially form a K-ary number system and each pig represents the 'digit' in the K-ary system.

The first pig goes to the troughs in the order of 1,2,3,...,n and takes its sips along with the rest of the pigs. The second pig would be a step behind and drink from the troughs in the order of 2,1,2,3,...,n. The ith pig will start drinking after i minutes.

This way, we can figure out exactly which bucket has the poison. This works because if the i-th pig dies at the j-th minute, it means that the poison was in the bucket which is the i-th bucket in j-th order.

Now let's take a look at the time complexity of the solution. Notice that the code does nothing more than a few mathematical operations. Hence, both the time and space complexity for this problem are O(1).

One key thing to keep in mind while solving problems like these is the re-orientation of the problem statement. Instead of the common question "How many buckets can x pigs test?“, we need to ask "What's the maximum number of buckets that x pigs can test?" and code accordingly.

It is often a good idea to write down a few examples and identify patterns before attempting to code the solution. It is also important to understand the mathematical principles that underlie the problem. For this problem, you would need a good understanding of logarithms and K-ary number systems. Finally, while the problem could be simply asking for the minimum number of pigs required, the real challenge lies in programming an efficient solution.

This problem can be a fun coding challenge to tackle and can help refine your problem-solving skills in computer science. Make sure to understand the logic behind the solution and try to implement it yourself in other languages. Understanding these concepts can also be very useful in preparing for coding interviews as they often demand a strong foundation of problem-solving skills.

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