Minimum number of pigs required to find the poisonous bucket

0 0 4 minutes read

Given an integer N denoting the number of buckets, and an integer M, denoting the minimum time in minutes required by a pig to die after drinking poison, the task is to find the minimum number of pigs required to figure out which bucket is poisonous within P minutes, if there is exactly one bucket with poison, while the rest is filled with water.

Examples:

Input: N = 1000, M = 15, P = 60
Output: 5
Explanation: Minimum number of pigs required to find the poisonous bucket is 5.

Input: N = 4, M = 15, P = 15
Output: 2
Explanation: Minimum number of pigs required to find the poisonous bucket is 2.

Approach: The given problem can be solved using the given observations:

A pig can be allowed to drink simultaneously on as many buckets as one would like, and the feeding takes no time.
After a pig has instantly finished drinking buckets, there has to be a cool downtime of M minutes. During this time, only observation is allowed and no feedings at all.
Any given bucket can be sampled an infinite number of times (by an unlimited number of pigs).

Now, P minutes to test and M minutes to die simply tells how many rounds the pigs can be used, i.e., how many times a pig can eat. Therefore, declare a variable called r = P(Minutes To Test) / M(Minutes To Die).

Consider the cases to understand the approach:

Case 1: If r = 1, i.e., the number of rounds is 1.
Example: 4 buckets, 15 minutes to die, and 15 minutes to test. The answer is 2. Suppose A and B represent 2 pigs, then the cases are:

Obviously, using the binary form to represent the solution as:

Conclusion: If there are x pigs, they can represent (encode) 2^x buckets.

Case 2: If r > 1, i.e. the number of rounds is more than 1. Let below be the following notations:

0 means the pig does not drink and die.
1 means the pig drinks in the first (and only) round.

Generalizing the above results(t means the pig drinks in the t round and die): If there are t attempts, a (t + 1)-based number is used to represent (encode) the buckets. (That’s also why the first conclusion uses the 2-based number)

Example: 8 buckets, 15 buckets to die, and 40 buckets to test. Now, there are 2 (= (40/15).floor) attempts, as a result, 3-based number is used to encode the buckets. The minimum number of pigs required are 2 (= Math.log(8, 3).ceil).

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum number of pigs
// required to find the poisonous bucket
void poorPigs(int buckets,
              int minutesToDie,
              int minutesToTest)
{
    // Print the result
    cout << ceil(log(buckets)
                 / log((minutesToTest
                        / minutesToDie)
                       + 1));
}
 
// Driver Code
int main()
{
    int N = 1000, M = 15, P = 60;
    poorPigs(N, M, P);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG 
{
 
  // Function to find the minimum number of pigs
  // required to find the poisonous bucket
  static void poorPigs(int buckets, int minutesToDie,
                       int minutesToTest)
  {
 
    // Print the result
    System.out.print((int)Math.ceil(
      Math.log(buckets)
      / Math.log((minutesToTest / minutesToDie)
                 + 1)));
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 1000, M = 15, P = 60;
    poorPigs(N, M, P);
  }
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python program for the above approach
import  math
 
# Function to find the minimum number of pigs
# required to find the poisonous bucket
def poorPigs(buckets, minutesToDie, minutesToTest):
   
    # Print the result
    print(math.ceil(math.log(buckets)\
                    // math.log((minutesToTest \
                                 // minutesToDie) + 1)));
 
# Driver Code
if __name__ == '__main__':
    N = 1000;
    M = 15;
    P = 60;
    poorPigs(N, M, P);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find the minimum number of pigs
  // required to find the poisonous bucket
  static void poorPigs(int buckets, int minutesToDie,
                       int minutesToTest)
  {
 
    // Print the result
    Console.WriteLine((int)Math.Ceiling(
      Math.Log(buckets)
      / Math.Log((minutesToTest / minutesToDie)
                 + 1)));
  }
 
  // Driver Code
  static public void Main()
  {
    int N = 1000, M = 15, P = 60;
    poorPigs(N, M, P);
  }
}
 
// This code is contributed by jana_sayantan.

Javascript

<script>
// Javascript program for the above approach
 
  // Function to find the minimum number of pigs
  // required to find the poisonous bucket
  function poorPigs(buckets, minutesToDie,
                       minutesToTest)
  {
  
    // Print the result
    document.write(Math.ceil(
      Math.log(buckets)
      / Math.log((minutesToTest / minutesToDie)
                 + 1)));
  }
 
    // Driver Code
    let N = 1000, M = 15, P = 60;
    poorPigs(N, M, P);
 
// This code is contributed by souravghosh0416.
</script>

Output:

Time Complexity: O(1)
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!