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) 2x 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 approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to find the minimum number of pigs// required to find the poisonous bucketvoid poorPigs(int buckets,              int minutesToDie,              int minutesToTest){    // Print the result    cout << ceil(log(buckets)                 / log((minutesToTest                        / minutesToDie)                       + 1));}Â
// Driver Codeint main(){Â Â Â Â int N = 1000, M = 15, P = 60;Â Â Â Â poorPigs(N, M, P);Â
    return 0;} |
Java
// Java program for the above approachimport java.io.*;Â
class GFG {Â
  // Function to find the minimum number of pigs  // required to find the poisonous bucket  static void poorPigs(int buckets, int minutesToDie,                       int minutesToTest)  {Â
    // Print the result    System.out.print((int)Math.ceil(      Math.log(buckets)      / Math.log((minutesToTest / minutesToDie)                 + 1)));  }Â
  // Driver Code  public static void main(String[] args)  {    int N = 1000, M = 15, P = 60;    poorPigs(N, M, P);  }}Â
// This code is contributed by Dharanendra L V. |
Python3
# Python program for the above approachimport mathÂ
# Function to find the minimum number of pigs# required to find the poisonous bucketdef poorPigs(buckets, minutesToDie, minutesToTest):       # Print the result    print(math.ceil(math.log(buckets)\                    // math.log((minutesToTest \                                 // minutesToDie) + 1)));Â
# Driver Codeif __name__ == '__main__':Â Â Â Â N = 1000;Â Â Â Â M = 15;Â Â Â Â P = 60;Â Â Â Â poorPigs(N, M, P);Â
# This code is contributed by 29AjayKumar |
C#
// C# program for the above approachusing System;class GFG{Â
  // Function to find the minimum number of pigs  // required to find the poisonous bucket  static void poorPigs(int buckets, int minutesToDie,                       int minutesToTest)  {Â
    // Print the result    Console.WriteLine((int)Math.Ceiling(      Math.Log(buckets)      / Math.Log((minutesToTest / minutesToDie)                 + 1)));  }Â
  // Driver Code  static public void Main()  {    int N = 1000, M = 15, P = 60;    poorPigs(N, M, P);  }}Â
// This code is contributed by jana_sayantan. |
Javascript
<script>// Javascript program for the above approachÂ
  // 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)));  }Â
    // Driver Code    let N = 1000, M = 15, P = 60;    poorPigs(N, M, P);Â
// This code is contributed by souravghosh0416.</script> |
5
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Time Complexity: O(1)
Auxiliary Space: O(1)
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