Poor Pigs

                                            
class Solution {
public:
    int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
        if (buckets == 1) return 0;
        int pigs = 0;
        int states = minutesToTest/minutesToDie + 1;
        while (pow(states, pigs) < buckets) {
            pigs++;
        }
        return pigs;
    }
};

                                            
class Solution {
    public int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
        int pigs = 0;
        int states = minutesToTest/minutesToDie + 1;
        while (Math.pow(states, pigs) < buckets) {
            pigs++;
        }
        return pigs;
    }
}

                                            
class Solution(object):
    def poorPigs(self, buckets, minutesToDie, minutesToTest):
        pigs = 0
        while (minutesToTest / minutesToDie + 1) ** pigs < buckets:
            pigs += 1
        return pigs

                                            
int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
    int pigs = 0;
    int states = minutesToTest/minutesToDie + 1;
    
    while (pow(states, pigs) < buckets) {
        pigs++;
    }
    
    return pigs;
}

                                            
function poorPigs(buckets: number, minutesToDie: number, minutesToTest: number): number {
    let pigs = 0;
    let states = Math.floor(minutesToTest/minutesToDie) + 1;
    
    while (Math.pow(states, pigs) < buckets) {
        pigs++;
    }
    
    return pigs;
};

                                            
class Solution {
    func poorPigs(_ buckets: Int, _ minutesToDie: Int, _ minutesToTest: Int) -> Int {
        var pigNumber = 0
        var states = (minutesToTest/minutesToDie) + 1
        while pow(Double(states), Double(pigNumber)) < Double(buckets) {
            pigNumber += 1
        }
        return pigNumber
    }
}

                                            
class Solution {
    fun poorPigs(buckets: Int, minutesToDie: Int, minutesToTest: Int): Int {
        var pigs = 0
        var testInterval = minutesToTest / minutesToDie + 1
        while (Math.pow(testInterval.toDouble(), pigs.toDouble()) < buckets) {
            pigs++
        }
        return pigs
    }
}

                                            
class Solution {
  int poorPigs(int buckets, int minutesToDie, int minutesToTest) {
    int pigs = 0;
    int rounds = minutesToTest ~/ minutesToDie + 1;
    while (pow(rounds, pigs) < buckets) {
      pigs++;
    }
    return pigs;
  }
}

                                            
func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
    pigs := 0
    base := minutesToTest/minutesToDie + 1
    for pow := 0; pow < buckets; pow++ {
        if power(base, pow) >= buckets {
            return pigs
        }
        pigs++
    }
    return pigs
}

func power(base, exp int) int {
    result := 1
    for exp > 0 {
        if exp%2 == 1 {
            result *= base
        }
        exp /= 2
        base *= base
    }
    return result
}

                                            
# @param {Integer} buckets
# @param {Integer} minutes_to_die
# @param {Integer} minutes_to_test
# @return {Integer}
def poor_pigs(buckets, minutes_to_die, minutes_to_test)
    pigs = 0
    while (minutes_to_test / minutes_to_die + 1) ** pigs < buckets
        pigs += 1
    end
    pigs
end

                                            
object Solution {
    def poorPigs(buckets: Int, minutesToDie: Int, minutesToTest: Int): Int = {
        var pigs = 0
        var tests = minutesToTest / minutesToDie + 1
        
        while (Math.pow(tests, pigs) < buckets) {
            pigs += 1
        }
        
        pigs
    }
}

                                            
public class Solution {
    public int PoorPigs(int buckets, int minutesToDie, int minutesToTest) {
        int pigs = 0;
        int states = minutesToTest / minutesToDie + 1;
        
        while (Math.Pow(states, pigs) < buckets) {
            pigs++;
        }
        
        return pigs;
    }
}

                                            
/**
 * @param {number} buckets
 * @param {number} minutesToDie
 * @param {number} minutesToTest
 * @return {number}
 */
var poorPigs = function(buckets, minutesToDie, minutesToTest) {
    let pigs = 0;
    while (Math.pow((Math.floor(minutesToTest/minutesToDie) + 1), pigs) < buckets) {
        pigs++;
    }
    return pigs;
};

To solve this coding challenge, the primary goal is to determine the minimum number of pigs required to identify the single poisonous bucket among several buckets of liquid within a certain time frame. Here’s a step-by-step explanation of how to approach this problem logically and derive the solution:

Explanation:

Understanding the Problem:

You have a number of buckets, with exactly one bucket containing poison.
We have a limited number of pigs and a set amount of time to determine which bucket is poisonous.
Each pig can drink from multiple buckets, but once a pig drinks from a poisonous bucket and dies, it cannot be reused.
You can perform tests in cycles, where each cycle takes
minutesToDie

minutes, and you can perform multiple such cycles within
minutesToTest

minutes.

Formulating the Problem:

The pigs, when drinking from the buckets, act as binary sensors (alive or dead).
Each pig can give us a bit of information depending on whether it survives after drinking from a set of buckets.
The total number of tests we can perform is
minutesToTest // minutesToDie

.
This means each pig can provide more than one bit of information through multiple cycles.

Deriving the Formula:

To determine the number of pigs required, we consider the number of possible states. Given
states = (minutesToTest // minutesToDie) + 1

, each pig can represent multiple states.
If we have
k

pigs and
states

states per pig, then the total number of combinations of states that the pigs can represent is
states^k

.
We need this number of combinations to be at least equal to the number of buckets to ensure we can distinguish each bucket.
We need to find the smallest number of pigs
k

such that
(minutesToTest // minutesToDie + 1)^k >= buckets

.

Solution Approach:

We incrementally increase the number of pigs
k

until the condition
(minutesToTest // minutesToDie + 1)^k >= buckets

is met.
The number of pigs
k

that satisfies this equation is the minimum number of pigs required.

Now, let’s illustrate this solution in pseudocode:

Pseudocode with Comments

                                            
# Function to calculate minimum number of pigs needed
function calculateMinimumPigs(buckets, minutesToDie, minutesToTest)
# Initialize the number of pigs to 0
numberOfPigs = 0

# Calculate the number of cycles we can perform
states = (minutesToTest // minutesToDie) + 1

# While the total number of buckets that can be tested by the current number of pigs is less than the required buckets
while states^numberOfPigs < buckets
# Increment the number of pigs
numberOfPigs += 1

# Return the minimum number of pigs found
return numberOfPigs

# Example Usage
buckets = 4
minutesToDie = 15
minutesToTest = 15
output = calculateMinimumPigs(buckets, minutesToDie, minutesToTest)
print(output)  # Expected output: 2

Step-by-Step Explanation / Detailed Steps in Pseudocode:

Initialize Variables:

Start with
numberOfPigs

set to 0.

Calculate States:

Calculate the number of states each pig can represent using the formula
states = (minutesToTest // minutesToDie) + 1

.

While Loop to Find Minimum Pigs:

Enter a loop that continues until the condition
(states^numberOfPigs) >= buckets

is met.
This loop ensures we keep adding pigs incrementally until their combined states are sufficient to identify the poisonous bucket.

Increment Number of Pigs:

Inside the loop, increment the
numberOfPigs

by 1 in each iteration.

Return Result:

Once the loop terminates, return the
numberOfPigs

as the minimum number required.

By following the above steps, we systematically find the minimum number of pigs needed to identify the poisonous bucket within the given constraints of time to test and time to die.