LeetCode 458: Poor Pigs Solution in Python – A Step-by-Step Guide

What Is LeetCode 458: Poor Pigs?

In LeetCode 458: Poor Pigs, you’re given three integers: buckets (number of buckets, one poisoned), minutesToDie (time for a pig to die after drinking poison), and minutesToTest (total time to test). Each pig can test buckets across multiple rounds (minutesToTest / minutesToDie), dying if it encounters the poison. Your task is to find the minimum number of pigs needed to identify the poisoned bucket. For example, with 4 buckets, 15 minutes to die, and 15 minutes to test (1 round), 2 pigs can do it. It’s like a deadly game of Guess Who, using pigs as binary detectors.

Problem Statement

• Input: buckets (int)—number of buckets; minutesToDie (int)—time to die; minutesToTest (int)—time to test.
• Output: int—minimum pigs needed.
• Rules:
• One bucket is poisoned.
• Pig dies in minutesToDie if it tests the poisoned bucket.
• Test rounds = minutesToTest / minutesToDie.
• Identify poisoned bucket with pig deaths.

Constraints

• 1 <= buckets <= 1000.
• 1 <= minutesToDie <= minutesToTest <= 100.

Examples to Get Us Started

• Input: buckets = 4, minutesToDie = 15, minutesToTest = 15
• Output: 2 (1 round, 2 pigs identify 4 buckets).
• Input: buckets = 4, minutesToDie = 15, minutesToTest = 30
• Output: 2 (2 rounds, 2 pigs handle 4 buckets).
• Input: buckets = 25, minutesToDie = 15, minutesToTest = 45
• Output: 3 (3 rounds, 3 pigs for 25 buckets).

Understanding the Problem: Decoding with Pigs

To solve LeetCode 458: Poor Pigs in Python, we need to find the smallest number of pigs whose test outcomes (alive or dead across rounds) can uniquely identify one poisoned bucket among buckets. Brute force simulation—testing all pig assignments—is exponential and infeasible. The key? Each pig’s state across rounds acts like a digit in a number system, where states = rounds + 1 (alive or die in one round). We’ll explore:

• Best Solution (Mathematical Base Encoding): O(1) time, O(1) space—fast and genius.
• Alternative Solution (Brute Force Simulation): O(2^n * buckets) time, O(n)—slow and impractical.

Let’s dive into the mathematical solution—it’s the pig-saving formula we need.

Best Solution: Mathematical Approach with Base Encoding

Why This Is the Best Solution

The mathematical base encoding approach is the top pick because it’s O(1) time and O(1) space, solving the problem with a single formula derived from information theory. It treats each pig as a digit in a base-(states) number system, where states = rounds + 1, and finds the smallest number of pigs (digits) to represent all buckets. It’s like using pigs as a living barcode scanner—mind-blowingly efficient!

How It Works

Here’s the strategy:

• Step 1: Calculate rounds = minutesToTest // minutesToDie.
• Step 2: States per pig = rounds + 1 (alive or die in round 1, 2, ..., or alive).
• Step 3: Minimum pigs = smallest x where (states)^x ≥ buckets.
• Step 4: Compute with log or iteration, return x.
• Why It Works:
• Each pig’s outcome (e.g., dies in round 1, alive) is a state.
• x pigs give (states)^x combinations, must cover all buckets.

Step-by-Step Example

Example: buckets = 4, minutesToDie = 15, minutesToTest = 15

• Rounds: 15 // 15 = 1.
• States: 1 + 1 = 2 (alive, die in round 1).
• Pigs Needed:
• 1 pig: 2¹ = 2 < 4.
• 2 pigs: 2² = 4 ≥ 4.
• Result: 2 pigs.
• Verify:
• Pig 1: [0,1], Pig 2: [0,1].
• 00 (both alive), 01, 10, 11 (unique for 4 buckets).

Example: buckets = 25, minutesToDie = 15, minutesToTest = 45

• Rounds: 45 // 15 = 3.
• States: 3 + 1 = 4 (alive, die in 1, 2, 3).
• Pigs:
• 2 pigs: 4² = 16 < 25.
• 3 pigs: 4³ = 64 ≥ 25.
• Result: 3.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, broken down clearly:

Copy codeclass Solution:
    def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
        # Step 1: Calculate rounds
        rounds = minutesToTest // minutesToDie

        # Step 2: States per pig
        states = rounds + 1  # Alive or die in each round

        # Step 3: Find minimum pigs
        pigs = 0
        while (states ** pigs) < buckets:
            pigs += 1

        return pigs

• Line 4: Compute rounds (e.g., 15 // 15 = 1).
• Line 7: States = rounds + 1 (e.g., 2 states for 1 round).
• Line 10-12: Increment pigs until states^pigs ≥ buckets.
• Line 14: Return minimum pigs.
• Time Complexity: O(log buckets)—while loop iterations.
• Space Complexity: O(1)—just variables.

It’s like a pig-powered math marvel!

Math Insight

• Base = states, need base^x ≥ buckets.
• x = ceil(log(buckets) / log(states)).
• Here, simple iteration avoids floating-point issues.

Alternative Solution: Brute Force Simulation

Why an Alternative Approach?

The brute force simulation tries pig assignments—O(2^n * buckets) time, O(n) space—simulating tests to find the minimum pigs. It’s wildly inefficient but shows the concept, like manually testing pig squads until one works. Good for intuition or tiny inputs!

How It Works

• Step 1: Start with 1 pig, simulate all test assignments.
• Step 2: Check if outcomes uniquely identify buckets.
• Step 3: Increment pigs until successful.

Step-by-Step Example

Example: buckets = 4, minutesToDie = 15, minutesToTest = 15

• 1 Pig, 1 Round:
• States: [alive, die], 2 outcomes < 4, fail.
• 2 Pigs:
• 00, 01, 10, 11 → 4 outcomes ≥ 4, success.
• Result: 2.

Code for Brute Force (Simplified)

Copy codeclass Solution:
    def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
        rounds = minutesToTest // minutesToDie
        states = rounds + 1

        def can_identify(pigs):
            # Simulate: count unique outcomes
            # This is conceptual; full simulation is complex
            return (states ** pigs) >= buckets

        pigs = 0
        while not can_identify(pigs):
            pigs += 1
        return pigs

• Line 4-5: Compute states.
• Line 7-9: Placeholder for simulation, uses math check.
• Line 11-13: Increment pigs until sufficient.
• Time Complexity: O(log buckets)—simplified, real brute force much worse.
• Space Complexity: O(1)—minimal.

It’s a slow pig tester!

Comparing the Two Solutions

• Mathematical (Best):
• Pros: O(1), fast, exact.
• Cons: Requires math insight.
• Brute Force (Alternative):
• Pros: O(2^n * buckets), intuitive.
• Cons: Impractical.

Math wins hands down.

Edge Cases and More Examples

• Input: buckets=1 → 0.
• Input: buckets=2, 15, 15 → 1.
• Input: buckets=1000, 15, 60 → 5.

Math handles all perfectly.

Complexity Recap

• Mathematical: Time O(1), Space O(1).
• Brute Force: Time O(2^n * buckets), Space O(n).

Math’s the champ.

Key Takeaways

• Mathematical: Encode with states.
• Brute Force: Simulate all.
• Python Tip: Math trumps trials—see [Python Basics](/python/basics).

Final Thoughts: Save Those Pigs

LeetCode 458: Poor Pigs in Python is a logic-packed farmyard puzzle. The mathematical solution is your pig-saving genius, while brute force is a slow slog. Want more brain teasers? Try LeetCode 287: Find the Duplicate Number or LeetCode 338: Counting Bits. Ready to test some buckets? Head to Solve LeetCode 458 on LeetCode (if unlocked) and solve it today—your coding skills are squealing with potential!