LeetCode 458 - Poor Pigs

The problem gives us a set of buckets where exactly one bucket contains poison. We need to determine which bucket is poisonous using the fewest number of pigs possible, under a strict time limit. Each pig can participate in multiple rounds of testing.

leetcodehardmathdynamic-programmingcombinatorics

LeetCode Problem 458

Difficulty: 🔴 Hard
Topics: Math, Dynamic Programming, Combinatorics

Solution

Problem Understanding

The problem gives us a set of buckets where exactly one bucket contains poison. We need to determine which bucket is poisonous using the fewest number of pigs possible, under a strict time limit.

Each pig can participate in multiple rounds of testing. A pig dies exactly minutesToDie minutes after drinking poison. The total testing window is minutesToTest, which means we can perform several rounds of experiments.

The key detail is that pigs can drink from multiple buckets in the same round, and buckets can be given to multiple pigs. This makes the problem fundamentally about encoding information efficiently.

The input parameters represent the following:

  • buckets is the number of possible poisonous buckets.
  • minutesToDie is how long it takes for a poisoned pig to die.
  • minutesToTest is the total amount of testing time available.

The output is the minimum number of pigs required to uniquely identify the poisonous bucket within the allotted time.

The constraints are relatively small, with at most 1000 buckets, but the challenge is conceptual rather than computational. A brute-force simulation of all possible feeding schedules would become extremely complicated very quickly. The intended solution relies on combinatorics and information theory.

Several edge cases are important:

  • If there is only one bucket, no pigs are needed because the poisonous bucket is already known.
  • If minutesToTest == minutesToDie, we only get one testing round.
  • If we have multiple rounds, each pig can encode more than just alive or dead information.
  • A naive binary-search-style interpretation is incorrect because pigs can participate in multiple rounds and therefore represent multiple states, not just two.

Approaches

Brute Force Approach

A brute-force approach would attempt to explicitly simulate all possible testing schedules. We could try every combination of which pigs drink from which buckets in each round, then verify whether the resulting death patterns uniquely identify every bucket.

This works because each testing strategy creates a mapping between buckets and observable pig outcomes. If every bucket maps to a unique outcome, the strategy is valid.

However, this becomes infeasible extremely quickly. The number of possible feeding configurations grows exponentially with both the number of pigs and the number of rounds. Even with only a handful of pigs, the state space becomes enormous.

The brute-force method is therefore impractical and unnecessary.

Optimal Insight

The key observation is that each pig can represent multiple distinct states, not just alive or dead.

Suppose we can perform:

$$\text{rounds} = \frac{\text{minutesToTest}}{\text{minutesToDie}}$$

testing rounds.

A pig can end in exactly rounds + 1 possible states:

  • Dies in round 1
  • Dies in round 2
  • ...
  • Dies in round rounds
  • Survives all rounds

This means one pig can encode rounds + 1 distinct outcomes.

If we use p pigs, the total number of distinguishable outcomes becomes:

$$(\text{rounds} + 1)^p$$

We need enough unique outcome combinations to identify every bucket:

$$(\text{rounds} + 1)^p \ge \text{buckets}$$

The task therefore reduces to finding the smallest integer p satisfying this inequality.

This transforms the problem from simulation into a compact mathematical computation.

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Simulates all feeding schedules and outcomes
Optimal O(log buckets) O(1) Uses combinatorial state counting

Algorithm Walkthrough

  1. Compute how many testing rounds are available.

The number of complete testing rounds is:

$$\text{rounds} = \frac{\text{minutesToTest}}{\text{minutesToDie}}$$

For example, if pigs die in 15 minutes and we have 60 minutes total, we can run 4 rounds. 2. Determine how many states a single pig can represent.

A pig can:

  • Die in round 1
  • Die in round 2
  • ...
  • Die in the final round
  • Survive completely

Therefore, one pig has:

$$\text{statesPerPig} = \text{rounds} + 1$$

possible outcomes. 3. Count how many buckets can be represented with p pigs.

If each pig has statesPerPig possible states, then p pigs together can encode:

$$(\text{statesPerPig})^p$$

unique combinations. 4. Find the minimum number of pigs.

Start from pigs = 0 and repeatedly increase it until:

$$(\text{statesPerPig})^{pigs} \ge \text{buckets}$$

The first valid value is the answer.

Why it works

Each bucket can be assigned a unique combination of pig outcomes. Conceptually, we are encoding bucket numbers in a positional numeral system where the base is statesPerPig.

Each pig acts like one digit in that numbering system. Since p pigs can produce exactly (statesPerPig)^p unique outcome patterns, having at least as many patterns as buckets guarantees every bucket can be uniquely identified.

Python Solution

class Solution:
    def poorPigs(self, buckets: int, minutesToDie: int, minutesToTest: int) -> int:
        rounds = minutesToTest // minutesToDie
        states_per_pig = rounds + 1
        
        pigs = 0
        
        while states_per_pig ** pigs < buckets:
            pigs += 1
        
        return pigs

The implementation first computes the number of testing rounds available. From that, it derives the number of states each pig can represent.

The variable pigs starts at zero because it is possible that no pigs are needed when there is only one bucket.

The loop repeatedly checks whether the current number of pigs can distinguish enough buckets. The expression:

states_per_pig ** pigs

represents the total number of unique outcome combinations.

As soon as this value becomes at least the number of buckets, we know the current number of pigs is sufficient, so we return it.

The implementation is extremely compact because the difficult part of the problem is the mathematical insight, not the coding itself.

Go Solution

func poorPigs(buckets int, minutesToDie int, minutesToTest int) int {
    rounds := minutesToTest / minutesToDie
    statesPerPig := rounds + 1

    pigs := 0

    for power(statesPerPig, pigs) < buckets {
        pigs++
    }

    return pigs
}

func power(base int, exp int) int {
    result := 1

    for exp > 0 {
        result *= base
        exp--
    }

    return result
}

The Go implementation follows the same logic as the Python version. Since Go does not have a built-in integer exponentiation operator like Python's **, we implement a small helper function named power.

Integer overflow is not a concern here because the constraints are small. Even in the largest case, the computed powers remain comfortably within Go's integer range.

Worked Examples

Example 1

Input:

buckets = 4
minutesToDie = 15
minutesToTest = 15

First compute the number of rounds:

$$\frac{15}{15} = 1$$

So each pig has:

$$1 + 1 = 2$$

possible states.

Pigs Distinguishable Buckets
0 1
1 2
2 4

When pigs = 2, we can distinguish 4 buckets, so the answer is:

2

Example 2

Input:

buckets = 4
minutesToDie = 15
minutesToTest = 30

Compute rounds:

$$\frac{30}{15} = 2$$

Each pig now has:

$$2 + 1 = 3$$

possible states:

  • Die in round 1
  • Die in round 2
  • Survive
Pigs Distinguishable Buckets
0 1
1 3
2 9

With 2 pigs we can distinguish 9 buckets, which is enough for 4 buckets.

Answer:

2

Complexity Analysis

Measure Complexity Explanation
Time O(log buckets) Each iteration increases pigs until the exponential capacity exceeds buckets
Space O(1) Only a few integer variables are used

The number of loop iterations is very small because the distinguishable bucket count grows exponentially with the number of pigs. Even for the maximum constraint of 1000 buckets, only a handful of iterations are needed.

Test Cases

solution = Solution()

assert solution.poorPigs(4, 15, 15) == 2  # Example 1
assert solution.poorPigs(4, 15, 30) == 2  # Example 2

assert solution.poorPigs(1, 1, 1) == 0  # Only one bucket
assert solution.poorPigs(2, 1, 1) == 1  # Single pig enough
assert solution.poorPigs(8, 15, 15) == 3  # One round, binary states

assert solution.poorPigs(9, 15, 30) == 2  # Two pigs, three states each
assert solution.poorPigs(27, 15, 45) == 3  # Three rounds -> four states

assert solution.poorPigs(1000, 15, 60) == 5  # Large constraint case
assert solution.poorPigs(625, 15, 60) == 4  # Exact power boundary
assert solution.poorPigs(626, 15, 60) == 5  # Just above power boundary

assert solution.poorPigs(16, 15, 15) == 4  # Pure binary encoding
assert solution.poorPigs(17, 15, 15) == 5  # Needs one more pig
Test Why
(4, 15, 15) Validates the first example
(4, 15, 30) Validates multiple testing rounds
(1, 1, 1) Tests zero-pig edge case
(2, 1, 1) Smallest nontrivial input
(8, 15, 15) Single-round binary encoding
(9, 15, 30) Exact capacity boundary
(27, 15, 45) Multiple rounds with larger state space
(1000, 15, 60) Maximum bucket stress test
(625, 15, 60) Exact power threshold
(626, 15, 60) Just above threshold
(16, 15, 15) Pure power-of-two case
(17, 15, 15) Requires additional pig after threshold

Edge Cases

One important edge case occurs when there is only one bucket. In this scenario, no experimentation is necessary because the poisonous bucket is already known. A buggy implementation might incorrectly return one pig simply because it assumes testing is always required. The current implementation handles this naturally because the loop condition fails immediately when buckets == 1.

Another important case occurs when there is only one testing round available, meaning minutesToTest == minutesToDie. In this situation, each pig only has two states, alive or dead. The problem effectively becomes binary encoding. Implementations that incorrectly assume multiple rounds are always available may overestimate the information capacity of each pig.

A subtle boundary case appears when the number of buckets is exactly a power of (rounds + 1). For example, if each pig has 5 states and there are exactly 625 buckets, then 4 pigs are sufficient because:

$$5^4 = 625$$

Using a loop condition like <= instead of < would incorrectly add an unnecessary pig. The implementation avoids this bug by using the correct strict comparison condition.