How do you solve Poor Pigs on LeetCode?

Poor Pigs (LeetCode #458) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The number of pigs needed grows logarithmically with the number of buckets when using the optimal approach.

What is the brute force solution for Poor Pigs?

The brute force approach for Poor Pigs is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can test each bucket individually by feeding a pig from each bucket. This is straightforward but inefficient, as it requires multiple rounds of testing and can take a long time. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Poor Pigs?

Poor Pigs uses the following concepts: Math, Dynamic Programming, Combinatorics. The recommended patterns to study are: Mathematical reasoning, Exponential growth patterns.

What are the interview tips for Poor Pigs?

Always clarify the problem constraints and edge cases before diving into solutions. Think about how you can reduce the problem size or complexity through mathematical relationships. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Poor Pigs?

Confusing the number of tests with the number of pigs required. Failing to account for the time constraint properly.

#458

Poor Pigs

Hard

MathDynamic ProgrammingCombinatoricsMathematical reasoningExponential growth patterns

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the mathematical relationship between the number of pigs, the number of tests, and the number of buckets. By calculating how many unique outcomes can be generated with a given number of pigs and tests, we can efficiently determine the minimum number of pigs required.

⚙️

Algorithm

3 steps

1Step 1: Calculate the number of tests available as minutesToTest / minutesToDie.
2Step 2: Use the formula (tests + 1) ^ pigs to determine how many buckets can be tested with a given number of pigs.
3Step 3: Increment the number of pigs until the total number of buckets can be covered.

solution.py9 lines

1def poor_pigs(buckets, minutesToDie, minutesToTest):
2    if minutesToTest == minutesToDie:
3        return buckets - 1
4    tests = minutesToTest // minutesToDie
5    pigs = 0
6    while (tests + 1) ** pigs < buckets:
7        pigs += 1
8    return pigs
9

ℹ

Complexity note: The complexity is linear because we are incrementing the number of pigs until we can cover all buckets, which is a straightforward calculation.

1The number of pigs needed grows logarithmically with the number of buckets when using the optimal approach.
2Understanding the relationship between tests and outcomes is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.