Leetcode #1226: The Dining Philosophers
In this guide, we solve Leetcode #1226 The Dining Philosophers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Five silent philosophers sit at a round table with bowls of spaghetti. Forks are placed between each pair of adjacent philosophers.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Concurrency
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: n = 1
Output: [[3,2,1],[3,1,1],[3,0,3],[3,1,2],[3,2,2],[4,2,1],[4,1,1],[2,2,1],[2,1,1],[1,2,1],[2,0,3],[2,1,2],[2,2,2],[4,0,3],[4,1,2],[4,2,2],[1,1,1],[1,0,3],[1,1,2],[1,2,2],[0,1,1],[0,2,1],[0,0,3],[0,1,2],[0,2,2]]
Explanation:
n is the number of times each philosopher will call the function.
The output array describes the calls you made to the functions controlling the forks and the eat function, its format is:
output[i] = [a, b, c] (three integers)
- a is the id of a philosopher.
- b specifies the fork: {1 : left, 2 : right}.
- c specifies the operation: {1 : pick, 2 : put, 3 : eat}.
Python Solution
# TODO: add Python solution
Complexity
The time complexity is O(n). The space complexity is O(1) to O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.