How do you solve The Dining Philosophers on LeetCode?

The Dining Philosophers (LeetCode #1226) 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: Using a specific order for picking forks can prevent deadlock.

What is the brute force solution for The Dining Philosophers?

The brute force approach for The Dining Philosophers is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, each philosopher will try to pick up the left fork first and then the right fork. If they can't pick both, they will put down the fork and try again. This can lead to deadlock if all philosophers pick up one fork simultaneously. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve The Dining Philosophers?

The Dining Philosophers uses the following concepts: Concurrency. The recommended patterns to study are: Mutex, Locking mechanisms.

What are the interview tips for The Dining Philosophers?

Explain your thought process clearly while discussing concurrency. Consider edge cases, such as what happens if all philosophers try to eat at once. Discuss potential improvements or alternative approaches.

What are common mistakes when solving The Dining Philosophers?

Not handling the case where a philosopher cannot pick both forks. Forgetting to release forks after eating, leading to potential deadlocks.

#1226

The Dining Philosophers

Medium

ConcurrencyMutexLocking mechanisms

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)

To avoid deadlock, we can enforce a rule where philosophers pick up the forks in a specific order. For example, if all philosophers pick up the right fork first, it ensures that at least one philosopher can always eat, preventing starvation.

⚙️

Algorithm

5 steps

1Step 1: Each philosopher will always pick up the right fork first.
2Step 2: If they can pick up the right fork, they then try to pick up the left fork.
3Step 3: If both forks are acquired, they eat.
4Step 4: After eating, they put down both forks.
5Step 5: Repeat the process.

solution.py20 lines

1import threading
2
3class DiningPhilosophers:
4    def __init__(self):
5        self.forks = [threading.Lock() for _ in range(5)]
6
7    def wantsToEat(self, philosopher, pickLeftFork, pickRightFork, eat, putLeftFork, putRightFork):
8        left = philosopher
9        right = (philosopher + 1) % 5
10        while True:
11            self.forks[right].acquire()
12            self.forks[left].acquire()
13            pickRightFork()
14            pickLeftFork()
15            eat()
16            putRightFork()
17            putLeftFork()
18            self.forks[left].release()
19            self.forks[right].release()
20            break

ℹ

Complexity note: The time complexity is O(n) because each philosopher can potentially eat without waiting indefinitely, ensuring that they can continue their cycle of thinking and eating without deadlock.

1Using a specific order for picking forks can prevent deadlock.
2Concurrency problems often require careful resource management.

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