How do you solve Fizz Buzz Multithreaded on LeetCode?

Fizz Buzz Multithreaded (LeetCode #1195) 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: Understanding thread synchronization is crucial for multithreaded problems.

What is the brute force solution for Fizz Buzz Multithreaded?

The brute force approach for Fizz Buzz Multithreaded is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves creating a series of tasks for each number from 1 to n, checking divisibility for each number, and printing the appropriate output. This method is straightforward but inefficient due to the overhead of managing multiple threads. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fizz Buzz Multithreaded?

Fizz Buzz Multithreaded uses the following concepts: Concurrency. The recommended patterns to study are: Concurrency, Thread Synchronization.

What are the interview tips for Fizz Buzz Multithreaded?

Explain your thought process clearly while coding. Be ready to discuss potential edge cases and how your solution handles them. Practice writing multithreaded code to become comfortable with synchronization.

What are common mistakes when solving Fizz Buzz Multithreaded?

Not using locks properly, leading to race conditions. Failing to check the shared counter condition correctly, causing infinite loops.

#1195

Fizz Buzz Multithreaded

Medium

ConcurrencyConcurrencyThread Synchronization

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 uses a single shared counter and manages the printing of 'fizz', 'buzz', 'fizzbuzz', and numbers in a synchronized manner. This reduces the overhead of multiple threads and ensures that the output is in the correct order.

⚙️

Algorithm

3 steps

1Step 1: Initialize a shared counter to 1.
2Step 2: Create four threads for fizz, buzz, fizzbuzz, and number.
3Step 3: Each thread checks the counter and prints the appropriate output while incrementing the counter in a synchronized manner.

solution.py43 lines

1import threading
2
3class FizzBuzz:
4    def __init__(self, n):
5        self.n = n
6        self.current = 1
7        self.lock = threading.Lock()
8
9    def fizz(self, printFizz):
10        while True:
11            with self.lock:
12                if self.current > self.n:
13                    return
14                if self.current % 3 == 0 and self.current % 5 != 0:
15                    printFizz()
16                    self.current += 1
17
18    def buzz(self, printBuzz):
19        while True:
20            with self.lock:
21                if self.current > self.n:
22                    return
23                if self.current % 5 == 0 and self.current % 3 != 0:
24                    printBuzz()
25                    self.current += 1
26
27    def fizzbuzz(self, printFizzBuzz):
28        while True:
29            with self.lock:
30                if self.current > self.n:
31                    return
32                if self.current % 3 == 0 and self.current % 5 == 0:
33                    printFizzBuzz()
34                    self.current += 1
35
36    def number(self, printNumber):
37        while True:
38            with self.lock:
39                if self.current > self.n:
40                    return
41                if self.current % 3 != 0 and self.current % 5 != 0:
42                    printNumber(self.current)
43                    self.current += 1

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the numbers once, and each thread checks the condition in constant time.

1Understanding thread synchronization is crucial for multithreaded problems.
2Managing shared state between threads effectively avoids race conditions.

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