LeetCode #1195 — MEDIUM

Fizz Buzz Multithreaded

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

Solve on LeetCode
The Problem

Problem Statement

You have the four functions:

  • printFizz that prints the word "fizz" to the console,
  • printBuzz that prints the word "buzz" to the console,
  • printFizzBuzz that prints the word "fizzbuzz" to the console, and
  • printNumber that prints a given integer to the console.

You are given an instance of the class FizzBuzz that has four functions: fizz, buzz, fizzbuzz and number. The same instance of FizzBuzz will be passed to four different threads:

  • Thread A: calls fizz() that should output the word "fizz".
  • Thread B: calls buzz() that should output the word "buzz".
  • Thread C: calls fizzbuzz() that should output the word "fizzbuzz".
  • Thread D: calls number() that should only output the integers.

Modify the given class to output the series [1, 2, "fizz", 4, "buzz", ...] where the ith token (1-indexed) of the series is:

  • "fizzbuzz" if i is divisible by 3 and 5,
  • "fizz" if i is divisible by 3 and not 5,
  • "buzz" if i is divisible by 5 and not 3, or
  • i if i is not divisible by 3 or 5.

Implement the FizzBuzz class:

  • FizzBuzz(int n) Initializes the object with the number n that represents the length of the sequence that should be printed.
  • void fizz(printFizz) Calls printFizz to output "fizz".
  • void buzz(printBuzz) Calls printBuzz to output "buzz".
  • void fizzbuzz(printFizzBuzz) Calls printFizzBuzz to output "fizzbuzz".
  • void number(printNumber) Calls printnumber to output the numbers.

Example 1:

Input: n = 15
Output: [1,2,"fizz",4,"buzz","fizz",7,8,"fizz","buzz",11,"fizz",13,14,"fizzbuzz"]

Example 2:

Input: n = 5
Output: [1,2,"fizz",4,"buzz"]

Constraints:

  • 1 <= n <= 50

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You have the four functions: printFizz that prints the word "fizz" to the console, printBuzz that prints the word "buzz" to the console, printFizzBuzz that prints the word "fizzbuzz" to the console, and printNumber that prints a given integer to the console. You are given an instance of the class FizzBuzz that has four functions: fizz, buzz, fizzbuzz and number. The same instance of FizzBuzz will be passed to four different threads: Thread A: calls fizz() that should output the word "fizz". Thread B: calls buzz() that should output the word "buzz". Thread C: calls fizzbuzz() that should output the word "fizzbuzz". Thread D: calls number() that should only output the integers. Modify the given class to output the series [1, 2, "fizz", 4, "buzz", ...] where the ith token (1-indexed) of the series is: "fizzbuzz" if i is divisible by 3 and 5, "fizz" if i is divisible by 3 and not 5, "buzz"

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

15

Example 2

5

Related Problems

  • Fizz Buzz (fizz-buzz)
  • Print Zero Even Odd (print-zero-even-odd)
Step 02

Core Insight

What unlocks the optimal approach

  • No official hints in dataset. Start from constraints and look for a monotonic or reusable state.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1195: Fizz Buzz Multithreaded
class FizzBuzz {
    private int n;

    public FizzBuzz(int n) {
        this.n = n;
    }

    private Semaphore fSema = new Semaphore(0);
    private Semaphore bSema = new Semaphore(0);
    private Semaphore fbSema = new Semaphore(0);
    private Semaphore nSema = new Semaphore(1);

    // printFizz.run() outputs "fizz".
    public void fizz(Runnable printFizz) throws InterruptedException {
        for (int i = 3; i <= n; i = i + 3) {
            if (i % 5 != 0) {
                fSema.acquire();
                printFizz.run();
                nSema.release();
            }
        }
    }

    // printBuzz.run() outputs "buzz".
    public void buzz(Runnable printBuzz) throws InterruptedException {
        for (int i = 5; i <= n; i = i + 5) {
            if (i % 3 != 0) {
                bSema.acquire();
                printBuzz.run();
                nSema.release();
            }
        }
    }

    // printFizzBuzz.run() outputs "fizzbuzz".
    public void fizzbuzz(Runnable printFizzBuzz) throws InterruptedException {
        for (int i = 15; i <= n; i = i + 15) {
            fbSema.acquire();
            printFizzBuzz.run();
            nSema.release();
        }
    }

    // printNumber.accept(x) outputs "x", where x is an integer.
    public void number(IntConsumer printNumber) throws InterruptedException {
        for (int i = 1; i <= n; i++) {
            nSema.acquire();
            if (i % 3 == 0 && i % 5 == 0) {
                fbSema.release();
            } else if (i % 3 == 0) {
                fSema.release();
            } else if (i % 5 == 0) {
                bSema.release();
            } else {
                printNumber.accept(i);
                nSema.release();
            }
        }
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.