Multithreaded FizzBuzz

The single-threaded FizzBuzz is a warm-up question. The multithreaded version is a real concurrency exercise: four separate threads must cooperate to produce the FizzBuzz sequence for 1..n in order, even though no single thread sees every number.

Implement FizzBuzz(n) with four methods, each run on its own thread. Each method is handed a callback and must invoke it at exactly the right moments, in ascending order:

• fizz(printFizz) — call printFizz() for every i divisible by 3 but not 5
• buzz(printBuzz) — call printBuzz() for every i divisible by 5 but not 3
• fizzbuzz(printFizzBuzz) — call printFizzBuzz() for every i divisible by both 3 and 5
• number(printNumber) — call printNumber(i) for every i divisible by neither

The four threads run concurrently, so you must synchronize them: the combined sequence of callback invocations must be exactly the FizzBuzz output for 1..n, with no gaps, duplicates, or out-of-order emissions.

Example (n = 15): the callbacks together must produce 1, 2, fizz, 4, buzz, fizz, 7, 8, fizz, buzz, 11, fizz, 13, 14, fizzbuzz

Constraints:

• 1 <= n <= 1000
• Use threading primitives only (Condition, Lock, Event, Semaphore)
• Each method returns once it has emitted all of its numbers
• No busy-waiting that spins the CPU (use a condition variable or equivalent to sleep until it is your turn)