Print in Order.md

Solution 1 - Semaphore

Use a Semaphore. Here is a good explanation of Semaphores (read their explanation, skip their code).

"Semaphore is a bowl of marbles. If you need a marble, and there are none, you wait. You wait until there is one marble and then you take it. If you release(), you will add one marble to the bowl (from thin air). If you release(100), you will add 100 marbles to the bowl (from thin air)." - from this post

Code

class Foo {
    private Semaphore run2 = new Semaphore(0);
    private Semaphore run3 = new Semaphore(0);

    public void first(Runnable printFirst) throws InterruptedException {
        printFirst.run();
        run2.release();
    }

    public void second(Runnable printSecond) throws InterruptedException {
        run2.acquire();
        printSecond.run();
        run3.release();
    }

    public void third(Runnable printThird) throws InterruptedException {
        run3.acquire();
        printThird.run();
    }
}

Time/Space Complexity

• Time Complexity: O(1)
• Space Complexity: O(1)

Solution 2 - CountDownLatch

Use a CountDownLatch

Code

class Foo {
    private CountDownLatch latch2 = new CountDownLatch(1);
    private CountDownLatch latch3 = new CountDownLatch(1);

    public void first(Runnable printFirst) throws InterruptedException {
        printFirst.run();
        latch2.countDown();
    }

    public void second(Runnable printSecond) throws InterruptedException {
        latch2.await();
        printSecond.run();
        latch3.countDown();
    }

    public void third(Runnable printThird) throws InterruptedException {
        latch3.await();
        printThird.run();
    }
}

Time/Space Complexity

• Time Complexity: O(1)
• Space Complexity: O(1)

Links

• github.com/RodneyShag