BuildingH2O.java

package by.andd3dfx.multithreading;

import java.util.concurrent.Semaphore;

/**
 * <pre>
 * <a href="https://leetcode.com/problems/building-h2o/description/">Task description</a>
 *
 * There are two kinds of threads: oxygen and hydrogen. Your goal is to group these threads to form water molecules.
 *
 * There is a barrier where each thread has to wait until a complete molecule can be formed. Hydrogen and oxygen
 * threads will be given releaseHydrogen and releaseOxygen methods respectively, which will allow them to pass the
 * barrier. These threads should pass the barrier in groups of three, and they must immediately bond with each other
 * to form a water molecule. You must guarantee that all the threads from one molecule bond before any other threads
 * from the next molecule do.
 *
 * In other words:
 *     If an oxygen thread arrives at the barrier when no hydrogen threads are present, it must wait for
 *     two hydrogen threads.
 *     If a hydrogen thread arrives at the barrier when no other threads are present, it must wait for
 *     an oxygen thread and another hydrogen thread.
 *
 * We do not have to worry about matching the threads up explicitly; the threads do not necessarily know which
 * other threads they are paired up with. The key is that threads pass the barriers in complete sets; thus, if we
 * examine the sequence of threads that bind and divide them into groups of three, each group should contain one
 * oxygen and two hydrogen threads.
 *
 * Write synchronization code for oxygen and hydrogen molecules that enforces these constraints.
 *
 * Example 1:
 * Input: water = "HOH"
 * Output: "HHO"
 * Explanation: "HOH" and "OHH" are also valid answers.
 *
 * Example 2:
 * Input: water = "OOHHHH"
 * Output: "HHOHHO"
 * Explanation: "HOHHHO", "OHHHHO", "HHOHOH", "HOHHOH", "OHHHOH", "HHOOHH", "HOHOHH" and "OHHOHH" are also valid
 * answers.
 *
 * Constraints:
 *     3 * n == water.length
 *     1 <= n <= 20
 *     water[i] is either 'H' or 'O'.
 *     There will be exactly 2 * n 'H' in water.
 *     There will be exactly n 'O' in water.
 *
 * Initial code:
 * class H2O {
 *     public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
 *         // releaseHydrogen.run() outputs "H". Do not change or remove this line.
 *         releaseHydrogen.run();
 *     }
 *
 *     public void oxygen(Runnable releaseOxygen) throws InterruptedException {
 *         // releaseOxygen.run() outputs "O". Do not change or remove this line.
 * 		   releaseOxygen.run();
 *     }
 * }
 * </pre>
 *
 * @see <a href="https://youtu.be/7S9e_vXuVFE">Video solution</a>
 */
public class BuildingH2O {

    private Semaphore hydrogenSemaphore = new Semaphore(2);
    private Semaphore oxygenSemaphore = new Semaphore(2);

    public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
        oxygenSemaphore.acquire();

        // releaseHydrogen.run() outputs "H". Do not change or remove this line.
        releaseHydrogen.run();
        hydrogenSemaphore.release();
    }

    public void oxygen(Runnable releaseOxygen) throws InterruptedException {
        hydrogenSemaphore.acquire(2);

        // releaseOxygen.run() outputs "O". Do not change or remove this line.
        releaseOxygen.run();
        oxygenSemaphore.release(2);
    }
}