LeetCode 1117: Building H2O Solution

Master LeetCode problem 1117 (Building H2O), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1117. Building H2O

Medium

Concurrency

Problem Explanation

Explanation

To solve this problem, we can use two semaphores - one for hydrogen and one for oxygen. We also need a counter to keep track of the number of hydrogen molecules formed so far. Each hydrogen thread will first acquire the semaphore for hydrogen. If the required number of hydrogen molecules are already formed, it will release the semaphore for oxygen to allow an oxygen thread to pass. If not, it will wait. Similarly, an oxygen thread will acquire the semaphore for oxygen and wait until the required number of hydrogen molecules are formed. Once the required number of hydrogen molecules are formed, the oxygen thread will release the semaphore for hydrogen to allow two hydrogen threads to pass. This way, we ensure that the threads pass the barrier in groups of three to form water molecules.

Time Complexity: O(n), where n is the number of water molecules. Space Complexity: O(1)

Solution Code

import java.util.concurrent.Semaphore;

class H2O {
    Semaphore hSemaphore, oSemaphore;
    int hCount;

    public H2O() {
        hSemaphore = new Semaphore(2);
        oSemaphore = new Semaphore(1);
        hCount = 0;
    }

    public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
        hSemaphore.acquire();
        releaseHydrogen.run();
        synchronized(this) {
            hCount++;
            if(hCount == 2) {
                hCount = 0;
                oSemaphore.release();
            }
        }
    }

    public void oxygen(Runnable releaseOxygen) throws InterruptedException {
        oSemaphore.acquire();
        releaseOxygen.run();
        hSemaphore.release(2);
    }
}

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