What is the time complexity of Building H2O?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Building H2O?

This problem uses Concurrency. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Building H2O?

Insight 1: Use semaphores to control the number of available hydrogen and oxygen atoms. Initialize hydrogen semaphore with 2 and oxygen with 1 to ensure the 2:1 ratio. Insight 2: Employ a barrier to ensure that all three atoms (2 hydrogen, 1 oxygen) are ready before any of them proceed and release their respective elements, hence forming a molecule. This guarantees the completion of the molecule.

Which companies ask LeetCode 1117. Building H2O in interviews?

This problem is frequently asked at Rubrik, Tesla.

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Building H2O - LeetCode 1117 Solution

Building H2O - Complete Solution Guide

Building H2O is LeetCode problem 1117, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem requires us to synchronize hydrogen and oxygen threads to form water molecules (H2O). We need to ensure that threads are released in the correct ratio (2 hydrogens and 1 oxygen) and that a full molecule is formed before the next one starts. The threads call `releaseHydrogen()` and `releaseOxygen()` respectively to indicate their readiness. The critical aspect is implementing the synchronization logic so threads wait for the necessary partners before releasing the hydrogen/oxygen.

Solution Approach

The provided solution uses semaphores to manage the availability of hydrogen and oxygen atoms, and a barrier to synchronize the threads before releasing the elements. The `hydrogen` method acquires a hydrogen semaphore, waits at the barrier, and then calls `releaseHydrogen`. The `oxygen` method acquires an oxygen semaphore, waits at the barrier, calls `releaseOxygen`, and finally releases two hydrogen semaphores and one oxygen semaphore, allowing the next water molecule to be formed.

Step-by-Step Algorithm

Step 1: Initialize two semaphores: `h_sem` for hydrogen (initial count 2) and `o_sem` for oxygen (initial count 1). This limits the availability of hydrogen and oxygen to the correct ratio.
Step 2: Initialize a barrier with a count of 3. This ensures that two hydrogen threads and one oxygen thread must all reach the barrier before any of them can proceed.
Step 3: In the `hydrogen` method, acquire `h_sem`. This decrements the hydrogen count. If the count is 0, the thread waits until a hydrogen semaphore is released.
Step 4: The `hydrogen` thread waits at the barrier using `barrier.wait()`. It blocks until two hydrogen and one oxygen thread has reached the same point.
Step 5: The `hydrogen` thread then calls `releaseHydrogen()` after the barrier is opened.
Step 6: In the `oxygen` method, acquire `o_sem`. This decrements the oxygen count. If the count is 0, the thread waits until an oxygen semaphore is released.
Step 7: The `oxygen` thread waits at the barrier using `barrier.wait()`. It blocks until two hydrogen and one oxygen thread has reached the same point.
Step 8: The `oxygen` thread calls `releaseOxygen()` after the barrier is opened.
Step 9: The `oxygen` thread then releases two hydrogen semaphores using `h_sem.release(2)` and one oxygen semaphore using `o_sem.release()`. This makes two hydrogen and one oxygen available for the next water molecule.

Key Insights

Insight 1: Use semaphores to control the number of available hydrogen and oxygen atoms. Initialize hydrogen semaphore with 2 and oxygen with 1 to ensure the 2:1 ratio.
Insight 2: Employ a barrier to ensure that all three atoms (2 hydrogen, 1 oxygen) are ready before any of them proceed and release their respective elements, hence forming a molecule. This guarantees the completion of the molecule.
Insight 3: The barrier is crucial for synchronizing the threads *after* the semaphores have been acquired, preventing a situation where the semaphores have been obtained but the threads are not yet ready to form a molecule.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Concurrency.

Companies

Asked at: Rubrik, Tesla.

Building H2O | LeetCode 1117

Medium

Concurrency

Time: O(1)Space: O(1)

Solution Code

Available in:

import java.util.concurrent.Semaphore;
import java.util.concurrent.CyclicBarrier;

class H2O {

    private Semaphore h_sem;
    private Semaphore o_sem;
    private CyclicBarrier barrier;

    public H2O() {
        h_sem = new Semaphore(2);
        o_sem = new Semaphore(1);
        barrier = new CyclicBarrier(3);
    }

    public void hydrogen(Runnable releaseHydrogen) throws InterruptedException {
        h_sem.acquire();
        try {
            barrier.await();
        } catch (Exception e) {
            e.printStackTrace();
        }
        // releaseHydrogen.run() outputs "H". Do not change or remove this line.
        releaseHydrogen.run();
    }

    public void oxygen(Runnable releaseOxygen) throws InterruptedException {
        o_sem.acquire();
        try {
            barrier.await();
        } catch (Exception e) {
            e.printStackTrace();
        }
        // releaseOxygen.run() outputs "O". Do not change or remove this line.
		releaseOxygen.run();
        o_sem.release();
        h_sem.release(2);
    }
}

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Problem Info

Description

Problem 1117: Building H2O - Detailed explanation and solution approaches.

Topics

Concurrency

Companies

RubrikTesla

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1117

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: Initialize two semaphores: `h_sem` for hydrogen (initial count 2) and `o_sem` for oxygen (initial count 1). This limits the availability of hydrogen and oxygen to the correct ratio.
2
Step 2: Initialize a barrier with a count of 3. This ensures that two hydrogen threads and one oxygen thread must all reach the barrier before any of them can proceed.
3
Step 3: In the `hydrogen` method, acquire `h_sem`. This decrements the hydrogen count. If the count is 0, the thread waits until a hydrogen semaphore is released.
4
Step 4: The `hydrogen` thread waits at the barrier using `barrier.wait()`. It blocks until two hydrogen and one oxygen thread has reached the same point.
5
Step 5: The `hydrogen` thread then calls `releaseHydrogen()` after the barrier is opened.
6
Step 6: In the `oxygen` method, acquire `o_sem`. This decrements the oxygen count. If the count is 0, the thread waits until an oxygen semaphore is released.
7
Step 7: The `oxygen` thread waits at the barrier using `barrier.wait()`. It blocks until two hydrogen and one oxygen thread has reached the same point.
8
Step 8: The `oxygen` thread calls `releaseOxygen()` after the barrier is opened.
9
Step 9: The `oxygen` thread then releases two hydrogen semaphores using `h_sem.release(2)` and one oxygen semaphore using `o_sem.release()`. This makes two hydrogen and one oxygen available for the next water molecule.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The operations within the hydrogen and oxygen methods, such as semaphore acquisition/release and barrier wait, are all constant-time operations. The space used is fixed regardless of the input size (number of water molecules), as it only consists of a few semaphores and a barrier.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using Locks and Condition Variables

Instead of semaphores and barriers, one could use locks and condition variables. A lock would protect the shared state (counts of available H and O), and condition variables would signal when enough H and O are available to form a molecule.

Time: O(1)Space: O(1)

Trade-offs: This approach can be more verbose than using semaphores and barriers, but it offers more fine-grained control over the synchronization process. It might be preferable when complex synchronization logic is needed.

✨

Key Insights

1Insight 1: Use semaphores to control the number of available hydrogen and oxygen atoms. Initialize hydrogen semaphore with 2 and oxygen with 1 to ensure the 2:1 ratio.
2Insight 2: Employ a barrier to ensure that all three atoms (2 hydrogen, 1 oxygen) are ready before any of them proceed and release their respective elements, hence forming a molecule. This guarantees the completion of the molecule.
3Insight 3: The barrier is crucial for synchronizing the threads *after* the semaphores have been acquired, preventing a situation where the semaphores have been obtained but the threads are not yet ready to form a molecule.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: The initial order of threads arriving (e.g., all oxygens first). The semaphores ensure that even if all oxygen threads arrive before any hydrogen threads, the oxygen threads will wait until hydrogen threads are available.
Edge case 2: A large number of hydrogen threads arriving before any oxygen threads. The barrier ensures that these threads will wait, and releaseHydrogen is not called out of order. The hydrogen semaphores prevent more than two hydrogens from proceeding before an oxygen.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to release the semaphores after releasing the elements. This would lead to deadlock as threads would be waiting for semaphores that are never released.

✗Common mistake 2: Not using a barrier to synchronize the threads after semaphore acquisition. Without a barrier, it is possible that a hydrogen thread acquires its semaphore and calls releaseHydrogen before another hydrogen and an oxygen have acquired their semaphores.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Resource management in operating systems. Semaphores are used to control access to limited resources, ensuring that only a certain number of processes can access a resource at any given time.

🏦Real-world scenario 2: Database connection pooling. Semaphores can limit the number of concurrent connections to a database, preventing overload and ensuring that connections are released back to the pool when they are no longer needed.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 8, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Building H2O"?

For this medium-level Concurrency problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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