loading...
Cover image for 12 Concurrency Interview Questions To Know Before System Design Interview
FullStack.Cafe

12 Concurrency Interview Questions To Know Before System Design Interview

aershov24 profile image Alex πŸ‘¨πŸΌβ€πŸ’»FullStack.Cafe Originally published at fullstack.cafe ・9 min read

Concurrency and multithreading are some of the most advanced topics brought up in interviews. Concurrency happens when two or more tasks can start, run, and complete in overlapping time periods. It doesn't necessarily mean they'll ever both be running at the same instant. Follow along to clarify 12 Most Common Concurrency Interview Questions To Know Before Your Next System Design Interview.

Originally published on FullStack.Cafe - Kill Your Next Tech Interview

Q1: Explain the difference between Asynchronous and Parallel programming? β˜†β˜†

Topics: Concurrency

Answer:

When you run something asynchronously it means it is non-blocking, you execute it without waiting for it to complete and carry on with other things.
Parallelism means to run multiple things at the same time, in parallel. Parallelism works well when you can separate tasks into independent pieces of work. Async and Callbacks are generally a way (tool or mechanism) to express concurrency i.e. a set of entities possibly talking to each other and sharing resources.

Take for example rendering frames of a 3D animation. To render the animation takes a long time so if you were to launch that render from within your animation editing software you would make sure it was running asynchronously so it didn't lock up your UI and you could continue doing other things. Now, each frame of that animation can also be considered as an individual task. If we have multiple CPUs/Cores or multiple machines available, we can render multiple frames in parallel to speed up the overall workload.


πŸ”— Source: stackoverflow.com

Q2: What is a Deadlock? β˜†β˜†

Topics: Concurrency

Answer:

  • A lock occurs when multiple processes try to access the same resource at the same time. One process loses out and must wait for the other to finish.

  • A deadlock occurs when the waiting process is still holding on to another resource that the first needs before it can finish.

So, an example:

Resource A and resource B are used by process X and process Y

  • X starts to use A.
  • X and Y try to start using B
  • Y 'wins' and gets B first
  • now Y needs to use A
  • A is locked by X, which is waiting for Y
Thread 1               Thread 2

Lock1->Lock();         Lock2->Lock();
WaitForLock2();        WaitForLock1();   <-- Oops!

The best way to avoid deadlocks is to avoid having processes cross over in this way. Reduce the need to lock anything as much as you can. In databases avoid making lots of changes to different tables in a single transaction, avoid triggers and switch to optimistic/dirty/nolock reads as much as possible.


πŸ”— Source: stackoverflow.com

Q3: Explain Deadlock to 5 years old β˜†β˜†β˜†

Topics: Concurrency

Answer:

Jack and Jill share a sparse kitchen that has only one of everything. They both want to make a sandwich at the same time. Each needs a slice of bread and each needs a knife, so they both go to get the loaf of bread and the knife from the kitchen.

Jack gets the knife first, while Jill gets the loaf of bread first. Now Jack tries to find the loaf of bread and Jill tries to find the knife, but both find that what they need to finish the task is already in use. If they both decide to wait until what they need is no longer in use, they will wait for each other forever. Deadlock.


πŸ”— Source: stackoverflow.com

Q4: Is there any difference between a Binary Semaphore and Mutex? β˜†β˜†β˜†

Topics: Concurrency

Answer:

  • A mutex (or Mutual Exclusion Semaphores) is a locking mechanism used to synchronize access to a resource. Only one task (can be a thread or process based on OS abstraction) can acquire the mutex. It means there will be ownership associated with mutex, and only the owner can release the lock (mutex).

  • Semaphore (or Binary Semaphore) is signaling mechanism (β€œI am done, you can carry on” kind of signal). For example, if you are listening songs (assume it as one task) on your mobile and at the same time your friend called you, an interrupt will be triggered upon which an interrupt service routine (ISR) will signal the call processing task to wakeup. A binary semaphore is NOT protecting a resource from access. Semaphores are more suitable for some synchronization problems like producer-consumer.

Short version:

  • A mutex can be released only by the thread that had acquired it.
  • A binary semaphore can be signaled by any thread (or process).

πŸ”— Source: stackoverflow.com

Q5: What is a Race Condition? β˜†β˜†β˜†

Topics: Concurrency

Answer:

A race condition is a situation on concurrent programming where two concurrent threads or processes compete for a resource and the resulting final state depends on who gets the resource first.

Because the thread scheduling algorithm can swap between threads at any time, you don't know the order in which the threads will attempt to access the shared data. Therefore, the result of the change in data is dependent on the thread scheduling algorithm, i.e. both threads are "racing" to access/change the data.

Problems often occur when one thread does a "check-then-act" (e.g. "check" if the value is X, then "act" to do something that depends on the value being X) and another thread does something to the value in between the "check" and the "act". E.g:

if (x == 5) // The "Check"
{
   y = x * 2; // The "Act"

   // If another thread changed x in between "if (x == 5)" and "y = x * 2" above,
   // y will not be equal to 10.
}

The point being, y could be 10, or it could be anything, depending on whether another thread changed x in between the check and act. You have no real way of knowing.

In order to prevent race conditions from occurring, you would typically put a lock (Mutex or Semaphores) around the shared data to ensure only one thread can access the data at a time. This would mean something like this:

// Obtain lock for x
if (x == 5)
{
   y = x * 2; // Now, nothing can change x until the lock is released. 
              // Therefore y = 10
}
// release lock for x

πŸ”— Source: stackoverflow.com

Q6: What is the meaning of the term β€œThread-Safe”? β˜†β˜†β˜†

Topics: Concurrency

Problem:

Does it mean that two threads can't change the underlying data simultaneously? Or does it mean that the given code segment will run with predictable results when multiple threads are executing that code segment?

Solution:

  • Thread-safe code is code that will work even if many Threads are executing it simultaneously within the same process. This often means, that internal data-structures or operations that should run uninterrupted are protected against different modifications at the same time.
  • Another definition may be like - a class is** thread-safe** if it behaves correctly when accessed from multiple threads, regardless of the scheduling or interleaving of the execution of those threads by the runtime environment, and with no additional synchronisation or other coordination on the part of the calling code.

πŸ”— Source: stackoverflow.com

Q7: Write a function that guarantees to never return the same value twice β˜†β˜†β˜†

Topics: Concurrency Brain Teasers

Problem:

Write a function that is guaranteed to never return the same value twice. Assume that this function will be accessed by multiple machines concurrently.

Solution:

  1. Just toss a simple (threadsafe) counter behind some communication endpoint:
long x = long.MinValue;
public long ID(){
    return Interlocked.Increment(ref x);
}
  1. Let interviewer be the one that follow up with those problems:
  • Does it need to survive reboots?
  • What about hard drive failure?
  • What about nuclear war?
  • Does it need to be random?
  • How random?
    1. If they made it clear that it has to be unique across reboots and across different machines, I'd give them a function that calls into the standard mechanism for creating a new GUID, whatever that happens to be in the language being used. This is basically the problem that guids solve. Producing a duplicate Guid, no matter its format, is the most difficult lottery on the planet.

πŸ”— Source: softwareengineering.stackexchange.com

Q8: Two customers add a product to the basket in the same time whose the stock was only one (1). What will you do? β˜†β˜†β˜†β˜†

Topics: Concurrency Software Architecture

Problem:

What is the best practice to manage the case where two customers add in the same time a product whose the stock was only 1?

Solution:

There is no perfect answer for this question and all depends on details but you have some options:

  1. As a first 'defense line' I would try to avoid such situations at all, by simply not selling out articles that low if any possible.
  2. You reserve an item for the customer for a fix time say (20 minutes) after they have added it to the basket – after they time they have to recheck the stock level or start again. This is often used to ticket to events or airline seats
  3. For small jobs the best way is to do a final check right before the payment, when the order is actually placed. In worst case you have to tell you customer that you where running out of stock right now and offer alternatives or discount coupon.
  4. Try to fulfil both orders later - just cause you don't have stock right now, doesn't mean you can't find it in an emergency. If you can't then someone has to contact the user who lucked out and apologise.

Side note:

  1. The solution to do the double check when adding something to the basket isn't very good. People put a lot in baskets without ever actually placing an order. So this may block this article for a certain period of time.

πŸ”— Source: softwareengineering.stackexchange.com

Q9: What is Starvation? β˜†β˜†β˜†β˜†

Topics: Concurrency

Answer:

Starvation describes a situation where a thread is unable to gain regular access to shared resources and is unable to make progress. This happens when shared resources are made unavailable for long periods by "greedy" threads or threads with more "prioroty". For example, suppose an object provides a synchronized method that often takes a long time to return. If one thread invokes this method frequently, other threads that also need frequent synchronized access to the same object will often be blocked.

One more real live example may be this one. Imagine you're in a queue to purchase food at a restaurant, for which pregnant women have priority. And there's just a whole bunch of pregnant women arriving all the time. You'll soon be starving.


πŸ”— Source: docs.oracle.com

Q10: What's the difference between Deadlock and Livelock? β˜†β˜†β˜†β˜†

Topics: Concurrency

Answer:

  • In concurrent computing, a deadlock is a state in which each member of a group of actions, is waiting for some other member to release a lock

  • A livelock is similar to a deadlock, except that the states of the processes involved in the livelock constantly change with regard to one another, none progressing. Livelock is a special case of resource starvation; the general definition only states that a specific process is not progressing.

Livelock is a risk with some algorithms that detect and recover from deadlock. If more than one process takes action, the deadlock detection algorithm can be repeatedly triggered. This can be avoided by ensuring that only one process (chosen randomly or by priority) takes action.


πŸ”— Source: en.wikipedia.org

Q11: What happens if you have a "race condition" on the lock itself? β˜†β˜†β˜†β˜†β˜†

Topics: Concurrency

Problem:

For example, two different threads, perhaps in the same application, but running on different processors, try to acquire a lock at the exact same time. What happens then? Is it impossible, or just plain unlikely?

Solution:

Have a "race condition" on the lock itself is impossible. It can be implemented in different ways, e.g., via the Compare-and-swap where the hardware guarantees sequential execution. It can get a bit complicated in presence of multiple cores or even multiple sockets and needs a complicated protocol (MESI protocol) between the cores, but this is all taken care of in hardware.

Compare-and-swap (CAS) is an atomic instruction used in multithreading to achieve synchronization. It compares the contents of a memory location with a given value and, only if they are the same, modifies the contents of that memory location to a new given value. This is done as a single atomic operation. The atomicity guarantees that the new value is calculated based on up-to-date information; if the value had been updated by another thread in the meantime, the write would fail.


πŸ”— Source: softwareengineering.stackexchange.com

Q12: What is the difference between Race Condition and Data Races? Are they the same? β˜†β˜†β˜†β˜†β˜†

Topics: Concurrency

Answer:

  • A data race occurs when 2 instructions from different threads access the same memory location, at least one of these accesses is a write and there is no synchronization that is mandating any particular order among these accesses.

  • A race condition is a semantic error. It is a flaw that occurs in the timing or the ordering of events that leads to erroneous program behavior. Many race conditions can be caused by data races, but this is not necessary.

The Race Condition and Data Races are not the same thing. They are not a subset of one another. They are also neither the necessary, nor the sufficient condition for one another.


πŸ”— Source: stackoverflow.com

Thanks πŸ™Œ for reading and good luck on your interview! Please share this article with your fellow Devs if you like it! Check more FullStack Interview Questions & Answers on πŸ‘‰ www.fullstack.cafe

Discussion

pic
Editor guide