1114. Print in Order

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Problem Description

In this problem, we are given a class Foo with three methods: first(), second(), and third(). These methods are intended to be called by three different threads, let's call them Thread A, Thread B, and Thread C respectively. The challenge is to ensure that these methods are called in the strict sequence where first() is called before second() and second() is called before third(), regardless of how the threads are scheduled by the operating system.

This means that if Thread B tries to call second() before Thread A calls first(), Thread B should wait until first() has been called. Similarly, Thread C must wait for second() to complete before calling third().


The solution to this problem involves synchronization mechanisms that allow threads to communicate with each other about the state of the execution. One common synchronization mechanism provided by many programming languages, including Python, is the Semaphore.

A semaphore manages an internal counter which is decremented by each acquire() call and incremented by each release() call. The counter can never go below zero; when acquire() finds that it is zero, it blocks, waiting until some other thread calls release().

In the given solution:

  • We have three semaphores: self.a, self.b, and self.c.
  • self.a is initialized with a count of 1 to allow the first() operation to proceed immediately.
  • self.b and self.c are initialized with a count of 0 to block the second() and third() operations respectively until they are explicitly released.

The first() method:

  • Acquires self.a, ensuring no other operation is currently in progress.
  • Once the first() operation is complete, it releases self.b to allow second() to proceed.

The second() method:

  • Acquires self.b, which will only be available once self.a has been released by the first() method.
  • After completing its operation, it releases self.c to allow third() method execution.

The third() method:

  • Acquires self.c, which will only be available after self.b is released by the second() method.
  • After completing its operation, it releases self.a. This is optional in the context where only one cycle of operations (first(), second(), third()) is being performed but might be included for scenarios where the sequence may restart.

Using these semaphores, the solution effectively enforces a strict order of execution as required by the problem statement, even though the threads might be scheduled in any order by the operating system.

Solution Approach

The solution approach effectively utilizes semaphores, which are synchronization primitives that control access to a common resource by multiple threads in a concurrent system. Here's a step-by-step breakdown of how the solution is implemented:

  1. Class Initialization:

    • Inside the Foo class constructor, three semaphores are initialized. Semaphores self.b and self.c are initialized with a count of 0 to ensure that second() and third() methods are blocked initially. Semaphore self.a is initialized with a count of 1 to allow first() to proceed without blocking.
  2. Executing first():

    • The first method starts by calling self.a.acquire(). Since self.a was initialized to 1, first() is allowed to proceed as acquire() will decrement the counter to 0, and no blocking occurs.
    • The method performs its intended operation printFirst().
    • After completion, it calls self.b.release(). This increments the counter of semaphore self.b from 0 to 1, thereby allowing a blocked second() operation to proceed.
  3. Executing second():

    • The second method calls self.b.acquire(). If first() has already completed and called self.b.release(), the semaphore self.b counter would be 1, and second() can proceed (the acquire operation decrements it back to 0). If first() has not completed, second() will be blocked.
    • It then executes printSecond().
    • Upon successful completion, it calls self.c.release(), increasing the counter of the semaphore self.c from 0 to 1, and thus unblocking the third() method.
  4. Executing third():

    • The third method begins with self.c.acquire(). Up until second() calls self.c.release(), third() will be blocked. Once self.c counter is 1, third() can proceed (the acquire operation decrements it to 0).
    • It executes printThird().
    • Optionally, it can call self.a.release() which is omitted in the given code snippet because it's not necessary unless the sequence of operations is intended to repeat.

Each semaphore acts as a turnstile, controlling the flow of operations. The use of semaphores ensures that no matter the order in which threads arrive, they will be forced to execute in the necessary order: first() then second() then third().

This implementation exemplifies a classic synchronization pattern where the completion of one action triggers the availability of the next action in a sequence. The semaphores act like gates that open once the previous operation has signaled that it's safe for the next operation to start.

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Example Walkthrough

Imagine we have three simple functions that need to be executed in order: printFirst(), printSecond(), and printThird(). They simply print "first", "second", and "third" respectively to the console. Now, let's see how the solution approach described earlier ensures these functions are called in the strict sequence required.

Let's assume the three threads are started almost simultaneously by the operating system, but due to some randomness in thread scheduling, the actual order of invocation is Thread B (second), Thread C (third), and finally, Thread A (first).

  1. Thread B (second) arrives first:

    • Calls second() method and tries to acquire semaphore self.b with an acquire() call.
    • Since self.b was initialized with 0, it is blocked as the counter is already at 0, and acquire() cannot decrement it further. Thread B will now wait for self.b to be released by another operation (specifically, the first() operation).
  2. Thread C (third) arrives next:

    • It calls the third() method which tries to acquire semaphore self.c with acquire().
    • As with self.b, self.c was initialized with 0, so Thread C is blocked because the semaphore's counter is not greater than 0. It must wait for self.c to be released by the second() operation.
  3. Thread A (first) arrives last:

    • It proceeds to call the first() method which attempts to acquire semaphore self.a with an acquire() call.
    • Since self.a was initialized with 1, the acquire() will succeed, the counter will decrement to 0, and the first() method will proceed.
    • The printFirst() function is executed, outputting "first".
    • Upon completion, the first() method calls self.b.release(), which increments semaphore self.b's counter to 1.
  4. Thread B (second) resumes:

    • With self.b's counter now at 1, Thread B can proceed as acquire() successfully decrements it back to 0.
    • The printSecond() function is called, printing "second" to the console.
    • Upon finishing its operation, Thread B calls self.c.release(), incrementing self.c's counter to 1, allowing the third operation to proceed.
  5. Thread C (third) resumes:

    • Similar to the previous steps, with the self.c counter now at 1, the third() method proceeds as acquire() brings the counter down to 0.
    • printThird() is executed, and "third" is printed to the console.

In this example, even though the threads arrived out of order, the use of semaphores forced them to wait for their turn, ensuring the desired sequence of "first", "second", "third" in the console output.

Solution Implementation

1from threading import Semaphore
2from typing import Callable
4class Foo:
5    def __init__(self):
6        # Initialize semaphores to control the order of execution.
7        # Semaphore 'first_done' allows 'first' method to run immediately.
8        self.first_done = Semaphore(1)
9        # Semaphore 'second_done' starts locked, preventing 'second' method from running.
10        self.second_done = Semaphore(0)
11        # Semaphore 'third_done' starts locked, preventing 'third' method from running.
12        self.third_done = Semaphore(0)
14    def first(self, print_first: Callable[[], None]) -> None:
15        # Acquire semaphore to enter 'first' method.
16        self.first_done.acquire()
17        # Execute the print_first function to output "first".
18        print_first()
19        # Release semaphore to allow 'second' method to run.
20        self.second_done.release()
22    def second(self, print_second: Callable[[], None]) -> None:
23        # Wait for the completion of 'first' method.
24        self.second_done.acquire()
25        # Execute the print_second function to output "second".
26        print_second()
27        # Release semaphore to allow 'third' method to run.
28        self.third_done.release()
30    def third(self, print_third: Callable[[], None]) -> None:
31        # Wait for the completion of 'second' method.
32        self.third_done.acquire()
33        # Execute the print_third function to output "third".
34        print_third()
35        # This line could re-enable the flow for 'first', in case of repeated calls.
36        self.first_done.release()
1import java.util.concurrent.Semaphore;
3public class Foo {
4    private Semaphore firstJobDone = new Semaphore(1); // Semaphore for first job, initially available
5    private Semaphore secondJobDone = new Semaphore(0); // Semaphore for second job, initially unavailable
6    private Semaphore thirdJobDone = new Semaphore(0); // Semaphore for third job, initially unavailable
8    public Foo() {
9        // Constructor
10    }
12    // Method for the first job; prints "first"
13    public void first(Runnable printFirst) throws InterruptedException {
14        firstJobDone.acquire(); // Wait for the first job's semaphore to be available
15        printFirst.run(); // Run the printFirst task; this should print "first"
16        secondJobDone.release(); // Release the second job semaphore, allowing the second job to run
17    }
19    // Method for the second job; prints "second"
20    public void second(Runnable printSecond) throws InterruptedException {
21        secondJobDone.acquire(); // Wait for the second job's semaphore to be available
22        printSecond.run(); // Run the printSecond task; this should print "second"
23        thirdJobDone.release(); // Release the third job semaphore, allowing the third job to run
24    }
26    // Method for the third job; prints "third"
27    public void third(Runnable printThird) throws InterruptedException {
28        thirdJobDone.acquire(); // Wait for the third job's semaphore to be available
29        printThird.run(); // Run the printThird task; this should print "third"
30        firstJobDone.release(); // Release the first job semaphore, allowing the cycle of jobs to be restarted (if necessary)
31    }
1#include <mutex>
2#include <condition_variable>
3#include <functional>
5class Foo {
7    std::mutex mtx; // Mutex to protect condition variable
8    std::condition_variable cv; // Condition variable for synchronization
9    int count; // Counter to keep track of the order
12    Foo() {
13        count = 1; // Initialize count to 1 to ensure first is executed first
14    }
16    void first(std::function<void()> printFirst) {
17        std::unique_lock<std::mutex> lock(mtx); // Acquire the lock
18        // printFirst() outputs "first". Do not change or remove this line.
19        printFirst();
20        count = 2; // Update the count to allow second to run
21        cv.notify_all(); // Notify all waiting threads
22    }
24    void second(std::function<void()> printSecond) {
25        std::unique_lock<std::mutex> lock(mtx); // Acquire the lock
26        cv.wait(lock, [this] { return count == 2; }); // Wait until first is done
27        // printSecond() outputs "second". Do not change or remove this line.
28        printSecond();
29        count = 3; // Update the count to allow third to run
30        cv.notify_all(); // Notify all waiting threads
31    }
33    void third(std::function<void()> printThird) {
34        std::unique_lock<std::mutex> lock(mtx); // Acquire the lock
35        cv.wait(lock, [this] { return count == 3; }); // Wait until second is done
36        // printThird() outputs "third". Do not change or remove this line.
37        printThird();
38        // No need to update count or notify, as no further actions are dependent on third
39    }
1// Counter to keep track of the order
2let count = 1;
3// Promises to control the execution order
4let firstSecondControl: (value: void | PromiseLike<void>) => void;
5let secondThirdControl: (value: void | PromiseLike<void>) => void;
7// Promise that will resolve when it's okay to run `second`
8const canRunSecond = new Promise<void>((resolve) => {
9  firstSecondControl = resolve;
12// Promise that will resolve when it's okay to run `third`
13const canRunThird = new Promise<void>((resolve) => {
14  secondThirdControl = resolve;
17async function first(printFirst: () => void): Promise<void> {
18  // printFirst() outputs "first".
19  printFirst();
20  // Update the count to allow second to run
21  count = 2;
22  // Resolve the promise to unblock the second function
23  firstSecondControl();
26async function second(printSecond: () => void): Promise<void> {
27  // Wait until the first function has completed
28  if (count !== 2) {
29    await canRunSecond;
30  }
31  // printSecond() outputs "second".
32  printSecond();
33  // Update the count to allow third to run
34  count = 3;
35  // Resolve the promise to unblock the third function
36  secondThirdControl();
39async function third(printThird: () => void): Promise<void> {
40  // Wait until the second function has completed
41  if (count !== 3) {
42    await canRunThird;
43  }
44  // printThird() outputs "third".
45  printThird();
46  // After third, there are no more actions, so nothing more to do

Time and Space Complexity

The time complexity of the Foo class methods first, second, and third is O(1) for each call. This is because each method performs a constant amount of work: acquiring and releasing a semaphore. The use of semaphores is to control the order of execution but does not add any significant computational overhead.

The space complexity of the Foo class is O(1) as well. The class has three semaphores as its member variables, and the number of semaphores does not increase with the input size. Hence, the memory used by an instance of the class is constant.

Learn more about how to find time and space complexity quickly using problem constraints.

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