Welcome to Subscribe On Youtube

1115. Print FooBar Alternately

Description

Suppose you are given the following code:

class FooBar {
  public void foo() {
    for (int i = 0; i < n; i++) {
      print("foo");
    }
  }

  public void bar() {
    for (int i = 0; i < n; i++) {
      print("bar");
    }
  }
}

The same instance of FooBar will be passed to two different threads:

• thread A will call foo(), while
• thread B will call bar().

Modify the given program to output "foobar" n times.

 

Example 1:

Input: n = 1
Output: "foobar"
Explanation: There are two threads being fired asynchronously. One of them calls foo(), while the other calls bar().
"foobar" is being output 1 time.

Example 2:

Input: n = 2
Output: "foobarfoobar"
Explanation: "foobar" is being output 2 times.

 

Constraints:

• 1 <= n <= 1000

Solutions

Solution 1: Multithreading + Semaphore

We use two semaphores $f$ and $b$ to control the execution order of the two threads, where $f$ is initially set to $1$ and $b$ is set to $0$, indicating that thread $A$ executes first.

When thread $A$ executes, it first performs the $acquire$ operation on $f$, which changes the value of $f$ to $0$. Thread $A$ then gains the right to use $f$ and can execute the $foo$ function. After that, it performs the $release$ operation on $b$, changing the value of $b$ to $1$. This allows thread $B$ to gain the right to use $b$ and execute the $bar$ function.

When thread $B$ executes, it first performs the $acquire$ operation on $b$, which changes the value of $b$ to $0$. Thread $B$ then gains the right to use $b$ and can execute the $bar$ function. After that, it performs the $release$ operation on $f$, changing the value of $f$ to $1$. This allows thread $A$ to gain the right to use $f$ and execute the $foo$ function.

Therefore, we only need to loop $n$ times, each time executing the $foo$ and $bar$ functions, first performing the $acquire$ operation, and then the $release$ operation.

The time complexity is $O(n)$, and the space complexity is $O(1)$.

• Java
• C++
• Python

• class FooBar {
      private int n;
      private Semaphore f = new Semaphore(1);
      private Semaphore b = new Semaphore(0);
  
      public FooBar(int n) {
          this.n = n;
      }
  
      public void foo(Runnable printFoo) throws InterruptedException {
          for (int i = 0; i < n; i++) {
              f.acquire(1);
              // printFoo.run() outputs "foo". Do not change or remove this line.
              printFoo.run();
              b.release(1);
          }
      }
  
      public void bar(Runnable printBar) throws InterruptedException {
          for (int i = 0; i < n; i++) {
              b.acquire(1);
              // printBar.run() outputs "bar". Do not change or remove this line.
              printBar.run();
              f.release(1);
          }
      }
  }
  

• #include <semaphore.h>
  
  class FooBar {
  private:
      int n;
      sem_t f, b;
  
  public:
      FooBar(int n) {
          this->n = n;
          sem_init(&f, 0, 1);
          sem_init(&b, 0, 0);
      }
  
      void foo(function<void()> printFoo) {
          for (int i = 0; i < n; i++) {
              sem_wait(&f);
              // printFoo() outputs "foo". Do not change or remove this line.
              printFoo();
              sem_post(&b);
          }
      }
  
      void bar(function<void()> printBar) {
          for (int i = 0; i < n; i++) {
              sem_wait(&b);
              // printBar() outputs "bar". Do not change or remove this line.
              printBar();
              sem_post(&f);
          }
      }
  };
  

• from threading import Semaphore
  
  
  class FooBar:
      def __init__(self, n):
          self.n = n
          self.f = Semaphore(1)
          self.b = Semaphore(0)
  
      def foo(self, printFoo: "Callable[[], None]") -> None:
          for _ in range(self.n):
              self.f.acquire()
              # printFoo() outputs "foo". Do not change or remove this line.
              printFoo()
              self.b.release()
  
      def bar(self, printBar: "Callable[[], None]") -> None:
          for _ in range(self.n):
              self.b.acquire()
              # printBar() outputs "bar". Do not change or remove this line.
              printBar()
              self.f.release()
  
  #############
  
  class FooBar:
      def __init__(self, n):
          self.n = n
          self.fooLock = threading.Lock()
          self.barLock = threading.Lock()
          self.barLock.acquire()
  
      def foo(self, printFoo: 'Callable[[], None]') -> None:
          for i in range(self.n):
              self.fooLock.acquire()
              printFoo()
              self.barLock.release()
  
      def bar(self, printBar: 'Callable[[], None]') -> None:
          for i in range(self.n):
              self.barLock.acquire()
              printBar()
              self.fooLock.release()
  
  

All Problems

All Solutions