LeetCode 1115. Print FooBar Alternately Solution in Java, C++ & Python | Explanation + Code

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).

PythonJavaC++

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()(code-box)

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); } } }(code-box)

#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); } } };(code-box)