[leetcode][Python][Concurrency][Medium] 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.

Idea

An example for output

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

Using a flag to switch printFoo() and printBar() when acquire the lock.

Solution

from threading import Condition

class FooBar:
    def __init__(self, n):
        self.n = n
        self._lock = Condition()
        self._is_printed_foo = False

    def foo(self, printFoo: 'Callable[[], None]') -> None:
        
        for i in range(self.n):
            with self._lock:
                while self._is_printed_foo:
                    self._lock.wait()
                # printFoo() outputs "foo". Do not change or remove this line.
                printFoo()
                self._is_printed_foo = True
                self._lock.notify_all()


    def bar(self, printBar: 'Callable[[], None]') -> None:
        
        for i in range(self.n):
            with self._lock:
                while not self._is_printed_foo:
                    self._lock.wait()
                # printBar() outputs "bar". Do not change or remove this line.
                printBar()
                self._is_printed_foo = False
                self._lock.notify_all()