1115. Print FooBar Alternately

Problem

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

Solution (Java)

class FooBar {
    private int n;

    private final Semaphore fooSemaphore;
    private final Semaphore barSemaphore;

    public FooBar(int n) {
        this.n = n;
        fooSemaphore = new Semaphore(1);
        barSemaphore = new Semaphore(1);
        try {
            barSemaphore.acquire();
        } catch (InterruptedException ignored) {
            // ignored
        }
    }

    public void foo(Runnable printFoo) throws InterruptedException {
        for (int i = 0; i < n; i++) {
            fooSemaphore.acquire();
            // printFoo.run() outputs "foo". Do not change or remove this line.
            printFoo.run();
            barSemaphore.release();
        }
    }

    public void bar(Runnable printBar) throws InterruptedException {
        for (int i = 0; i < n; i++) {
            barSemaphore.acquire();
            // printBar.run() outputs "bar". Do not change or remove this line.
            printBar.run();
            fooSemaphore.release();
        }
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).