# [Programming Problem] Print FooBar Alternately

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.

Why use Semaphore and not ReentrantLock?

A semaphore initialized to one, and which is used such that it only has at most one permit available, can serve as a mutual exclusion lock. This is more commonly known as a binary semaphore, because it only has two states: one permit available, or zero permits available. When used in this way, the binary semaphore has the property (unlike many Lock implementations), that the “lock” can be released by a thread other than the owner (as semaphores have no notion of ownership). This can be useful in some specialized contexts, such as deadlock recovery.

```class FooBar { private int n; private final Semaphore mutexFoo = new Semaphore(1); private final Semaphore mutexBar = new Semaphore(1);   public FooBar(int n) { this.n = n; try { mutexBar.acquire(); } catch ( Exception e ) {} }   public void foo(Runnable printFoo) throws InterruptedException {   for (int i = 0; i < n; i++) {   // printFoo.run() outputs "foo". Do not change or remove this line. mutexFoo.acquire(); printFoo.run(); mutexBar.release(); } }   public void bar(Runnable printBar) throws InterruptedException {   for (int i = 0; i < n; i++) { // printBar.run() outputs "bar". Do not change or remove this line. mutexBar.acquire(); printBar.run(); mutexFoo.release(); } } }```

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