# 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 $O(1)$.

• 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()