Reusable barriers using semaphore

Question

I was reviewing the discussion on reusable barriers in "The Little Book of Semaphores". I wrote down this solution (below) before checking the answer in the book. Since the solution in the book is a bit more complex (two turnstiles) I am guessing I missed something obvious here. Can someone help me point out any problems with it?

# count = 0;
# mutex = semaphore(1);
# barrier = semaphore(0);

# rendezvous point

mutex.wait();
    count = count + 1;
    if (count == n) barrier.signal();
mutex.signal();

barrier.wait();

mutex.wait();
    count = count - 1;
    if (count != 0) barrier.signal();
mutex.signal();

# critical point

For reference, here's the solution proposed in the book using two turnstiles (barrier1 and barrier2)

# count = 0; 
# mutex = semaphore(1); 
# barrier1 = semaphore(0); 
# barrier2 = semaphore(1); 

# rendezvous point  

mutex.wait(); 
    count = count + 1; 
    if (count == n) 
        barrier2.wait(); 
        barrier1.signal(); 
mutex.signal(); 

barrier1.wait(); 
barrier1.signal(); 

# critical point

mutex.wait(); 
    count = count – 1; 
    if (count == 0) 
        barrier1.wait(); 
        barrier2.signal(); 
mutex.signal(); 

barrier2.wait(); 
barrier2.signal();