Stack and Queue (reading code)

Question

I want to solve this problem WITHOUT putting it into an Java IDE. (If it came up in an exam I would not be able to solve it.)

So what I know so far.

I know that stack is referred to as LIFO and queue is referred to as FIFO. But I do not understand what this code would pop out (Or if I'm correct).

 public static void main (String[] args) {
    Queue<String> q = new Queue<String> ();
    q.enqueue ("one");
    q.enqueue ("two");
    q.enqueue ("four");
    q.enqueue ("six");
    String s = "";
    int i = 0;
    while (!q.isEmpty()) {
      s = s + q.dequeue().substring(i);
      i++; 
    }
    StdOut.print (s);
  }

Since it is a queue it would be FIFO and the substring would cause the output to be: onewour? Since s.substring() for six would be (3) so no value would be present.

And lastly I found two problems in a Princeton CS class but do not know how the answer came to be

Suppose that a client performs an intermixed sequence of (queue) enqueue and dequeue operations. The enqueue operations put the integers 0 through 9 in order onto the queue; the dequeue operations print out the return value. Which of the following sequence(s) could not occur?

(a)  0 1 2 3 4 5 6 7 8 9

(b)  4 6 8 7 5 3 2 9 0 1 

(c)  2 5 6 7 4 8 9 3 1 0

(d)  4 3 2 1 0 5 6 7 8 9

Answer: (b), (c), and (d).

Suppose that an intermixed sequence of (stack) push and pop operations are performed. The pushes push the integers 0 through 9 in order; the pops print out the return value. Which of the following sequence(s) could not occur?

(a)  4 3 2 1 0 9 8 7 6 5

(b)  4 6 8 7 5 3 2 9 0 1

(c)  2 5 6 7 4 8 9 3 1 0

(d)  4 3 2 1 0 5 6 7 8 9

(e)  1 2 3 4 5 6 9 8 7 0

(f)  0 4 6 5 3 8 1 7 2 9

(g)  1 4 7 9 8 6 5 3 0 2

(h)  2 1 4 3 6 5 8 7 9 0

Answer: (b), (f), and (g).

Answer 1

question 3)

Look at b)

4 6 8 7 5 3 2 9 0 1

When it popped and printed 9, it means all push operations were finished at that point
(due to the ordering of pushes 0,1,...,9). OK, now it is reading 9. Then it is reading 0,
this means it is reading the very first number which was ever pushed to the stack.
Then there's no way it could pop and print 1 because it already popped the last
possible number (there's no way 1 was pushed after 0 was popped because as said
all push operations were done at the time 9 was popped).

You need to use similar logical observations to see why f) and g) are not possible too.
When reading all these sequences of popped numbers just try to imagine the sequence
of pushes which happens in between the pops (i.e. what was pushed since the last pop).
Whichever sequence of pops leads to a hiccup: that one is not possible i.e. could not occur.

question 2) This one is trivial, anything but a) is not possible. Right?
Because as with a queue in a shop, the customers are served in the
order in which they came to the cash desk. Otherwise it would not be a queue.

question 1) You will learn more if you actually type this one, compile it and run it.

Answer 2

Ques 3) "Intermixed push and pop operations of integers 0 through 9 in order" I think this means the order is maintained starting anywhere from 0 and 9. Rotation is allowed. Ie: for example: pushing 8,9,0,1,2 in the same order, is valid. Otherwise the option 'e' is not possible. The option 'g' here is also possible. Here are the operations you have to perform to get the sequence given in option 'g'

Stack r = new Stack(); 
//Not providing the implementation details of stack here. 
r.push(1);
System.out.print(r.pop());
r.push(2);
r.push(3);
r.push(4);
System.out.print(r.pop());
r.push(5);
r.push(6);
r.push(7);
System.out.print(r.pop());
r.push(8);
r.push(9);
System.out.print(r.pop());
System.out.print(r.pop());
System.out.print(r.pop());
System.out.print(r.pop());
System.out.print(r.pop());
r.push(0);
System.out.print(r.pop());
System.out.print(r.pop());

Only 'b' and 'f' cannot occur. Correct me if I am wrong.