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GO Classes DPP | GATE DA | Python & DSA | Queue Reversal

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1 1 vote

Consider the following Python code fragment:

stack = []

while len(q) > 0:
    stack.append(q.pop(0))

while len(stack) > 0:
    q.append(stack.pop())

Here, $\texttt{q.pop(0)}$ removes the front element of the queue, and $\texttt{q.append(x)}$ inserts $\texttt{x}$ at the rear of the queue.

What does this code fragment do to the queue $\texttt{q}$?

  1. It leaves the queue unchanged.
  2. It reverses the items in the queue.
  3. It removes all items from the queue permanently.
  4. It sorts the queue in increasing order.

2 Answers

1 1 vote

The first loop removes items from the front of the queue and pushes them onto the stack.

Suppose the queue initially is:

$\texttt{q = ["A", "B", "C"]}$

After the first loop:

$\texttt{stack = ["A", "B", "C"]}$
$\texttt{q = []}$

Now the second loop pops from the stack. Since stack follows LIFO order, the popped order is:

$\texttt{C, B, A}$

These are appended back to the queue.

So the final queue becomes:

$\texttt{q = ["C", "B", "A"]}$

Therefore, the queue is reversed.

Correct Option: B

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