Question

How can a stack be implemented using two queues? Explain a solution that maintains a constant time pop operation but allows a linear time push operation (with respect to the number of elements in the stack).

a) Enqueue all elements into Queue 1.
b) For push operation, enqueue the element into Queue 2 and then transfer all elements from Queue 1 to Queue 2.
c) For pop operation, dequeue an element from Queue 1.
d) Repeat steps b and c as needed.

Answer 1

To implement a stack using two queues in such a way that pop is O(1) and push is O(n), one queue is used to reverse the order of elements during the push operation, ensuring the newest element is always at the front for popping.

Step-by-step explanation:

To implement a stack using two queues while maintaining a constant time 'pop' operation but allowing a linear time 'push' operation, the procedure outlined can be slightly modified. Let's consider two queues: Queue 1 and Queue 2.

For the push operation, the new element is always enqueued into Queue 2. Then, each element in Queue 1 is dequeued and enqueued into Queue 2. This way, the newest element stays at the front of Queue 2, thus maintaining the LIFO (Last In, First Out) property of the stack for the next pop operation. After transferring all elements, we swap the names of Queue 1 and Queue 2, so that the next pop operation dequeues from the queue with the latest added element.

For the pop operation, you simply perform a dequeue operation on Queue 1 since the latest added element is at the beginning of Queue 1 due to the push operation process.

This implementation ensures that the pop operation always runs in O(1) time, while the push operation runs in O(n) time, where n is the number of elements in the stack.