To add on to the previous answer, both can be done in O(1) if a special element (end-of-queue) is allowed to be introduced. Push - push to the queue that has the end-of-queue element as the first element. Pop - Dequeue both queue, one of them is end-of-queue element, so serve the result of the other queue.

You only need 1 queue to implement a stack. queues are more powerful. If there are N elements in the queue and you need to pop the last element in order to simulate the stack LIFO behavior, just pop the first N-1 elements and immediately push them back into the queue. Then pop the Nth element and return it. Done. You are basically cycling through the queue to get the element you want without disrupting the rest of the elements. Note O(N) time for pop(). This is what valentino said above but you don't 2 queues, just 1, to accomplish it.

totally Agree with bob, you only need one queue, I guess using the 2 queues answer is based on the question "2 queues" but once again this can be solve with just one queue.

The question asks to implement a stack with 2 queues. Your solution has 2 stacks and is implementing a queue.

One the blog below, there is a detailed solution about how to implement a queue with two stacks: http://codercareer.blogspot.com/2011/10/no-17-queue-implemented-with-two-stacks.html It is quite similar to simulate a stack with two queues.

it is pretty similar, on Queue check which queue is not empty and enqueue in that one, if both are empty pick one. On Deque, start dequeue from the queue that is not empty and queue the element to the other queue until you reach the last element.