# 3.4. Queue via stacks

> Implement a MyQueue class which implements a queue using two stacks

## First idea

A stack is LIFO, first element to get out is the one inserted last. And queue is FIFO.

## Hints

* How to remove the oldest item from a stack if you only have access to the newest item?
  * Delete the one opposite in the array, so the last one...
  * The trick must be inverting something because I need two stacks...
* We can remove the oldest elem in a stack by repeatedly removing the newest item (inserting those in the temporary stack, stack \#2) until we get down to one element, then put all the elems back. This requires O(n). Can we optimize for scenarios where we might do several pops in a row?
  * If we could check if an elem is present in stack \#2 yes, but there's no such function

## Solution

We need to modify `peek` and `pop` to work in reverse order. We can use stack \#2 to reverse the order of the elements by popping s1 and pushing the elements on to s2. In each `peek` and `pop` operation, we pop everything from s1 onto s2, perform `peek` and `pop` and then push everything back.

If two `peek`/`pop` must be performed, we need to move everything again. In a lazy approach, all elemenst remain in s2 until they must absolutely be reversed.

stack_newest has the newest elements on top and stack_oldest has the oldest elements on top. When we dequeue an element, we dequeue from stack_oldest. If empty, transfer all elemens from stack_newest into stack_oldest in reverse order.

[Code in Github](https://github.com/careercup/CtCI-6th-Edition/blob/master/Java/Ch%2003.%20Stacks%20and%20Queues/Q3_04_Queue_via_Stacks/MyQueue.java)