Implementing FIFO using LIFO

Question

Looking at some algorithm exercices on the net, I found an interesting one :

How would you implement a FIFO using a LIFO ?

I tried myself but I ended up wit

Answer 1

You can get amortized time complexity of O(1) per OP FIFO [queue] using 2 LIFOs [stacks].

Assume you have stack1, stack2:

insert(e):
   stack1.push(e)

take():
   if (stack2.empty()):
      while (stack1.empty() == false):
            stack2.push(stack1.pop())
   return stack2.pop() //assume stack2.pop() handles empty stack already

example:

push(1)

|1|  | |
|-|  |-|

push(2)
|2|  | |
|1|  | |
|-|  |-|

pop()
push 2 to stack2 and pop it from stack1:
|1|  |2|
|-|  |-|
push 1 to stack2 and pop it from stack2:
| |  |1|
| |  |2|
|-|  |-|
pop1 from stack2 and return it:
| |  |2|
|-|  |-|

To get real O(1) [not amortized], it is much more complicated and requires more stacks, have a look at some of the answers in this post

EDIT: Complexity analysis:

1. each insert() is trivaially O(1) [just pushing it to stack1]
2. Note that each element is push()ed and pop()ed at most twice, once from stack1 and once from stack2. Since there is no more ops then these, for n elements, we have at most 2n push()s and 2n pop()s, which gives us at most 4n * O(1) complexity [since both pop() and push() are O(1)], which is O(n) - and we get amortized time of: O(1) * 4n / n = O(1)