Keep 2 stacks, let's call them stack1 and stack2.
- Push the new element onto stack1
- If outbox is empty, refill it by popping each element from stack1 and pushing it onto stack2
- Pop and return the top element from stack2
Using this method, each element will be in each stack exactly once - meaning each element will be pushed twice and popped twice, giving amortized constant time operations.