The space cost is O(n). The runtime cost of enqueue is O(1); the runtime cost of
dequeue is O(n). Using the "accounting method", the cost of an item in the
system is O(1). Here's why:
+------------+----------------------------+------+
| enqueue | push onto lhs | O(1) |
+------------+----------------------------+------+
| lhs -> rhs | pop off lhs; push onto rhs | O(1) |
+------------+----------------------------+------+
| dequeue | pop off rhs | O(1) |
+------------+----------------------------+------+
The bottom-up solution run in O(n) time instead of O(2^n) time, which the
recursive solution runs as:
```
def fib(n):
return fib(n - 2) + fib(n - 1)
```
Remember that exponential algorithms are usually recursive algorithms with
multiple sibling calls to itself.
Write a predicate for checking if a linked-list contains a cycle. For additional
practice, I also implemented a function that accepts a linked-list containing a
cycle and returns the first element of that cycle.
InterviewCake.com has a section on Facebook's interview, so I'm attempting to
solve all of the problems on there even if that means I'm resolving
problems. The more practice, the better. Right?
URL: interviewcake.com/facebook-interview-questions
