jennieohyoung / interview_practice Goto Github PK

in your circular_flag solution, you can only ever call that once... you'll need to clear all the flags before you go and use them for circular check.

It occurred to me that we never covered this. In question 2, you're solving for Fibonacci sequence recursively.

e.g. fib(n) = fib(n - 1) + fib(n -2)

which is a straight up implementation of the mathematical formula.

However, as briefly mentioned before, this is ~O(fib(n)). Can you do this in a non-recursive fashion? What is that complexity?

Can you do this as a fixed form constant time expression (e.g. O(1))?

Again, as in 5.py, you're not handling your self.tail.

Your solution is incomplete:

1. you don't have a case for if self.head is None.
2. you don't ever write to self.tail
3. your solution could be made shorter by using self.tail if you're going to include it in the data structure. presumably self.tail will point at the last element in the linked list, so you could just

self.tail.next = node;
self.tail = node;

or something like that.