Okay, so today I had a phone interview for a job and I feel like it went fairly well. I am notoriously bad over the phone so I was a bit nervous, but I think I answered well for all the JavaScript questions. At the end I was given a question which would apply to programming in C. This was not surprising because I listed experience with C, C++, and other languages I have used a fair amount in the past.

I probably didn't do so well on this last question, but I learned about an algorithm I didn't know about before so there's my silver lining. One of my problems was that I was having trouble understanding the question over the phone at first, but eventually I understood it as:

How would you write a function that takes a linked list as an argument and should return true if the linked list contains a loop, and return false otherwise.

After I understood this I thought about it. So basically, it's a function that takes a linked list and tells you whether or not it's circular? Not quite. My first attempt was a description of something like this. Looking back I was obviously nervous and rushing it.

/** Return true if a loop is found */ int detectLoop(Node * list) { Node * current = list; while (current = current->next) { if (current == list) return 1; } return 0; }

The problem with this one as my interviewer pointed out, is that it only detects circular linked lists, or lists that link back to the first node. I realized he was right and tried again. After a minute of fumbling around with the problem I came up with an inefficient solution (which I believe is quadratic) using a nested loop:

int detectLoop(Node * list) { if (!list->next) return 0; Node * current = list->next; int i = 0; do { Node * scan = list; int j = 0; do { if (scan = current) return 1; j++; } while (j < i && scan = scan->next); i++; } while (current = current->next); return 0; }

I had no access to a computer and was trying to come up with something on my own. I had never written a function to do this before. I think my answer was a bit complicated to describe over the phone and I think the interviewer probably had trouble understanding me completely. The interviewer told me we were out of time after I gave this answer. With a single google search of "loops in linked lists" I was able to figure out exactly the answer I should have given. I am a huge fan of StackOverflow and always find good information there that sometimes have been vetted by hundreds of programmers. Such was the case with this question that pointed me to Floyd's Cycle-Finding Algorithm, also known as the tortoise and hare algorithm. The algorithm is named for Robert W. Floyd, a very accomplished computer sceintist and professor at Stanford, who invented it in late 60s.

The algorithm is absolutely brilliant and I had to write about it. The general idea is to have two pointers which you advance through the list at different rates. It can be shown that if there is a loop, then eventually both pointers will point to the same node. Not only will they eventually point to the same node, they will do so in O(n) steps. Here is my C implementation:

int detectLoop(Node * list) { if (!list) return 0; Node * tortoise = list; Node * hare = list; do { if (!(hare = hare->next) || !(hare = hare->next)) return 0; tortoise = tortoise->next; } while (tortoise != hare); return 1; }

This is by no means a formal proof, but here is how I think about the above algorithm:

If the list does not contain a loop, we will eventually hit NULL and return false. If incremented first, the hare pointer will always be NULL before the tortoise pointer.

and return false. If incremented first, the hare pointer will always be before the tortoise pointer. If the list does contain a loop, then both pointers will end up in the loop eventually. It is not possible for the pointers to meet before they enter the loop. Assume the list does contain a loop of length L. At this point, I think the simplest way to think about the algorithm at this point is to consider only the loop. Take the starting location of the loop to be node 0 . When the tortoise pointer enters the loop at node 0 , the hare pointer will have already entered the loop and may be located at any node in the loop (ranging from node 0 to node L-1 ). When the tortoise pointer enters the loop at node 0 we will consider the unknown location of the hare pointer to be an unknown number of steps p behind the tortoise pointer in the loop and will eventually catch up to it. Once entering the loop at node 0 , after t iterations the tortoise pointer will be located at node t mod L and the hare pointer will be located at the node (2t - p) mod L .



Now if we create an equation using the above values, we can easily see that the hare pointer will catch up and meet the tortoise pointer after t = p iterations (after the tortoise pointer enters the loop).

2t - p = t --> 2t = t + p --> t = p

The total number of steps required is the number of steps (or number of nodes) prior to entering the loop plus the number of steps after entering the loop p . Although we do not know the value of p , we do have an upper bound (which is the important thing) and that is the length of the loop L.