[nojvek]

> Implementation would be another matter as there are a number of complexity gotchas.

I interviewed at an internet darling company. They had the main interviewer (who was a bit rude) and another shadow interviewer who takes notes on everything you do an say. All interviews are like this.

The coding gotchas quickly make you super nervous as everything you do and say is recorded. You have 15 minutes. Plus whiteboard is not an editor. I don't code linearly like writing a text book.

I have come to a realization that fresh out of university me performs better at interviews than 7 year experienced me.

[tripzilch]

I just took a shot at solving this puzzle for the /usr/share/dict wordlist (70k words when filtered to strictly alpha), using Python.

Assuming the "first build a graph" approach is the most efficient one, it's definitely taking me longer than 15 minutes to get a reasonably efficient solution. Just for the graph-building part. I'm taking a two-step approach, first connect the matching words that differ in length by one, so filtering for (a == b[1:] or a == b[:-1]), assuming a is always the shorter word of the pair. Checking all words of length N vs length N+1 that way took a few minutes on my (not very powerful) netbook. Not entirely happy with it, but no way I could optimize that further in 15 minutes anyway (if there is an even better algorithm), and that part of the calculation is done.

edit: just thought of a better way

    aa = sorted([words of length N])
    bb = [words of length N+1]

    # now make a sorted list of the longer words, truncated on either side,
    # (paired with the original for retrieval later)
    bb1 = sorted([(w[1:], w) for w in bb] 
        + [(w[:-1], w) for w in bb])

    # now we can do a merge operation on aa and bb1, and find 
    # the matches in O(len(aa)+len(bb1)) instead of O(len(aa)*len(bb))
    #
    # a merge operation that I can't be arsed to write out here, but I
    # promise it'll involve some super clever usage of the itertools module
    #

Now for step 2, words of equal length...