Thanks for the reply. Levenshtein distance would be especially useful
for when I need to compare general strings, with a general amount of
edits between them.

Unfortunately, the problem isn't so much determining whether any two
strings are one edit apart: that can be done just by checking if the
first two, first and third, or last two characters of one string
equals those of the other.

The problem is which strings you compare with to determine which ones
are one edit apart. Right now, I have to compare every string with
every other string, but it seems really wasteful, seemingly being in
O(n^2) time. I just can't figure out how to make it more efficient,
though.

I'm hoping that this a common problem with a common solution, like
using some data structure I haven't thought of.

On May 30, 1:01 pm, Laurent PETIT <laurent.pe...@gmail.com> wrote:
> Hi,
>
> there's the levenshtein distance algorithm which will help you
> determine which string is one "edit" close to another (since all your
> strings are of length 3, then the distance will inevitably be a single
> replacement if of size one).
>
> Don't know if that helps, anyway here's a compact functional
> implementation:https://gist.github.com/828413
>
> 2011/5/30 joshua-choi <rbysam...@gmail.com>:
>
>
>
>
>
>
>
> > Let's say that I have a set of strings, each three English letters
> > long.
>
> > How can I determine which strings differ only at one location (e.g.
> > "xxe" and "xbe")?
>
> > Right now, I'm writing a loop that sequentially compares every string
> > to every other string. I think that there's a better way, but I don't
> > know where to start.
>
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