Written by Roy van Rijn (royvanrijn.com) on
Feb 7, 2012 14:54:54
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Levenshtein Distance Challenge: Causes
Today I’ve been playing around with the Levenshtein distance. The Levenshtein distance is a number which measures the ‘distance’ between two strings. For example, the distance between “test” and “rest” is one.
A Levenshtein distance of one is the key element in a challenge I’ve been reading about. I first encountered it on williamedwardscoder’s blog.
The problem description:
Two words are friends if they have a Levenshtein distance of 1. That is, you can add, remove, or substitute exactly one letter in word X to create word Y. A word’s social network consists of all of its friends, plus all of their friends, and all of their friends’ friends, and so on. Write a program to tell us how big the social network for the word “causes” is, using this word list. Have fun!
Java solution (8.1 sec)
After some Googling and tweaking I decided to make an implementation based on the Trie structure. How this helps is excellently described by Steve Hanov. I’ve also had a peek in another Java based Trie implementation by Ximus.
I’ve been able to get the code below run in 8.1 seconds, which is pretty good. But I’ve read that there are Java implementations running in just 4 seconds…!? Maybe based on Levenshtein Automata?