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I have been looking for an advanced levenshtein distance algorithm, and the best I have found so far is O(n*m) where n and m are the lengths of the two strings. The reason why the algorithm is at this scale is because of space, not time, with the creation of a matrix of the two strings such as this one:
Is there a publicly-available levenshtein algorithm which is better than O(n*m)? I am not averse to looking at advanced computer science papers & research, but haven't been able to find anything. I have found one company, Exorbyte, which supposedly has built a super-advanced and super-fast Levenshtein algorithm but of course that is a trade secret. I am building an iPhone app which I would like to use the Levenshtein distance calculation. There is an objective-c implementation available, but with the limited amount of memory on iPods and iPhones, I'd like to find a better algorithm if possible.
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The Levenshtein distance is a string metric for measuring difference between two sequences. Informally, the Levenshtein distance between two words is the minimum number of single-character edits (i.e. insertions, deletions or substitutions) required to change one word into the other.
Different definitions of an edit distance use different sets of string operations. Levenshtein distance operations are the removal, insertion, or substitution of a character in the string. Being the most common metric, the term Levenshtein distance is often used interchangeably with edit distance.
In linguistics, the Levenshtein distance is used as a metric to quantify the linguistic distance, or how different two languages are from one another.
The Levenshtein distance used as a metric provides a boost to accuracy of an NLP model by verifying each named entity in the entry. The vector search solution does a good job, and finds the most similar entry as defined by the vectorization.
Are you interested in reducing the time complexity or the space complexity ? The average time complexity can be reduced O(n + d^2), where n is the length of the longer string and d is the edit distance. If you are only interested in the edit distance and not interested in reconstructing the edit sequence, you only need to keep the last two rows of the matrix in memory, so that will be order(n).
If you can afford to approximate, there are poly-logarithmic approximations.
For the O(n +d^2) algorithm look for Ukkonen's optimization or its enhancement Enhanced Ukkonen. The best approximation that I know of is this one by Andoni, Krauthgamer, Onak
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