algorithms for fast string approximate matching

My gut instinct is just to iterate over each String n, maintaining a counter of how many characters are different than s, but I'm not claiming it is the most efficient solution. However it would be O(n) so unless this is a known performance problem, or an academic question, I'd go with that.

Sedgewick in his book "Algorithms" writes that Ternary Search Tree allows "to locate all words within a given Hamming distance of a query word". Article in Dr. Dobb's

Given that the strings are fixed length, you can compute the Hamming distance between two strings to determine the similarity; this is O(n) on the length of the string. So, worst case is that your algorithm is O(nm) for comparing your string against m words.

As an alternative, a fast solution that's also a memory hog is to preprocess your dictionary into a map; keys are a tuple (p, c) where p is the position in the string and c is the character in the string at that position, values are the strings that have characters at that position (so "the" will be in the map at {(0, 't'), "the"}, {(1, 'h'), "the"}, {(2, 'e'), "the"}). To query the map, iterate through query string's characters and construct a result map with the retrieved strings; keys are strings, values are the number of times the strings have been retrieved from the primary map (so with the query string "the", the key "thx" will have a value of 2, and the key "tee" will have a value of 1). Finally, iterate through the result map and discard strings whose values are less than K.

You can save memory by discarding keys that can't possibly equal K when the result map has been completed. For example, if K is 5 and N is 8, then when you've reached the 4th-8th characters of the query string you can discard any retrieved strings that aren't already in the result map since they can't possibly have 5 matching characters. Or, when you've finished with the 6th character of the query string, you can iterate through the result map and remove all keys whose values are less than 3.

If need be you can offload the primary precomputed map to a NoSql key-value database or something along those lines in order to save on main memory (and also so that you don't have to precompute the dictionary every time the program restarts).

Rather than storing a tuple (p, c) as the key in the primary map, you can instead concatenate the position and character into a string (so (5, 't') becomes "5t", and (12, 'x') becomes "12x").

Without knowing where in each input string the match characters will be, for a particular string, you might need to check every character no matter what order you check them in. Therefore it makes sense to just iterate over each string character-by-character and keep a sum of the total number of mismatches. If i is the number of mismatches so far, return false when i == k and true when there are fewer than k-i unchecked characters remaining in the string.

Note that depending on how long the strings are and how many mismatches you'll allow, it might be faster to iterate over the whole string rather than performing these checks, or perhaps to perform them only after every couple characters. Play around with it to see how you get the fastest performance.

My method if we're thinking out loud :P I can't see a way to do this without going through each n string, but I'm happy to be corrected. On that it would begin with a pre-process to save a second set of your n strings so that the characters are in ascending order.

The first part of the comparison would then be to check each n string a character at a time say n' to each character in s say s'.

If s' is less than n' then not equal and move to the next s'. If n' is less than s' then go to next n'. Otherwise record a matching character. Repeat this until k miss matches are found or the alternate matches are found and mark n accordingly.

For further consideration, an added pre-processing could be done on each adjacent string in n to see the total number of characters that differ. This could then be used when comparing strings n to s and if sufficient difference exist between these and the adjacent n there may not be a need to compare it?