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Just for the sake of completeness: A fast and elegant algorithm for this is a tricky use of suffix trees. One concatenates the two strings of length n and m, say abcdef%efgab$. It is possible to construct a suffix tree of this string in O(n+m) (Ukkonen algorithm). To find the common substrings, one has then to search for nodes that have exactly two (or the number of strings) leafs belonging to the different words. The resulting suffix tree for "abcdef" and "efgab":
| |(3:cdef%efgab$)|leaf |(1:ab)| | |(13:$)|leaf tree:| | |(3:cdef%efgab$)|leaf |(2:b)| | |(13:$)|leaf | |(3:cdef%efgab$)|leaf | |(4:def%efgab$)|leaf | | |(7:%efgab$)|leaf |(5:ef)| | |(10:gab$)|leaf | | |(7:%efgab$)|leaf |(6:f)| | |(10:gab$)|leaf | |(7:%efgab$)|leaf | |(10:gab$)|leaf |
So "ab" has two leafs in the different words (position <= 7 for leaf 1 and position > 7 for leaf 2). So have 'b', 'ef' and 'f'.

http://en.wikipedia.org/wiki/Longest_common_substring_problem

Update: Just found some perl code with google ... on perlmonks ;) Re: finding longest common substring


In reply to Re: Challenge: Fast Common Substrings by lima1
in thread Challenge: Fast Common Substrings by Limbic~Region

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