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Re: Re: finding longest common substring

by revdiablo (Prior)
on Nov 20, 2003 at 05:53 UTC ( #308515=note: print w/replies, xml ) Need Help??


in reply to Re: finding longest common substring
in thread finding longest common substring

I haven't looked at your solution in depth, but upon initial inspection it seems to do the same thing as pizza_milkshake's at Re: finding longest common substring. I must admit that -- at least to my eyes -- your version looks cleaner and is more understandable than his, but it doesn't seem like either of these solve the problem I was asking about. Perhaps my quick summary of the problem was unclear, or perhaps I am simply missing something. I was looking for the longest substring that is common to all elements of the list. In my example, I have 3 lists, but each one is analyzed independently. Both of your solutions seem to be searching two lists in parallel for the longest matching element. Maybe I'm just not looking at them from the right perspective?

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Re: Re: Re: finding longest common substring
by davido (Cardinal) on Nov 20, 2003 at 06:37 UTC
    :) Ok, now that I better understand what you're asking, here's my regex solution. I think that this improves upon some other methods by starting with the longest and working down to the shortest, (quitting as soon as a match is found).

    use strict; use warnings; my @array = ( "this is a string is", "a string this is", "tie a string together" ); my $string = join "|", @array; my $repeat = '\|.*?\1.*?' x ( scalar(@array) - 1 ); for my $n ( reverse ( 1 .. length( $string ) ) ) { next unless $string =~ /(.{$n}).*?$repeat/; print $1, "\n"; last; }

    Ob-Update: This assumes the substrings don't overlap. And I've tinkered with the code to make it keep track of what the original substrings looked like. In so doing, I used the | character as a delimiter, which means it shouldn't appear in the original substrings. Thanks danger for the polite nudge. ;)


    Dave


    "If I had my life to live over again, I'd be a plumber." -- Albert Einstein

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