You want the Longest common subsequence, not substring. Below a simple modification of the wikipedia pseudocode of the dynamic programming algorithm (

http://en.wikipedia.org/wiki/Longest_common_subsequence_problem) to make it word based.

`use strict;
use warnings;
use Data::Dumper;
# word lists, first element a dummy
my @s1 = (q{}, split(/s+/, "Perlmonks is the best perl community"));
my @s2 = (q{}, split(/s+/, "Perlmonks is one of the best community of
+perl users"));
my @M;
#init dyn. prog. matrix
for my $i ( 0 .. $#s1) {
$M[$i][0] = 0;
}
for my $i ( 0 .. $#s2) {
$M[0][$i] = 0;
}
#calc lcs (word based)
for my $i ( 1 .. $#s1) {
for my $j ( 1 .. $#s2) {
if ($s1[$i] eq $s2[$j]) {
$M[$i][$j] = $M[$i-1][$j-1]+1;
}
else {
if ($M[$i][$j-1] > $M[$i-1][$j]) {
$M[$i][$j] = $M[$i][$j-1];
}
else {
$M[$i][$j] = $M[$i-1][$j];
}
}
}
}
#print Dumper \@M;
printDiff($#s1, $#s2);
sub printDiff {
my ($i, $j) = @_;
if ($i > 0 && $j > 0 and $s1[$i] eq $s2[$j]) {
printDiff($i-1, $j-1);
print " " . $s1[$i];
}
else {
if ($j > 0 && ($i == 0 || $M[$i][$j-1] >= $M[$i-1][$j]
+)) {
printDiff($i, $j-1);
print " <" . $s2[$j] . ">";
}
elsif ($i > 0 && ($j == 0 || $M[$i][$j-1] < $M[$i-1][$
+j])) {
printDiff($i-1, $j);
print " [" . $s1[$i] . "]";
}
}
}
`

This outputs:

` Perlmonks is <one> <of> the best [perl] community <of> <perl> <us
+ers>
`

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