Not sure if this is faster, but it might be, as it doesn't use nested loops.
sub score {
my ($word1, $word2) = @_;
my (%chars1, %chars2) = ();
$chars1{$_}++ for split '', $word1;
$chars2{$_}++ for split '', $word2;
# the minimum of the two hashes is the number in common for each l
+etter
my $sum = 0;
$sum += ($chars1{$_} < $chars2{$_}
? $chars1{$_}
: $chars2{$_}) for keys %chars1;
return $sum;
}
while (<DATA>) {
chomp;
print "$_: " . score(split /\s+/) . " in common\n";
}
__DATA__
perl monk
help temp
frood hoopy
bilbo baggins
jibber jaber
This prints:
perl monk: 0 in common
help temp: 2 in common
frood hoopy: 2 in common
bilbo baggins: 2 in common
jibber jabber: 5 in common
I notice your algorithms give 3 matches for 'bilbo' and 'baggins'. I think this is because both 'b's in bilbo match inside baggins. I'm not sure if this is correct behavior by your specifications or not.
Update: To speed up your score2 sub, consider using index($word2, $a) > 0 instead of the regex match. Changing this alone made it approximately as fast as your initial score sub for me.
blokhead
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