Two years behind the rest of the class, I needed to solve a very similar problem. It is in fact not so computationally complex -- with dynamic programming, it seems it can be solved in O(length(@arr1)).

The algorithm below is a bit like edit distance for strings +"edit distance" +"dynamic programming" with scores instead of costs and maxima searched for at any point.

#!/usr/bin/perl -w
use strict;
my @arr1 = qw( 123 a4 b c david e f g 8 h f g 8 X i j k l george );
my @arr2 = qw( c ¤¤ a4 b c d e f g 8 X l m k a4 b c david george );

print "array1: @arr1\n";
print "array2: @arr2\n";
my($longest, $longest_inds) = find_lccs(\@arr1, \@arr2);
print
    scalar @$longest_inds,
    " instance(s) of subsequence of length $longest, starting at index
+ ",
    (join ', ' => @$longest_inds), ":\n";
    
foreach (@$longest_inds) {
    print "@arr1[$_ .. $_ + $longest - 1]\n";
}


sub find_lccs {
    my ($arr1, $arr2) = @_;
    my %inds;
    my $longest = 0;
    my @longest_inds = ();

    for (my $i = 0; $i < @$arr1; $i++) {
        $inds{$arr1->[$i]}->{$i} = 0;
    }

    for (my $i = $#$arr2; $i >= 0; $i--) {
        foreach (keys %{ $inds{$arr2->[$i]} } ) {
            if (defined $arr2->[$i+1] && exists $inds{$arr2->[$i+1]}->
+{$_ + 1}) {
                $inds{$arr2->[$i]}->{$_} = $inds{$arr2->[$i+1]}->{$_ +
+ 1} + 1;
            } else {
                $inds{$arr2->[$i]}->{$_} =  1;
            }
            if ($inds{$arr2->[$i]}->{$_} > $longest ) {
                $longest = $inds{$arr2->[$i]}->{$_};
                @longest_inds = ($_);
            } elsif ($inds{$arr2->[$i]}->{$_} == $longest ) {
                push @longest_inds, $_;
            }
        }
    }

    @longest_inds = sort {$a <=> $b} @longest_inds;
    return ($longest, \@longest_inds);
}
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