I don't think the "way to do it" of your update really does it, if I understand correctly what "it" is ... that way would result in two copies of any element that was twice in @temp2, whether it was once, twice, or three or more times in @temp.

Provided I understand correctly what "it" is, I'd count up and down:

my %count; $count{$_}++ for @temp;
my @non_unique_intersection = grep { $count{$_}-->0 } @temp2;
[download]

This will preserve the order of elements of @temp2, until the count (initially the frequency in @temp of that element's string equivalence class) has been exhausted.

Update: Might as well make a function, while I'm at it:

# my @non_unique_intersection = nonunique_intersect( \@temp, \@temp2 )
+;

sub nonunique_intersect {
  my ($x, $y) = @_;
  my %count; $count{$_}++ for @$x;
  my @return = grep { $count{$_}-->0 } @$y;
  return wantarray ? @return : \@return;
}
[download]

Update 2: For your next challenge, the non-unique intersection, leftovers from @temp, and leftovers from @temp2; assuming that the bag of ( @temp, @temp2 ) is supposed to be the same as the bag of ( (@non_unique_intersection) x 2, @leftovers, @leftovers2 ):

my %count; $count{$_}++ for @temp;
my @non_unique_intersection = grep { $count{$_}-->0 } @temp2;
my (@leftovers, @leftovers2);
push @{$count{$_}>0 ? \@leftovers : \@leftovers2}, ($_) x abs($count{$
+_}) for keys %count;
[download]

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