in reply to How to hash without the hash- basic

What kind of surreal contest is that?

Anyway, a rather quick solution would be to sort, and then look through the array, checking if the current element is identical to the previous or next. If no, it's unique.

Update: now with code: 

# intentionally no warnings used.
my @array = qw(1 3 4 55 4 3 4 22 1 3 4 3 2 2 3 34);
@array = sort @array;

my $prev;
for (0..$#array) {
    my $i = $array[$_];
    print "$i\n" if $i != $prev && $i != $array[$_+1];
    $prev = $i;
}

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Second update: Another idea. This one is O(n^2) in theory, but the regex engine being fast makes up for that in practise for low-ish values of n: 

my @array = qw(1 3 4 55 4 3 4 22 1 3 4 3 2 2 3 34);
my $template = join ' ', @array;
study $template;

for (@array) {
    my @matches = ($template =~ /\b$_\b/g);
    print "$_\n" if 1 == @matches;

}

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Perl 6 - second systems done right

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Re^2: How to hash without the hash- basic
by Util (Priest) on Oct 26, 2011 at 19:58 UTC 
    use v6;
my @array = < 1 3 4 55 4 3 4 22 1 3 4 3 2 2 3 34 >;

my @y = @array.classify({$_})\
              .categorize({ .value.elems==1 || Nil }).[0].value».key;

say "{@y.sort.join(',')} are the numbers that appear only once" if @y;

#my @y = @array.classify({$_}).grep({ .value.elems == 1 })».key;

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