Not that there is anything wrong with the other solutions, but here is another one. I sort the keys by values, then use a while loop to figure out where the highest value ends in the list of sorted keys, @keys. Once I know that, I slice @keys to get the right keys.

#!/usr/bin/perl

my %h = (
    red    => 2,
    pink   => 1,
    orange => 4,
    black  => 3,
    blue   => 4,
    green  => 3,
    );
    
my $i = 0;

my @keys    = sort { $h{$b} <=> $h{$a} } keys %h;

1 while( $h{$keys[++$i]} == $h{$keys[0]} );

my @largest = @keys[0..$i-1];

print qq|Largest are "@largest"\n|;
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I had a solution that used grep, but that's stupid since I don't need to go through the rest of the elements once I know I've seen the highest ones.

#!/usr/bin/perl
use strict;

my %h = (
    red    => 2,
    pink   => 1,
    orange => 4,
    black  => 3,
    blue   => 4,
    );
    
my @keys    = sort { $h{$b} <=> $h{$a} } keys %h;

my @largest = grep { $h{$_} == $h{$keys[0]} } @keys;

print qq|Largest are "@largest"\n|;
[download]