TL;DR

On with TASK #2 from The Weekly Challenge #187. Enjoy!

# The challenge

You are given a list of positive numbers, @n, having at least 3 numbers.

Write a script to find the triplets (a, b, c) from the given list that satisfies the following rules.

1. a + b > c
2. b + c > a
3. a + c > b
4. a + b + c is maximum.


In case, you end up with more than one triplets having the maximum then pick the triplet where a >= b >= c.

Example 1

Input: @n = (1, 2, 3, 2);
Output: (3, 2, 2)


Example 2

Input: @n = (1, 3, 2);
Output: ()


Example 3

Input: @n = (1, 1, 2, 3);
Output: ()


Example 4

Input: @n = (2, 4, 3);
Output: (4, 3, 2)


# The questions

From a very generic point of view, Iâ€™d argue that the disambiguation about the triple to select might leave out corner cases where there might be two triangles with the same perimeter but different side lenghts. On the other hand, I had a hard time coming up with an example where this perimeter is maximal, so I guess this condition might not appear practically. Anywayâ€¦ at least a hat tip would have helped!

# The solution

I guess thereâ€™s a ton to optimize but weâ€™ll assume a small bunch of numbers here and no up-front optimization..

The basic approach will be the following:

• find out all possible triplets, drawing three elements at any time;
• verify each triple is indeed a triangle and calculate its perimeter;
• select the triple with the highest sum

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

my @result = magical_triplets(@ARGV);
say '(', join(', ', @result), ')';

sub magical_triplets (@n) {
my $it = combinations_iterator(3, @n); my ($best, $best_score); while (my ($combination, $complement) =$it->()) {
my $score = is_triangle($combination->@*) or next;
($best,$best_score) = ($combination,$score)
if (! defined $best) || ($best_score < $score); } return reverse sort {$a <=> $b} ($best // [])->@*;
}

sub is_triangle ($A,$B, $C) { return 0 if$A >= $B +$C
|| $B >=$C + $A ||$C >= $A +$B;
return $A +$B + $C; } sub combinations_iterator ($k, @items) {
my @indexes = (0 .. ($k - 1)); my$n = @items;
return sub {
return unless @indexes;
my (@combination, @remaining);
my $j = 0; for my$i (0 .. ($n - 1)) { if ($j < $k &&$i == $indexes[$j]) {
push @combination, $items[$i];
++$j; } else { push @remaining,$items[$i]; } } for my$incc (reverse(-1, 0 .. ($k - 1))) { if ($incc < 0) {
@indexes = (); # finished!
}
elsif ((my $v =$indexes[$incc]) <$incc - $k +$n) {
$indexes[$_] = ++$v for$incc .. ($k - 1); last; } } return (\@combination, \@remaining); } }  The translation into Raku is pretty much literal, with a couple exceptions: • finding out combinations is part of the language, yay! • I couldnâ€™t figure out how to pass an arrayâ€™s content as three individual variables to a sub, so I had to unpack the input array in a very old-Perl5-style. #!/usr/bin/env raku use v6; sub MAIN (*@args) { put magical-triplets(@args) } sub magical-triplets (@n) { my ($best, $best-score); for @n.combinations(3) ->$comb {
my $score = is-triangle($comb) or next;
($best,$best-score) = ($comb,$score)
if (! defined $best) || ($best-score < $score); }$best //= [];
return $best.sort({$^a <=> $^b}).reverse; } sub is-triangle ($x) {
my ($A,$B, $C) = @$x;
return 0 if
$A >=$B + $C ||$B >= $C +$A
|| $C >=$A + $B; return$A + $B +$C;
}


I like that the solution comes out so compact anyway.

Stay safe!

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