TL;DR

Here we are with TASK #1 from The Weekly Challenge #218. Enjoy!

The challenge

You are given a list of 3 or more integers.

Write a script to find the 3 integers whose product is the maximum and return it.

Example 1

Input: @list = (3, 1, 2)
Output: 6

1 x 2 x 3 => 6

Example 2

Input: @list = (4, 1, 3, 2)
Output: 24

2 x 3 x 4 => 24

Example 3

Input: @list = (-1, 0, 1, 3, 1)
Output: 3

1 x 1 x 3 => 3

Example 4

Input: @list = (-8, 2, -9, 0, -4, 3)
Output: 216

-9 × -8 × 3 => 216

The questions

From a high-level, generic standpoint I’d probably ask whether there’s any limit on the inputs, e.g. to know whether we have to use a big integer library or not.

But well, yeah… nothing serious.

The solution

If there’s three numbers only, the solution is trivial as there is only one possible value.

Otherwise, for negative numbers to make sense, they should come into pairs to get to a positive value. In this case, the two lowest values and the highest value would produce the biggest result.

The other candidate is that taking the three highest values.

As a matter of fact, these two alternatives are the same when there are only three values, so we don’t really have to check for that condition, right?

Raku first:

#!/usr/bin/env raku
use v6;
sub MAIN (*@args) { say maximum-product(@args) }

sub maximum-product (@args) {
   my @sorted = @args».Int.sort;
   my $below = @sorted[0] * @sorted[1] * @sorted[* - 1];
   my $above = [*] @sorted.reverse[0..2];
   return ($below, $above).max;
}

Perl now:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';

say maximum_product(@ARGV);

sub maximum_product (@list) {
   @list = sort { $a <=> $b } @list;
   my $below = $list[0] * $list[1] * $list[-1];
   my $above = $list[-3] * $list[-2] * $list[-1];
   return $below > $above ? $below : $above;
}

Cheers!