AoC day 5. Part 1 was straightforward; the naive solution that checks every ID against every range is plenty fast. For part 2 I went with a three-step approach: Read provided ranges, merge adjacent/overlapping ranges, sum up the remaining (non-overlapping) ranges.

use v5.36;<br>use List::Util qw(sum0 min);<br><br>my @ranges;<br>while (readline) {<br>    chomp;<br>    last if $_ eq '';<br><br>    my ($lo, $hi) = m{^ (\d+) - (\d+) \z}xa<br>        or die "bad input: '$_'";<br><br>    push @ranges, [$lo, $hi + 1];<br>}<br><br>@ranges = sort { $a->[1] <=> $b->[1] } @ranges;<br><br>for my $i (reverse 1 .. $#[url=https://infosec.exchange/tags/ranges]ranges[/url]) {<br>    if ($ranges[$i - 1][1] >= $ranges[$i][0]) {<br>        $ranges[$i - 1][1] = $ranges[$i][1];<br>        $ranges[$i - 1][0] = min $ranges[$i - 1][0], $ranges[$i][0];<br>        splice @ranges, $i, 1;<br>    }<br>}<br><br>say sum0 map $_->[1] - $_->[0], @ranges;<br>

#AdventOfCode #perl