This is a solution to Advent of Code 2023 day 1, written in Raku.
https://adventofcode.com/2023/day/1
Part One
Input has the following format:
two1nine eightwothree abcone2threexyz xtwone3four 4nineeightseven2 zoneight234 7pqrstsixteen
On each line, the calibration value can be found by combining the first digit and the last digit (in that order) to form a single two-digit number.
Consider your entire calibration document. What is the sum of all of the calibration values?
my $total = 0;
for '1-input.txt'.IO.lines -> $l {
my @digits = $l.comb(/\d/);
my $j = @digits[0,*-1].join;
$total += $j
}
say $total;54597
Part Two
It looks like some of the digits are actually spelled out with letters: one, two, three, four, five, six, seven, eight, and nine also count as valid "digits" …
What is the sum of all of the calibration values?
my %numbers = 'one' => 1, 'two' => 2, 'three' => 3,
'four' => 4, 'five' => 5, 'six' => 6,
'seven' => 7, 'eight' => 8, 'nine' => 9;
my @words = %numbers.keys;
my $total = 0;
for '1-input.txt'.IO.lines -> $l {
# sadly an epically slow regex
my $m = $l.match(/(@words | \d)/, :global, :overlap);
my ($first, $last) = $m[0,*-1]>>.Str.map({ .Int // %numbers{$_} });
$total += $first * 10 + $last;
}
say $total;
say now - ENTER now;54504 1.589480831
The hand-written equivalent regex is considerably faster. Need to figure out why.
my $m = $l.match(/('one'|'two'|'three'|'four'|'five'|'six'|'seven'|'eight'|'nine'|\d)/, :global, :overlap);