TL;DR

Here we are with TASK #1 from The Weekly Challenge #237. Enjoy!

The challenge

Given a year, a month, a weekday of month, and a day of week (1 (Mon) .. 7 (Sun)), print the day.

Example 1

Input: Year = 2024, Month = 4, Weekday of month = 3, day of week = 2
Output: 16

The 3rd Tue of Apr 2024 is the 16th

Example 2

Input: Year = 2025, Month = 10, Weekday of month = 2, day of week = 4
Output: 9

The 2nd Thu of Oct 2025 is the 9th

Example 3

Input: Year = 2026, Month = 8, Weekday of month = 5, day of week = 3
Output: 0

There isn't a 5th Wed in Aug 2026

The questions

I have to admit that I had a hard time understanding what a weekday of month means, so I hope I got it right from the examples (i.e. that $n$ means the $n$-th such day in the specific month).

Another thing that is not clear is what to return if such a day does not exist, except that there’s an explicit example about it so I get that printing 0 is fine.

I would also like to know if there are limits that can be assumed on the dates range, just to avoid pitfalls like when different countries switched to the Gregorian calendar.

Speaking of which, I assume we’re actually talking about the Gregorian calendar or anything people living in the so-called “west” are used to.

The solution

We will start with Raku, which provides an excellent Date class out of the box that allows doing intuitive arithmetics with integers representing days. I’ll leave the algorithm to the comments:

#!/usr/bin/env raku
use v6;
sub MAIN (Int $year, Int $month, Int $weekday_in_month, Int $weekday) {
   my $date = Date.new(year => $year, month => $month, day => 1);

   # how much should we advance to find the first occurrence of the
   # target $weekday?
   my $delta = ($weekday + 7 - $date.day-of-week) % 7;

   # advance that much to land on the first, then additional weeks to
   # land on the target $weekday_in_month
   $date += $delta + ($weekday_in_month - 1) * 7;

   # print the result making sure it's still in the same year & month
   put($date.year == $year && $date.month == $month ?? $date.day !! 0);
}

It’s still easy to translate this into Perl, although we have to play at a lower level of abstraction if we want to stick with stuff in CORE (there are excellent libraries but they come with a toll).

So here our dates will be represented by the epoch of its mid-day in Greenwich Mean Time, and all transformations will take into account that a day has exactly $24 * 3600$ seconds. We are not worried by leap seconds here, though, because we’re anyway focusing on mid-day and we’re not using anything below the day granularity anyway.

Again, I’ll leave it to the comments to explain the gory details:

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

say seize_the_day(@ARGV);

sub seize_the_day ($year, $month, $weekday_in_month, $weekday) {
   my $date = timegm_modern(0, 0, 12, 1, $month - 1, $year);
   my $date_day_of_week = (gmtime($date))[6];

   # how many days should we advance to find the first occurrence of the
   # target $weekday?
   my $delta = ($weekday + 7 - $date_day_of_week) % 7;

   # advance that many days to land on the first, then additional weeks
   # to land on the target $weekday_in_month.
   $date += ($delta + ($weekday_in_month - 1) * 7) * 24 * 3600;

   # get back the year and month of the date we landed on
   my (undef, undef, undef, $day, $newm, $newy) = gmtime($date);
   $newm += 1; # apply offset for month
   $newy += 1900; # apply offset for year

   # return making sure we're in the same year & month
   return $year == $newy && $month == $newm ? $day : 0;
}

This said… stay safe and have a good time!