TL;DR

Here we are with TASK #1 from The Weekly Challenge #175. Enjoy!

The challenge

Write a script to list Last Sunday of every month in the given year.

For example, for year 2022, we should get the following:

2022-01-30
2022-02-27
2022-03-27
2022-04-24
2022-05-29
2022-06-26
2022-07-31
2022-08-28
2022-09-25
2022-10-30
2022-11-27
2022-12-25

The questions

I always get shivers when I see a challenge involving dates (or times) with such a broad span. In the given year screams for a question about which limits should (or could) be assumed.

In the same vein, I’d also ask where should we anchor the answer to.

I’m no expert about dates and times, but Socratically I know that I don’t know enough and I’ll probably never will. I got alive through a single read of UTC is enough for everyone …right? and that convinced me to avoid date and time stuff (at least with a broad span) as much as possible.

Hence, I’ll assume that whatever applies to Roma (Italy) since the seventies should be correct, everything else is nice to have. Anyway… these are test challenges, not production stuff right?!?

The solution

As usual, we’ll start with Raku, which supports Date handling natively. I’ll blindly assume that it’s fine for the constraints I’ve set in the questions!

#!/usr/bin/env raku
use v6;
sub MAIN (Int:D $year = 2022) { .put for sundays-in($year) }

sub sundays-in (Int:D $year) {
   my $year-start = Date.new(:$year);

   # find the first sunday in the year
   my $cursor = $year-start;
   $cursor++ while $cursor.day-of-week % 7;

   # find all last sundays in the year
   return gather loop {

      # we will compare a candidate sunday against the next one
      my $candidate = $cursor;
      $cursor += 7;

      # a jump in the month for the "next" sunday means that our
      # $candidate was the last one in its month, so take it
      take $candidate if $cursor.month != $candidate.month;

      # a jump in the year means we've taken the last one in the
      # requested year, so we can just say goodbye
      last if $cursor.year > $candidate.year;
   };
}

The approach is the following:

• find the first Sunday in the target year, this will be our starting point (initializing variable $cursor);
• loop to do the same:
  • advance the cursor ahead one week (assuming that 7 days are one week);
  • compare whether this advanced date stepped onto the next month with respect to the starting value
  • if advanced, the previous value is a Last Sunday and gets taken
  • stop when the year of the new value is the next one (we have it January of the next year).

I still cross my finger thinking of applying this to any year.

In Perl, we can leverage the venerable DateTime but otherwise more or less translate the solution from Raku:

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

use DateTime;

my $year = shift // 2022;
say for sundays_in($year);

sub sundays_in ($year) {
   my $year_start = DateTime->new(
      year => $year,
      month => 1,
      day  => 1,
      hour => 12,
      time_zone => 'floating',
   );

   # find the first sunday in the year
   my $cursor = $year_start->clone;
   $cursor->add(days => 1) while $cursor->day_of_week % 7;

   # find all last sundays in the year
   my @retval;
   while ($cursor->year == $year) {

      # we will compare a candidate sunday against the next one
      my $candidate = $cursor->clone;
      $cursor->add(days => 7);

      # a jump in the month for the "next" sunday means that our
      # $candidate was the last one in its month, so take it
      push @retval, $candidate->ymd('-')
         if $cursor->month != $candidate->month;
   }
   return @retval;
}

I know I said something about being valid for Roma (Italy), but this is only because I know that nothing specific has happened since the seventies in the last century. To be on the safe side (e.g. for leap stuff) I’m setting in the middle of the day at 12 o’clock.

So there we have, all the Last Sundays in a year, in a hopefully working solution!