Advent Calendar 2025

|   Day 8   |   Day 9   |   Day 10   |

The gift is presented by Roger Bell_West. Today he is talking about his solutioni to The Weekly Challenge - 311. This is re-produced for Advent Calendar 2025 from the original post.

I’ve been doing the Weekly Challenges. The latest involved case swapping and character grouping. (Note that this ends today.)

Task 1: Upper Lower

You are given a string consists of english letters only.

Write a script to convert lower case to upper and upper case to lower in the given string.

There’s an obvious logic to this:

if char.is_upper() {
  lowercase char
} else {
  uppercase char
}

but instead I can take advantage of ASCII's encoding, such that flipping bit 5 (0x10) of a letter swaps its case.

Raku:

sub upperlower($a) {
    my $out = '';
    for $a.comb -> $c {
        $out ~= chr(ord($c) +^ 32);
    }
    $out;
}

And I rather like clarity of the PostScript version, though it does rely on some of my library functions (converting string to array and back, and map to apply its function to all elements in an array).

/upperlower {
    s2a
    { 32 xor } map
    a2s
} bind def

Task 2: Group Digit Sum

You are given a string, $str, made up of digits, and an integer, $int, which is less than the length of the given string.

Write a script to divide the given string into consecutive groups of size $int (plus one for leftovers if any). Then sum the digits of each group, and concatenate all group sums to create a new string. If the length of the new string is less than or equal to the given integer then return the new string, otherwise continue the process.

Stating the problem clearly pretty much lays out the route to the solution. In Crystal:

def groupdigitsum(a, sz)

Make a working copy of the string, and initialise the accumulator.

  s = a
  n = 0

Infinitely loop.

  while true

Initialise output string.

    t = ""

Iterate over characters of input string, with an index value.

    s.chars.each_with_index do |c, i|

Add the digit value to the accumulator.

      n += c.to_i

If we’ve hit the end of the string or a group boundary, flush the accumulator to the output string and zero it.

      if i == s.size - 1 || (i + 1) % sz == 0
        t += n.to_s
        n = 0
      end
    end

Copy output string to next input.

    s = t

If it’s short enough, break out of the lopo and return it.

    if s.size <= sz
      break
    end
  end
  s
end

Full code on github.

If you have any suggestion then please do share with us perlweeklychallenge@yahoo.com.

|   Advent Calendar 2025   |