RogerBW's Blog

The Weekly Challenge 255: Most Frequently Odd Character 11 February 2024

I’ve been doing the Weekly Challenges. The latest involved a lot of counting. (Note that this ends today.)

Task 1: Odd Character

You are given two strings, $s and $t. The string $t is generated using the shuffled characters of the string$s with an additional character.

Write a script to find the additional character in the string $t.

My approach is to build a table of character and count in the first string, then subtract characters in the second string from it one by one until I underflow. With Rust I have a Counter class, and with most of the other languages I at least have the ability to add a default for a hash value. Ruby:

def oddcharacter(s, t)

Count characters in s

  ss =
  s.chars.each do |c|
    ss[c] += 1

…then iterate the characters of t.

  t.chars.each do |c|
    if ss.has_key?(c) and ss[c] > 0 then
      ss[c] -= 1
      return c

We should never get here (but typed languages need a typed return).

  return "@"

Task 2: Most Frequent Word

You are given a paragraph $p and a banned word $w.

Write a script to return the most frequent word that is not banned.

The least portable part here was the split on non-alpha characters (which in PostScript I had to write from scratch, and a version of that may end up in the library). But every other language had some way of doing it. Raku:

sub mostfrequentword($para, $banned) {
    my %words;
    map {%words{$_}++}, $para.comb(/<[A..Za..z]>+/);

As in task 1, %words now contains a list of words and frequencies. We find the frequency of the commonest word(s)…

    my $m = %words.values.max;

…filter the list of words to include only those…

    my @v = %words.keys.grep({%words{$_} == $m}).sort;

…and return one of them.

    return @v[0];

Full code on github.

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