RogerBW's Blog

The Weekly Challenge 266: X, the Uncommon 28 April 2024

I’ve been doing the Weekly Challenges. The latest involved word list comparison and matrix validation. (Note that this ends today.)

Task 1: Uncommon Words

You are given two sentences, $line1 and $line2.

Write a script to find all uncommmon words in any order in the given two sentences. Return ('') if none found.

A word is uncommon if it appears exactly once in one of the sentences and doesn't appear in other sentence.

For me this turned out to hinge on destructuring assignments. In Perl, for example:

For each input sentence, I'll want two data structures: a list of words that appear only once, and a set (or hash, in languages that don't have a set type) of words that appear at all.

sub getlistset($a) {

A list of words that appear at all:

  my @la = split ' ', $a;

A hash of word counts:

  my %ca;
  map {$ca{$_}++} @la;

Use the hash to filter the list to only those words that appear exactly once.

  @la = grep {$ca{$_} == 1} @la;

Then return a tuple of word list and hash.

  return (\@la, \%ca);

For the main function, I run this on each input sentence.

sub uncommonwords($a, $b) {
  my ($la, $sa) = getlistset($a);
  my ($lb, $sb) = getlistset($b);

Then initialise the output list.

  my @out;

Do a two-way comparison: unique words in sentence A vs all words in sentence B, then unique words in B vs all words in A.

  foreach my $r ([$la, $sb], [$lb, $sa]) {
    my ($wl, $t) = @{$r};

For each word,

    foreach my $w (@{$wl}) {

if it doesn't appear at all in the other sentence,

      unless (exists $t->{$w}) {

add it to the output list.

        push @out, $w;
  return \@out;

In Rust I have some more advanced tools, like the Counter and HashSet classes, but the basic logic is the same:

fn getlistset(a: &str) -> (Vec<&str>, HashSet<&str>) {
    let mut la = a.split(' ').collect::<Vec<&str>>();
    let ca = la.iter().copied().collect::<Counter<&str>>();
    la.retain(|k| *ca.get(k).unwrap() == 1);
    (la, ca.keys().copied().collect::<HashSet<&str>>())

fn uncommonwords(a: &str, b: &str) -> Vec<String> {
    let (la, sa) = getlistset(a);
    let (lb, sb) = getlistset(b);
    let mut out = Vec::new();
    for (&ref wl, &ref t) in [(&la, &sb), (&lb, &sa)] {
        for w in wl {
            if !t.contains(w) {

Task 2: X Matrix

You are given a square matrix, $matrix.

Write a script to find if the given matrix is X Matrix.

A square matrix is an X Matrix if all the elements on the main diagonal and antidiagonal are non-zero and everything else are zero.

By contrast this is a much easier problem. The constraint is a little bit like a diagonal matrix, or two of them superimposed. But it's simply a matter of looking at each value, working out from its coordinates whether it should be zero or non-zero, and verifying.


sub xmatrix(@a) {
  my $order = @a.elems - 1;
  my $valid = True;
  for (0 .. @a.end) -> $y {
    for (0 .. @a[$y].end) -> $x {

If we're on the main diagonal or antidiagonal, fail if the element is zero.

      if ($x == $y || $x == $order - $y) {
        if (@a[$y][$x] == 0) {
          $valid = False;

Otherwise fail if it's not zero.

      } else {
        if (@a[$y][$x] != 0) {
          $valid = False;
      unless ($valid) {
    unless ($valid) {
  return $valid;

With languages that provide an enumerate-type function this is easier to express. (Yes, Raku does this with kv but it's hard to find if you don't already know what they've chosen to call it, and I didn't spot it in time.) JavaScript:

function xmatrix(a) {
    const order = a.length - 1;
    let valid = true;
    a.forEach((row, y) => {
        row.forEach((value, x) => {
            if (x == y || x == order - y) {
                if (value == 0) {
                    valid = false;
            } else {
                if (value != 0) {
                    valid = false;
    return valid;

Full code on github.

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