Perl Weekly Challenge 130: quick

One way to let me improve my knowledge about Raku (aka Perl 6) is to implement programs in it. Unluckily, I don’t have any production code to implement in Raku yet (sob!). So, why not try solving the Perl Weekly Challenge tasks?

In the following, the assigned tasks for Challenge 110.

PWC 130 - Task 1

The first task was quite simple: it required to find out a number in a list of arguments that appears an odd number of times. It is simple enough to solve it by means of grep and for:

sub MAIN( *@values where { @values.elems > 1 && @values.grep( * ~~ Int ).elems == @values.elems  } ) {
    $_.say and exit if  @values.grep( $_ ).elems !%% 2 for @values;

It must be read from right to left: the for loop values $_ to every single element of the array, then the if tests if the $_ is found in the array, and elems counts how many time it appears. If the number of times is odd, that is is not even (!%% 2), the value can be printed out. This is do with a say, and then the program exits with a low precedence and exit, because the task states that only one number can appear an odd even of times.

PWC 130 - Task 2

The second task was about verifying if a tree is a balanced binary tree. I hate trees with a passion!
I implemented a class Node that has the current value, and its descendants, as well as two methods:
  • is-BST returns True if this node respects the rules according to its descendants;
  • is-BST-from-here recursively tests all the tree and its subtree.

Therefore, testing the tree is as simple as invoking the latter method on the root:

class Node {
    has Int $.value;
    has Node $.left is rw;
    has Node $.right is rw;

    method is-BST() {
        my $ok-left  = ! $!left  || ( $!left && $!left.value < $!value );
        my $ok-right = ! $!right || ( $!right && $!right.value >= $!value );
        return $ok-left && $ok-right;

    method is-BST-from-here() {
        my $am-I-ok =;
        return False if ! $am-I-ok;

        my $ok-right = True;
        $ok-right = $! if $!right;
        my $ok-left  = True;
        $ok-left = $! if $!left;
        return $ok-right && $ok-left;

sub MAIN() {
    my $root         = value => 8 );
    $root.left       = value => 5 );
    $root.left.left  = value => 4 );
    $root.left.right = value => 6 );
    $root.right      = value => 9 );

    "1".say if $;

At the end, the program prints 1 as requested by the task only if the tree is a balanced one.

The article Perl Weekly Challenge 130: quick has been posted by Luca Ferrari on September 13, 2021