Perl Weekly Challenge 316: coming back from PostgreSQL OpenDay

This post presents my solutions to the Perl Weekly Challenge 316.
I keep doing the Perl Weekly Challenge in order to mantain my coding skills in good shape, as well as in order to learn new things, with particular regard to Raku, a language that I love.
This week, I solved the following tasks:

• PWC 316 - Task 1 - Raku
• PWC 316 - Task 2 - Raku

Raku Implementations

PWC 316 - Task 1 - Raku Implementation

The first task was about searching for in a list of words, to see if the last letter of a word is the first one of the next word in the list.

sub MAIN( *@words where { @words.elems > 0 } ) {

    for 0 ..^ @words.elems - 1 {
		my $ending = @words[ $_ ].comb[ * - 1 ].fc;
		my $beginning = @words[ $_ + 1 ].comb[ 0 ].fc;
		'False'.say and exit unless ( $beginning ~~ $ending );
    }

    'True'.say;
}

The idea is simple: I iterate over the list of words and extract the $ending letter of the current word and the $beginning letter of the next word. If the two letters do not match, I end the program. If the loop ends, then all words satisfy the constraint, so the program was sucessful.

PWC 316 - Task 2 - Raku Implementation

The second task was to find out if a given word can be a subsequence of another one, i.e., if all the letters of the first word are contained in the second word in the same order, even if interleaved with other chars.

sub MAIN( Str $original, Str $subsequence ) {
    my $index = 0;
    for $original.lc.comb  {
	'False'.say and exit unless ( $subsequence.comb[ $index .. * ].grep( $_ ) );
	$index = $subsequence.comb[ $index .. * ].grep( $_,  :k )[ 0 ];
    }

    'True'.say;

}

The solution is simple: I iterate over the letters of the first word and see if I can find it in a sub-array of the second word.