Perl Weekly Challenge 309: Perl and Raku mainly
This post presents my solutions to the Perl Weekly Challenge 309.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 309 - Task 1 - Raku
- PWC 309 - Task 2 - Raku
- PWC 309 - Task 1 in PostgreSQL PL/Perl
- PWC 309 - Task 2 in PostgreSQL PL/Perl
- PWC 309 - Task 1 in PostgreSQL PL/PgSQL
- PWC 309 - Task 2 in PostgreSQL PL/PgSQL
Raku Implementations
PWC 309 - Task 1 - Raku Implementation
Given a sorted array of integers, find out the integer that has the min differences from the previous one.sub MAIN( *@nums where { @nums.grep( * ~~ Int ).elems == @nums.elems } ) {
my %gaps;
for 1 ..^ @nums.elems {
%gaps{ @nums[ $_ ] - @nums[ $_ - 1 ] } = @nums[ $_ ];
}
%gaps{ %gaps.keys.min }.say;
}
I populate an hash of
%gaps with the difference between elements, and value as the element, so that it does just suffice to extract the min key of the has to get the value. Please note that duplicated differences are overwritten, so only the last smallest difference is kept.
PWC 309 - Task 2 - Raku Implementation
Given an array of integers, find out the smallest difference between any element of the array.sub MAIN( *@nums where { @nums.grep( * ~~ Int ).elems == @nums.elems } ) {
my $min = Inf;
for @nums -> $current {
my $found = @nums
.grep( * != $current )
.map( { abs( $current - $_.Int ) } )
.min;
$min = $found if $min > $found;
}
$min.say;
}
I iterate over element of the array, and then
grep for all other elements, mapping each one to the difference and extracting the min value.
If the min is less than the initial one, or previous one, I keep this, otherwise go to the next element.
PL/Perl Implementations
PWC 309 - Task 1 - PL/Perl Implementation
Essentially, this is the same implementation as in Raku.CREATE OR REPLACE FUNCTION
pwc309.task1_plperl( int[] )
RETURNS int
AS $CODE$
my ( $nums ) = @_;
my $gaps = {};
for my $index ( 1 .. $nums->@* - 1 ) {
$gaps->{ $nums->@[ $index ] - $nums->@[ $index - 1 ] } = $nums->@[ $index ];
}
return $gaps->{ ( sort( keys $gaps->%* ) )[0] };
$CODE$
LANGUAGE plperl;
PWC 309 - Task 2 - PL/Perl Implementation
Same solution as in Raku.CREATE OR REPLACE FUNCTION
pwc309.task2_plperl( int[] )
RETURNS int
AS $CODE$
my ( $nums ) = @_;
my $min;
for my $current ( $nums->@* ) {
my $current_min = (
sort
map { $_ > $current ? $_ - $current : $current - $_ }
grep { $_ != $current }
$nums->@* )[ 0 ];
if ( ! $min || $current_min < $min ) {
$min = $current_min;
}
}
return $min;
$CODE$
LANGUAGE plperl;
I use
sort on the resulting list to get out the min value.
PostgreSQL Implementations
PWC 309 - Task 1 - PL/PgSQL Implementation
A single query with window functions make the deal to find out, for each value, the difference with the next one, hence the min difference.CREATE OR REPLACE FUNCTION
pwc309.task1_plpgsql( nums int[] )
RETURNS int
AS $CODE$
WITH data AS (
SELECT v, v - lag( v, 1, v * -10 ) over () AS diff
FROM unnest( nums ) v
ORDER BY diff ASC
)
SELECT v
FROM data
LIMIT 1;
$CODE$
LANGUAGE sql;
PWC 309 - Task 2 - PL/PgSQL Implementation
In this implementation I iterate over the array elements and compute the differences.CREATE OR REPLACE FUNCTION
pwc309.task2_plpgsql( nums int[] )
RETURNS int
AS $CODE$
DECLARE
diff int;
current_diff int;
l int;
r int;
BEGIN
FOR l IN SELECT v FROM unnest( nums ) v LOOP
for r IN SELECT v FROM unnest( nums ) v LOOP
IF r = l THEN
CONTINUE;
END IF;
current_diff := r - l;
IF current_diff < 0 THEN
current_diff := current_diff * -1;
END IF;
IF diff IS NULL OR current_diff < diff THEN
diff := current_diff;
END IF;
END LOOP;
END LOOP;
RETURN diff;
END
$CODE$
LANGUAGE plpgsql;