This is my 43th week participating into the weekly challenge.
TASK #1 › Trailing Zeroes
Submitted by: Mohammad S Anwar
TASK #1 › Min Sliding Window
Submitted by: Mohammad S Anwar
You are given an array of integers @A and sliding window size $S.
Write a script to create an array of min from each sliding window.
Example
Input: @A = (1, 5, 0, 2, 9, 3, 7, 6, 4, 8) and $S = 3
Output: (0, 0, 0, 2, 3, 3, 4, 4)
[(1 5 0) 2 9 3 7 6 4 8] = Min (0)
[1 (5 0 2) 9 3 7 6 4 8] = Min (0)
[1 5 (0 2 9) 3 7 6 4 8] = Min (0)
[1 5 0 (2 9 3) 7 6 4 8] = Min (2)
[1 5 0 2 (9 3 7) 6 4 8] = Min (3)
[1 5 0 2 9 (3 7 6) 4 8] = Min (3)
[1 5 0 2 9 3 (7 6 4) 8] = Min (4)
[1 5 0 2 9 3 7 (6 4 8)] = Min (4)
I didn’t have time for the challenge this week.
For the first challenge it was just a matter of iterating through the array and and finding the min value. I used Perl’s List::Util and Raku’s native min function to do the dirty work.
Perl 5 solution
#!/usr/bin/perl
# Test: ./ch-1.pl
use Modern::Perl;
use List::Util qw /min/;
my @out;
my @A = (1, 5, 0, 2, 9, 3, 7, 6, 4, 8);
my $S = 3;
for my $i (2 .. scalar(@A) - 1) {
push @out, min $A[$i], $A[$i - 1], $A[$i - 2];
}
say '(' . (join ', ', @out) . ')';
Output: perl ./ch-1.pl
(0, 0, 0, 2, 3, 3, 4, 4)Raku solution
# Test: perl6 ch-1.p6
sub MAIN() {
my @out;
my @A = (1, 5, 0, 2, 9, 3, 7, 6, 4, 8);
my $S = 3;
for (2 .. @A.elems - 1) -> $i {
push @out, min @A[$i], @A[$i - 1], @A[$i - 2];
}
say '(' ~ @out.join(", ") ~ ')';
}Output perl6 ch-1.p6
(0, 0, 0, 2, 3, 3, 4, 4)TASK #2 › Smallest Neighbour
Submitted by: Mohammad S Anwar
You are given an array of integers @A.
Write a script to create an array that represents the smallest element to the left of each corresponding index. If none found then use 0.
Example 1
Input: @A = (7, 8, 3, 12, 10)
Output: (0, 7, 0, 3, 3)
For index 0, the smallest number to the left of $A[0] i.e. 7 is none, so we put 0.
For index 1, the smallest number to the left of $A[1] as compare to 8, in (7) is 7 so we put 7.
For index 2, the smallest number to the left of $A[2] as compare to 3, in (7, 8) is none, so we put 0.
For index 3, the smallest number to the left of $A[3] as compare to 12, in (7, 8, 3) is 3, so we put 3.
For index 4, the smallest number to the left of $A[4] as compare to 10, in (7, 8, 3, 12) is 3, so we put 3 again.
Example 2
Input: @A = (4, 6, 5)
Output: (0, 4, 4)
For index 0, the smallest number to the left of $A[0] is none, so we put 0.
For index 1, the smallest number to the left of $A[1] as compare to 6, in (4) is 4, so we put 4.
For index 2, the smallest number to the left of $A[2] as compare to 5, in (4, 6) is 4, so we put 4 again.
For this task I just iterated through the array and kept track of the smallest neighbor to the left of the array
Perl 5 solution
#!/usr/bin/perl
# Test: ./ch-2.pl
use Modern::Perl;
say smallest_neighbor(7, 8, 3, 12, 10);
say smallest_neighbor(4, 6, 5);
sub smallest_neighbor {
my @A = @_;
my @out;
my $smallest_so_far;
for my $i (0 .. scalar(@A) - 1) {
if ( defined($smallest_so_far) &&
$A[$i] > $smallest_so_far ) {
push @out, $smallest_so_far;
} else {
push @out, 0;
}
$smallest_so_far = $A[$i]
unless (defined($smallest_so_far));
$smallest_so_far = $A[$i]
if ($smallest_so_far > $A[$i]);
}
return '(' . (join ', ', @out) . ')';
}
Output ./ch-2.pl
(0, 7, 0, 3, 3)
(0, 4, 4)Raku solution
# Test: perl6 ch-2.p6
sub MAIN() {
say smallest-neighbor((7, 8, 3, 12, 10));
say smallest-neighbor((4, 6, 5));
}
sub smallest-neighbor(@A) {
my @out;
my $smallest_so_far;
for (0 .. @A.elems - 1) -> $i {
if ( defined($smallest_so_far) &&
@A[$i] > $smallest_so_far ) {
@out.push($smallest_so_far);
} else {
@out.push(0);
}
$smallest_so_far = @A[$i]
unless (defined($smallest_so_far));
$smallest_so_far = @A[$i]
if ($smallest_so_far > @A[$i]);
}
return '(' ~ @out.join(', ') ~ ')';
}
Output perl6 ch-2.p6
(0, 7, 0, 3, 3)
(0, 4, 4)
