oldtechaa

Hi everybody, we've got another two challenges this week, so let's dive into them!

Remove One

The goal is to see if there's any one number that can be removed to make the set sorted in increasing order. Here's the code:

#!/usr/bin/perl
use v5.36;

my $success = 0;
REMOVAL: for my $removal (0 .. $#ARGV) {
    my @modified = @ARGV;
    splice(@modified, $removal, 1);
    for my $scan (1 .. $#modified) {
        if($modified[$scan] <= $modified[$scan - 1]) {
            next REMOVAL;
        }
    }
    $success = 1;
    last;
}

say ($success ? 'true' : 'false');

We have a labelled outer loop for the numbers we choose to remove. $removal is set to the index of each number we attempt to remove, then we copy the array, remove that number, and scan the result to make sure they all are increasing. If they don't, we skip this number and move on.

If we succeed, we set our flag and exit the loops and print the result.

Duplicate Ones

The next one is an array shifting challenge. We want to insert a duplicate of each 0, shifting everything to the right, and popping off the list to keep the length the same.

Here's the code:

#!/usr/bin/perl
use v5.36;

my @ints = @ARGV;
for(my $i = 0; $i <= $#ints; $i++) {
    if($ints[$i] == 0) {
        splice(@ints, $i, 0, 0);
        pop(@ints);
        $i++;
    }
}

say('(', join(', ', @ints), ')');

This one's also really quite simple. We scan the array, use splice to insert a 0, pop the last number off the end of the array, and skip over the 0 we just inserted. It's that simple!

Both of this week's solutions make use of splice() to insert and remove array elements, something I haven't used a lot before.

Stay tuned for next week's challenge, which should come out Monday!

Hi everybody! Back this week with a (surprisingly long) solution to just Task 1 of the weekly challenge. Task 2 makes no sense to me at all because it seems like examples 1 and 3 disagree with each other. Just sticking to one challenge for that reason. Anyways, let's dive into it!

The goal here is to find the letters that all the provided words share. Here's the code:

#!/usr/bin/perl
use v5.36;

my @words;
my %result_chars;

for(@ARGV) {
    push(@words, [split(//, $_)])
}
@words = sort {$#{$a} <=> $#{$b}} @words;
$result_chars{$_}++ for @{$words[0]};

for my $word (1..$#words) {
    for my $key (keys(%result_chars)) {
        my $occurrences = grep(/$key/, @{$words[$word]});
        if($occurrences == 0) {
            delete($result_chars{$key});
            next;
        } elsif($occurrences <= $result_chars{$key}) {
            $result_chars{$key} = $occurrences;
        }
    }
}

for my $char (@{$words[0]}) {
    if($result_chars{$char}) {
        $result_chars{$char}--;
        say $char;
    }
}

First we make a 2D array of the characters in all the words. That way we only have to split the words up once, instead of repeatedly as we seek through them. It does mean a bit more complexity to deal with a matrix, unfortunately.

We also sort the words by length so the shortest one is first, then make a histogram of all the letters in it. Now it's important to keep all the letters and not remove duplicates, because we have to print duplicates as we see from the example that prints "e, l, l".

We loop through each word, then loop through each letter in the first word (keys of the histogram) and search the current word for that letter. If we don't find it, we delete the letter from the histogram. If we find fewer occurrences than in the histogram, we remove some from the histogram to show how many we actually can make in the current word. If that letter is in the histogram fewer or equal times to the occurrences in the word, we move on to the next letter.

Next, for printing, we would have to have multiple loops to loop through the histogram and remove one instance at a time, so instead I decided I should simply search the original word for characters that successfully passed the test of the other words, then print those characters.

When I first started, I thought this would be super easy, but I discovered complications numerous times through the challenge. Perhaps others will have some better ideas of solutions I can learn from. Unfortunately Flavio Poletti hasn't been doing his solutions recently, I always enjoyed them very much, but be sure to check out past solutions of his at github.polettix.it.

Hopefully I'll be back next week with more solutions!

Hello everybody! For this week's weekly challenge I thought the challenges looked really easy, but they both had a couple slight complicating factors. Also, this was the first time I've used sub signatures.

Similar Words

For this one, we're looking for words that share all characters. We print out each pair of countries.

#!/usr/bin/perl
use v5.36;

my @words = @ARGV;
my $matched = 0;
for (my $i = 0; $i <= $#ARGV - 1; $i++) {
    my $start_word = $words[$i];
    my %start_chars = map {$_ => 1} split(//, $start_word);

    for (my $j = $i + 1; $j <= $#ARGV; $j++) {
        my $match_word = $words[$j];
        my %match_chars = map {$_ => 1} split(//, $match_word);

        if (hashes_equal(\%start_chars, \%match_chars)) {
            say $start_word . ", " . $match_word;
            $matched = 1;
        }
    }
}
say 0 unless $matched;

sub hashes_equal ($start_ref, $match_ref) {
    my %start_chars = %{$start_ref};
    my %match_chars = %{$match_ref};

    if (scalar keys %start_chars == scalar keys %match_chars) {
        foreach (keys %start_chars) {
            if (!defined($match_chars{$_})) {
                return;
            }
        }
        return 1;
    } else {
        return;
    }
}

It essentially boils down to looping through all combinations of words in two loops, converting each word to a hash containing all unique characters. hashes_equal makes sure that the hashes have the same number of keys, then tries the keys and makes sure they have the same values.

Frequency Sort

For this one, we're sorting numbers by how frequently they occur, in increasing order, except when they share a frequency, then we sort them decreasing by value.

Here's the code:

#!/usr/bin/perl
use v5.36;

my %ints;
$ints{$_}++ foreach @ARGV;

my %ints_by_occurrence;
my @results;

foreach (keys %ints) {
    push @{$ints_by_occurrence{$ints{$_}}}, $_;
}
foreach (sort keys %ints_by_occurrence) {
    my $frequency = $_;
    foreach (sort {$b <=> $a} @{$ints_by_occurrence{$frequency}}) {
        my $number = $_;
        for (1..$frequency) {
            push @results, $number;
        }
    }
}
say $_ foreach @results;

This time we're making a hash of arrays, where %intsbyoccurrence uses frequencies as the key, and an array of numbers as the value. %ints contains the initial histogram which is reversed into %intsbyoccurrence. We sort once by frequency, then we sort each array of a given frequency by value, which is pushed onto the results array in the proper order.

Those are my solutions to this week's challenge! Hopefully I'll have more for both challenges next week. See you then.

Hi everybody! In this week's weekly challenge, we're searching for anything but the minimum or maximum in a dataset, and searching for senior citizens on a plane.

Min And Max

This challenge is a very interesting one, because obviously the easiest solution in terms of development is to sort and filter the first and last element. However, that is O(n log n) and it's very little added complexity to do the O(n) solution with a single-pass filter.

my %hist;
$hist{$_} = 1 for @ARGV;

say "You didn't provide two or more arguments." and exit if scalar keys %hist < 2;
say "-1" and exit if scalar keys %hist == 2;

my ($max, $min);
for (keys %hist) {
    $max = $_ if (!defined($max) || $_ > $max);
    $min = $_ if (!defined($min) || $_ < $min);
}
for (keys %hist) {
    say $_ if $_ != $max and $_ != $min;
}

First we make a histogram hash where duplicates are filtered out. Using the histogram, we exit with a failure or -1 if there are 1 or 2 unique numbers left. As we scan the histogram keys we keep track of the minimum/maximum values, then one more pass to print all of them out.

Senior Citizens

Contrary to the typical pattern of the second challenge being harder, this one's incredibly simple. We're looking for any senior citizens in a dataset made up of "9999999999A1122”, where 9 denotes the phone number, A the sex, 1 the age and 2 the seat number.

The simplest way I can think of is substr():

my $count;
for (@ARGV) {$count++ if substr($_, 11, 2) >= 60}
say $count // 0;

That's it! We just count up every time substr() comes up with someone 60 or over. Then we print out the result.

That's it for this week! Hopefully I'll be back with another blog post next week.

Hi everybody! I'm finally back with another PWC/TWC blog post for week 230.

Separate Digits

For the first challenge we want to split all the numbers in the array into single digits. Here's the code:

use v5.36;
my @nums;
push(@nums, split(//, $_)) for @ARGV;
say $_ for @nums;

It very simply splits anything in its arguments into individual characters and pushes them onto a new array.

Count Words

Our second challenge asks us to count the words that start with the given prefix. Here's a 4-liner (minus boilerplate) to help us out with this one:

use v5.36;
my $pattern = shift;
my $count;
for (@ARGV) {$count++ if $_ =~ /^$pattern/}
say $count // 0;

We take the pattern and then add to the count if our regex prefix matches the start of any of the strings in the arguments.

That's all for this week, two nice easy challenges! Hope I'll be able to post again next week maybe.