The Weekly Challenge: 399

TABLE OF CONTENTS

  1. HEADLINES

  2. SPONSOR

  3. RECAP

  4. PERL REVIEW

  5. RAKU REVIEW

  6. CHART

  7. NEW MEMBERS

  8. GUESTS

  9. TASK #1: Longest Repeating Character

10. TASK #2: Longest Absolute Path

HEADLINES

Welcome to the Week #399 of The Weekly Challenge.

Welcome aboard, Sanyam Brar, and thanks for your first contributions in Perl.

For the second week in a row, we have had more guest contributions than the regular contributions. Ideally, I want both to reach the highest mark. I am sure one day we will get there again.

Below is my contributions to the Week #398.

[Task #1 - Perl]: source code

sub divisible_average(@arr) {
    my @filtered = grep { $_ % 6 == 0 } @arr;
    return 0 unless @filtered;

    my $sum = 0;
    $sum += $_ for @filtered;

    return int($sum / @filtered + 0.5);
}

[Task #2 - Perl]: source code

sub odd_one_out(@words) {
    my @diff_keys = map {
        my @c = map { ord } split //;
        join ',', map { $c[$_ + 1] - $c[$_] } 0 .. $#c - 1;
    } @words;

    my %count;
    $count{$_}++ for @diff_keys;

    my ($unique_key) = grep { $count{$_} == 1 } keys %count;
    my ($idx) = grep { $diff_keys[$_] eq $unique_key } 0 .. $#words;

    return $words[$idx];
}

[Task #1 - Raku]: source code

sub divisible-average(@arr) {
    my @filtered = @arr.grep({ $_ % 6 == 0 });
    return 0 unless @filtered;

    my $sum = 0;
    $sum += $_ for @filtered;

    return (($sum / @filtered) + 0.5).Int;
}

[Task #2 - Raku]: source code

sub odd-one-out(*@words) {
    my @diff_keys = @words.map: {
        my @c = .split('', :skip-empty).map(*.ord);
        join ',', (0 .. @c.end - 1).map({ @c[$_ + 1] - @c[$_] });
    };

    my %count;
    %count{$_}++ for @diff_keys;

    my ($unique_key) = %count.keys.grep({ %count{$_} == 1 });
    my ($idx)        = (0 .. @words.end).grep({ @diff_keys[$_] eq $unique_key });

    return @words[$idx];
}

[Task #1 - Python]: source code

def divisible_average(arr):
    filtered = [x for x in arr if x % 6 == 0]
    if not filtered:
        return 0

    total_sum = sum(filtered)
    return math.floor(total_sum / len(filtered) + 0.5)

[Task #2 - Python]: source code

def odd_one_out(*words):
    diff_keys = []
    for word in words:
        c = [ord(ch) for ch in word]
        diffs = [c[i + 1] - c[i] for i in range(len(c) - 1)]
        diff_keys.append(",".join(map(str, diffs)))

    count = {}
    for key in diff_keys:
        count[key] = count.get(key, 0) + 1

    unique_key = next(k for k, v in count.items() if v == 1)
    idx = diff_keys.index(unique_key)

    return words[idx]

Thank you Team PWC, once again.

Happy Hacking!!

Last 5 weeks mainstream contribution stats. Thank you Team PWC for your support and encouragements.

Last 5 weeks guest contribution stats. Thank you each and every guest contributors for your time and efforts.

TOP 10 Guest Languages

Do you see your favourite language in the Top #10? If not then why not contribute regularly and make it to the top.

 1. Python     (4720)
 2. Rust       (1259)
 3. C          (1086)
 4. Haskell    (977)
 5. Ruby       (965)
 6. Lua        (947)
 7. C++        (766)
 8. Go         (740)
 9. JavaScript (664)
10. Java       (534)

Blogs with Creative Title

1. Divisible Difference by Arne Sommer.

2. Six of one, half-dozen of the other by Bob Lied.

3. Average Strings by Jorg Sommrey.

4. Sixes to Average and Strings to Buckets by Matthias Muth.

5. It makes no difference if it’s divisible or not… by Packy Anderson.

6. Divisible and difference by Peter Campbell Smith.

7. Difference Indivisible by Roger Bell_West.

8. A different average by Simon Green.

GitHub Repository Stats

1. Commits: 52,036 (+110)

2. Pull Requests: 14,937 (+34)

3. Contributors: 282 (+1)

4. Fork: 353 (+1)

5. Stars: 221

SPONSOR

With start of Week #355, we have a new sponsor Marc Perry until the end of year 2026. Having said we are looking for more sponsors so that we can go back to weekly winner. If anyone interested please get in touch with us at perlweeklychallenge@yahoo.com. Thanks for your support in advance. You can find more informations here.

RECAP

Quick recap of The Weekly Challenge - 398 by Mohammad Sajid Anwar.

PERL REVIEW

If you missed any past reviews then please check out the collection.

RAKU REVIEW

If you missed any past reviews then please check out the collection.

CHART

Please take a look at the charts showing interesting data.

I would like to THANK every member of the team for their valuable suggestions. Please do share your experience with us.

NEW MEMBERS

Sanyam Brar, an expert hacker in Perl, joined Team PWC.

Please find out How to contribute?, if you have any doubts.

Please try the excellent tool EZPWC created by respected member Saif Ahmed of Team PWC.

GUESTS

Please check out the guest contributions for the Week #398.

Please find past solutions by respected guests. Please share your creative solutions in other languages.

Task 1: Longest Repeating Character

Submitted by: Mohammad Sajid Anwar

You are given a string str and an integer k > 0.

Write a script to choose any character of str and change it to another character. You may perform this operation up to k times. Return the length of the longest reachable substring of a single letter.

Example 1

Input: $str = "AAAAA", $k = 2
Output: 5

Every character in the string is already the same letter "A".
The maximum operations allowed (k = 2) do not even need to be used.
The entire string of length 5 is already the longest valid substring.

Example 2

Input: $str = "ABCDE", $k = 4
Output: 5

All 5 characters in the string are distinct.
The most frequent character occurs only 1 time.
To turn the entire string into a single repeated character, we need to replace 5 - 1 = 4 characters.
Since k = 4, we can replace "B","C","D", and "E" with "A" to form "AAAAA".

Example 3

Input: $str = "AAABBBBAA", $k = 1
Output: 5

If we target "A", we could change the "B" in "AAAB" (k=1 replacement) to get length 4 ("AAAA").
If we target "B", we can pick the substring "BBBBA" and change the single trailing "A" to "B" (k=1 replacement) to form "BBBBB", which gives a length of 5.
Targeting "B" gives better result than targeting "A".

Example 4

Input: $str = "ABACADA", $k = 2
Output: 5

Look at the substring [0..4] which is "ABACA" (length 5).
In this window, "A" appears 3 times, and there are 2 non-"A" characters ("B" and "C").
With k = 2, we can replace "B" and "C" with "A" to form "AAAAA".

Look at the substring [2..6] which is "ACADA" (length 5).
Here, "A" appears 3 times, and non-"A" characters ("C" and "D") count to 2.
Replacing them gives "AAAAA".

In both cases, the maximum valid length is 5.

Example 5

Input: $str = "BAAABAAAA", $k = 1
Output: 8

The string has two groups of "A"s separated by a single "B" ("AAABAAAA").

We can replace the character 4 "B" with an "A" to get a sequence of 8 "A".

Task 2: Longest Absolute Path

Submitted by: Mohammad Sajid Anwar

You are given a filesystem as string.

Write a script to return the length of the longest absolute path to a file in the abstracted file system. If there is no file in the system, return 0.

dir
- subdir1
  - file1.ext
  - subsubdir1
- subdir2
  - subsubdir2
    - file2.ext

The above filesystem can be represented as below:

"dir\n\tsubdir1\n\t\tfile1.ext\n\t\tsubsubdir1\n\tsubdir2\n\t\tsubsubdir2\n\t\t\tfile2.ext"

Example 1

Input: $fs = "dir\n\tsubdir1\n\t\tsubsubdir1\n\tsubdir2"
Output: 0

There are no files in the system (a file must contain a . character separating a name and extension).
Since there are no files, the answer is 0.

Example 2

Input: $fs = "file1.txt\nfile2.ext\na.png"
Output: 9

All entries are located at root depth (0 tabs).
Found File Paths:
    "file1.txt" (length = 9)
    "file2.ext" (length = 9)
    "a.png"     (length = 5)
The longest absolute file path length is 9.

Example 3

Input: $fs = "my documents\n\tproject v1.0\n\t\tnotes v1.txt\n\timage.png"
Output: 38

Notice project v1.0 is a directory (it has children beneath it), while notes v1.txt and image.png are files.
Found File Paths:
    "my documents/project v1.0/notes v1.txt" (length = 38)
    "my documents/image.png" (length = 22)
The maximum absolute length is 38.

Example 4

Input: $fs = "root\n\ta\n\t\tb\n\t\t\tc.txt\n\td\n\t\te.txt"
Output: 14

Found File Paths:
    "root/a/b/c.txt" => 4 + 1 + 1 + 1 + 1 + 1 + 5 = 14
    "root/d/e.txt"   => 4 + 1 + 1 + 1 + 5 = 12
The longest file path is "root/a/b/c.txt" with length 14.

Example 5

Input: $fs = "a\n\tb\n\t\tc\n\t\t\td\n\t\t\t\te.txt\n\tvery_long_directory_name\n\t\tf.txt"
Output: 32

Found File Paths:
    "a/b/c/d/e.txt" => 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 5 = 13
    "a/very_long_directory_name/f.txt" => 1 + 1 + 24 + 1 + 5 = 32
The long directory name compensates for depth, giving a max length of 32.

By submitting a response to the challenge you agree that your name or pseudonym, any photograph you supply and any other personal information contained in your submission may be published on this website and the associated mobile app. Last date to submit the solution 23:59 (UK Time) Sunday 15th November 2026.