Another week blogging a Perl Weekly Challenge!

TASK #1 › Count Number

Submitted by: Mohammad S Anwar
You are given a positive number $N.

Write a script to count number and display as you read it.

Given the 80s Yes album 90125, we’d go “one zero, one one, one two, one five, one nine”, or 1011121519.

Which can get ambiguous if you start counting large numbers, because if there were 101112 1s and 51 9s, it would give the same output. But eh.

I didn’t do cleanup, because even if this was a negative number, we don’t have to worry about anything \D, so I just go through 0..9, get the number of instances and add to the output.

Notice how the examples don’t start out with 00 to indicate a lack of zeroes? if or grep to the rescue! I wrote a solution that used for and if and another that uses map and grep.

But we want to count the number of times a character is used, and splitting the string into an array is wasteful of memory. So, how do you get the count of the times a character is used? We’ll switch prog rock bands and go to Rush’s classic 2112. A first stop is $x = $N =~ /2/g, but that’ll just say “I found it!”, or 1 in boolean parlance. We’ll grab the 2 with $x = $N =~ /(2)/, but that’s just proving it was a 2, which we’re already sure of. We want every 2, so we run $x = $N =~ /(2)/g and get 22, but we want 2 meaning we have two 2s.

The solution is as follows $x = () = $N =~ /(2)/, which forces a list context and gives us the size of the list. =()= is sometimes called the goatse operator, and if you don’t know why, don’t look it up.

Let’s See The Code

#!/usr/bin/env perl

use strict;
use warnings;
use feature qw{ say signatures state };
no warnings qw{ experimental };

my @examples = qw{1122234 2333445 12345 90125 };

for my $example (@examples) {
    my $output = count_number($example);
    my $o2     = count_number2($example);
    say '  input: ' . $example;
    say ' output: ' . $output;
    say 'output2: ' . $o2;
    say '';

sub count_number( $input ) {
    my $output = '';
    for my $i ( 0 .. 9 ) {
        my $x = () = $input =~ /($i)/gmx;
        $output .= $x . $i if $x;
    return $output;

sub count_number2( $input ) {
    return join '', map {
        my $x = () = $input =~ /($_)/g;
        $x . $_; }
        grep { $input =~ /$_/ } 0 .. 9;

TASK #2 › Jump Game

Submitted by: Mohammad S Anwar
You are given an array of positive numbers @N, where value at each index determines how far you are allowed to jump further.

Write a script to decide if you can jump to the last index. Print 1 if you are able to reach the last index otherwise 0.

We’ll run through the two examples, first with 1, 2, 1, 2:

  • index is 0 and $n[0] is 1, so we add 1 to the index
  • index is 1 and $n[1] is 1, so we add 2 to the index
  • index is 3 and that’s the last index — 0,1,2,3 — and we have a win!

In contrast, with 2, 1, 1, 0, 2:

  • index is 0 and $n[0] is 2, so we add 2 to the index
  • index is 2 and $n[2] is 1, so we add 1 to the index
  • index is 3 and $n[3] is 0, so we add 0 to the index
  • index is 3 and $n[3] is 0, so we add 0 to the index
  • I think we’ve seen this one before…

We’ve hit one failure point: we can’t move the index to progress. (It says array of positive numbers, and zero is neither positive nor negative, but I’ll work with it.) The easier failure point is when we’ve gone past the end of the array. Going negative would introduce $n[-1], which is simply the end of the index, but we’re sure that array of positive numbers excludes negative numbers, so that won’t happen. We’d have to add a case to handle that if so, but otherwise, it’s simply

  • if this is the last index, win
  • if this is beyond the last index, lose
  • if the current value is zero, lose
  • if this is below the last index, add the current value to the current index and go again

(My first pass had code to extract the last index to test against, but I decided to rewrite without that value. It may be slightly clearer, but I like it better this way.)

Let’s See The Code

sub jump_game( @n ) {
    say join ' ', @n;
    my $i = 0;
    while (1) {
        if ( !defined $n[$i] )                         { last }
        if ( $n[$i] == 0 )                             { last }
        if ( defined $n[$i] && !defined $n[ $i + 1 ] ) { return 1 }
        $i += $n[$i];
    return 0;
1 2 1 2
2 1 1 0 2
1 9 9 2

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