TL;DR

On with TASK #2 from The Weekly Challenge #194. Enjoy!

# The challenge

You are given a string made of alphabetic characters only, a-z.

Write a script to determine whether removing only one character can make the frequency of the remaining characters the same.

Example 1:

Input: $s = 'abbc' Output: 1 since removing one alphabet 'b' will give us 'abc' where each alphabet frequency is the same.  Example 2: Input:$s = 'xyzyyxz'
Output: 1 since removing 'y' will give us 'xzyyxz'.


Example 3:

Input: $s = 'xzxz' Output: 0 since removing any one alphabet would not give us string with same frequency alphabet.  # The questions I guess that frequency is the amount of times that each character appears in the string. # The solution This is an interesting challenge, because it fooled me into thinking it’s utterly simple in the beginning. I mean, it’s simple, but not that simple. The requested condition is only true if all characters have the same exact number of occurrences, except exactly one single character which occurs one time more than the others. All in all, it’s a matter of counting things. So we’re doing things in two passes: • in the first pass, we count the occurrences of each character; • in the second pass, we count how many times each occurrence occurs. At the end of the second pass, our true condition is that: • there are only two different values for occurrences; • the two occurrences value differ by 1; • the higher value is counted only once (corresponding to one single character occurring that amount of times) OK, this is probably best expressed in Perl: #!/usr/bin/env perl use v5.24; use warnings; use experimental 'signatures'; no warnings 'experimental::signatures'; @ARGV = qw< abbc xyzyyxz xzxz > unless @ARGV; say frequency_equalizer($_) . " -> $_" for @ARGV; sub frequency_equalizer ($string) {
my (%first_counter, %second_counter);
++$first_counter{substr($string, $_, 1)} for 0 .. length($string) - 1;
++$second_counter{$_} for values %first_counter;
return 0 if scalar(keys %second_counter) != 2;
my ($k1,$v1, $k2,$v2) = %second_counter;
($k1,$v1, $k2,$v2) = ($k2,$v2, $k1,$v1) if $k1 >$k2;
return 1 if $v2 == 1 &&$k2 - $k1 == 1; return 0; }  I mean, it’s better than pseudocode. I cheated a bit and did the minimum amount of tweaks to adapt it to Raku: #!/usr/bin/env raku use v6; sub MAIN (*@ARGV) { @ARGV = < abbc xyzyyxz xzxz > unless @ARGV; put "{frequency-equalizer($_)} -> $_" for @ARGV; } sub frequency-equalizer ($string) {
my (%first_counter, %second_counter);
++%first_counter{$string.substr($_, 1)} for ^$string.chars; ++%second_counter{$_} for %first_counter.values;
return 0 if %second_counter.elems != 2;
my ($k1,$v1, $k2,$v2) = %second_counter.kv;
($k1,$v1, $k2,$v2) = $k2,$v2, $k1,$v1 if $k1 >$k2;
return 1 if $v2 == 1 &&$k2 - \$k1 == 1;
return 0;
}


I guess this is it, STRIKE THE EARTH!