# Freq Out, Man!

What is the frequency, Kenneth?

#### THE WEEKLY CHALLENGE – PERL & RAKU #194 Task 2

“It’s like, how much more black could this be? And the answer is none — none more black.”

— Nigel Tufnel

## Frequency Equalizer

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.
``````

#### ANALYSIS

Let’s start this analysis with a definition of terms. What, then, is being asked of us?

Well as stated, we are looking for a frequency distribution of letters that can be corrected by the removal of one element to make a set with equal distribution.

So we need a set of letters all occurring with a count of x, and one letter with a count of y, where y is not equal to x. So two values for frequencies, one of which only has one occurrence.

Or something like that. In fact, there seem to be several ways we could produce this result. Let’s backtrack a bit and think this through.

#### METHOD

In every case it seems we will need to make a frequency distibution of the letters, counting the occurences of the values. From there there are a number of ways to proceed, depending on the results. Our goal is to equalize the system so that there is only one frequency remaining.

If there are only two frequencies, and if one of those is for a single letter, we’re in. Removal of that letter will eliminate the category, which brings the total count to one and a flat distribution.

But wait! There is another possibility. This is when there are only two frequencies and this incidence of one is one more than the other, say two of one value and three of another. In this case removing one of the letters that there are three of will equilise the distribution at the lower value. So we’ll need to take that possibilty into account as well.

A third more exotic case is when there is only one category to start, containing more than one member, and that frequency is 1. In this situation the distribution is already equalized among a set of unique elements, and removal of any single letter changes the total count of the category but does not affect the overall frequency — that there is only one of each member within the multiset. There is still only one category: items with a frequency of 1. One might consider this an edge-case variant of the first scheme outlined above.

The complement to this is when there is only one frequency and all the letters are the same. In that case removing one letter won’t throw the system out of equilibrium. Is that then the last edge case?

I suppose there is one more pathological corner-case that should be considered, where there is only one character in the string. There will only be one frequency, which is promising, and if the single element is removed there will only be one frequency remaining. This is the same as the scenario above, however the frequency will be zero.

Which brings us back to our quote for today:

I will go out on a limb here and declare a frequency of zero does not satisfy the conditions, sidestepping by decree any debate as to whether a frequency of zero is in fact a frequency, or best considered a conspicuous void, notable in its absence.

Zip.

Nil.

None more black.

##### PERL 5 SOLUTION
``````use warnings;
use strict;
use utf8;
use feature ":5.26";
use feature qw(signatures);
no warnings 'experimental::signatures';

use constant VERBOSE => 1;

sub free_eq ( \$str ) {
{VERBOSE and say "\nstr  : \$str";}

## make frequency bag
my %freq;
\$freq{ substr \$str, \$_, 1 }++ for (0..length(\$str)-1);
{VERBOSE and say "freq : char { \$_ } freq \$freq{\$_}" for keys %freq;}

## make bag of occurrences of individual values in frequency bag
my %f_incidence;
\$f_incidence{ \$_ }++ for values %freq;
{VERBOSE and
say "f_incidence: freq { \$_ } => \$f_incidence{\$_} times"
for keys %f_incidence;}

my @counts  = sort {\$a<=>\$b} keys %f_incidence;
my @letters = keys %freq;

## CASE 1: single frequency only
if (@counts == 1) {

## all the letters are different
return 1 if \$counts == 1 and @letters > 1;

## all the letters are the same (but not only one letter)
return 1 if \$counts > 1  and \$counts == length( \$str );
}

## CASE 2: two frequencies
if (@counts == 2) {

## if at least one of the two frequency classes has only one member
## it can be removed
for (@letters) { return 1 if \$freq{\$_} == 1 }

## if one frequency incidence is one greater than the other and has
## exactly one more element in it
return 1 if  \$counts + 1           == \$counts
and \$f_incidence{\$counts} == 1;
}

return 0;

}

``````

The Perl Weekly Challenge, that idyllic glade wherein we stumble upon the holes for these sweet descents, is now known as

The Weekly Challenge – Perl and Raku

It is the creation of the lovely Mohammad Sajid Anwar and a veritable swarm of contributors from all over the world, who gather, as might be expected, weekly online to solve puzzles. Everyone is encouraged to visit, learn and contribute at

https://theweeklychallenge.org