This is my response to The Weekly Challenge #380.
Input: $str = "banana"
Output: 5
Vowel: "a" appears 3 times.
Consonant: "n" appears 2 times, "b" appears 1 time.
Max frequency of vowel: 3
Max frequency of consonant: 2
Input: $str = "teestett"
Output: 7
Vowel: "e" appears 3 times.
Consonant: "t" appears 4 times, "s" appears 1 time.
Max frequency of vowel: 3
Max frequency of consonant: 4
Input: $str = "aeiouuaa"
Output: 3
Vowel: "a" appears 3 times, "u" 2 times, "e", "i", "o" 1 time each.
Consonant: None.
Max frequency of vowel: 3
Max frequency of consonant: 0
Input: $str = "rhythm"
Output: 2
Vowel: None
Consonant: "h" appears 2 times, "r", "y", "t", "m" 1 time each.
Max frequency of vowel: 0
Max frequency of consonant: 2
Input: $str = "x"
Output: 1
Vowel: None
Consonant: "x" appears 1 time.
Max frequency of vowel: 0
Max frequency of consonant: 1
The letter y is an English vowel, some of the time.
See e.g.
The Truth About 'Y': It's Mostly a Vowel for more info. The fourth example
in our challenge, rythm, (wrongly) treats it as consonant when it really should
not, so I'll do the same.
#! /usr/bin/env raku
unit sub MAIN ($str where $str ~~ /^ <[a .. z]>+ $/,
:v(:$verbose));
my $bag = $str.comb.Bag;
my @vowels = $bag<a e i o u>;
my @consonants = $bag<b c d f g h j k l m n p q r s t v w x y z>;
my $max-vowel = @vowels.max;
my $max-consonants = @consonants.max;
if $verbose
{
say ": Vovel Count: @vowels[] -> $max-vowel";
say ": Consonant Count: @vowels[] -> $max-consonants";
}
say $max-vowel + $max-consonants;[3] Ensure English lowercase letters only.
[6] Count the occurene of each letter in the word, as a Bag.
See docs.raku.org/routine/Bag for more information about Bag.
[8] A list with the occurence of the given vowels. Note that the vowels are not returned, just a list of integers.
[9] Ditto for the consonants. (We could have started with a list of all the letters
(a..z) and removed the vowels, to avoid the long list of consonants.)
[11] The highest vowel frequency.
[12] Ditto for the consonants.
[20] Add the numbers from [11] and [12] and print the result.
Running it:
$ ./sum-of-frequencies banana
5
$ ./sum-of-frequencies teestett
7
$ ./sum-of-frequencies aeiouuaa
3
$ ./sum-of-frequencies rhythm
2
$ ./sum-of-frequencies x
1
Looking good.
With verbose mode:
$ ./sum-of-frequencies -v banana
: Vovel Count: 3 0 0 0 0 -> 3
: Consonant Count: 3 0 0 0 0 -> 2
5
$ ./sum-of-frequencies -v teestett
: Vovel Count: 0 3 0 0 0 -> 3
: Consonant Count: 0 3 0 0 0 -> 4
7
$ ./sum-of-frequencies -v aeiouuaa
: Vovel Count: 3 1 1 1 2 -> 3
: Consonant Count: 3 1 1 1 2 -> 0
3
$ ./sum-of-frequencies -v rhythm
: Vovel Count: 0 0 0 0 0 -> 0
: Consonant Count: 0 0 0 0 0 -> 2
2
$ ./sum-of-frequencies -v x
: Vovel Count: 0 0 0 0 0 -> 0
: Consonant Count: 0 0 0 0 0 -> 1
1
Input: $str = "z"
Output: 1
Reverse alphabet value of "z" is 1.
Position 1: 1 x 1
Sum of product: 1
Input: $str = "a"
Output: 26
Reverse alphabet value of "a" is 26.
Position 1: 1 x 26
Sum of product: 26
Input: $str = "bbc"
Output: 147
Reverse alphabet value of "b" is 25 and "c" is 24.
Position 1: 1 x 25
Position 2: 2 x 25
Position 3: 3 x 24
Sum of product: 25 + 50 + 72 => 147
Input: $str = "racecar"
Output: 560
Reverse alphabet value of "r" is 9, "a" is 26, "c" is 24 and "e" is 24.
Position 1: 1 x 9
Position 2: 2 x 26
Position 3: 3 x 24
Position 4: 4 x 22
Position 5: 5 x 24
Position 6: 6 x 26
Position 7: 7 x 9
Sum of product: 9 + 52 + 72 + 88 + 120 + 156 + 63
Input: $str = "zyx"
Output: 14
Reverse alphabet value of "z" is 1, "y" is 2 and "x" is 3.
Position 1: 1 x 1
Position 2: 2 x 2
Position 3: 3 x 3
Sum of product: 1 + 4 + 9
#! /usr/bin/env raku
unit sub MAIN ($str where $str ~~ /^ <[a .. z]>+ $/,
:v(:$verbose));
my %h = ('z' ... 'a').map({ state $i = 1; $_ => $i++});
my $total = 0;
for 1 .. $str.chars -> $i
{
my $letter = $str.substr($i -1, 1);
my $sum = $i * %h{$letter};
$total += $sum;
say ": Pos:$i Char:'$letter' Val:%h{$letter} Sum:$sum"
if $verbose;
}
say $total;[3] The challenge specifies only lowercase English letters in the input.
[6] Construct a hash with mappings from letter to value,
starting with z => 1 and ending with a => 26. Note the
state variable used as a counter for the right hand side.
See docs.raku.org/syntax/state for more information about the variable declarator state.
[8] The result will end up here.
[10] Iterate over the positions of the letters in the string, starting at 1.
[12] Get the character at that position. (Note the -1 as
indices are zero based.)
[13] Get the sum for this letter, the position ($i) multiplied
with the occurence (%h{$letter}).
[15] Add that sum to the result.
[21] Print the final result.
Running it:
$ ./reverse-degree z
1
$ ./reverse-degree a
26
$ ./reverse-degree bbc
147
$ ./reverse-degree racecar
560
$ ./reverse-degree zyx
14
Looking good.
With verbose mode:
$ ./reverse-degree -v z
: Pos:1 Char:'z' Val:1 Sum:1
1
$ ./reverse-degree -v a
: Pos:1 Char:'a' Val:26 Sum:26
26
$ ./reverse-degree -v bbc
: Pos:1 Char:'b' Val:25 Sum:25
: Pos:2 Char:'b' Val:25 Sum:50
: Pos:3 Char:'c' Val:24 Sum:72
147
$ ./reverse-degree -v racecar
: Pos:1 Char:'r' Val:9 Sum:9
: Pos:2 Char:'a' Val:26 Sum:52
: Pos:3 Char:'c' Val:24 Sum:72
: Pos:4 Char:'e' Val:22 Sum:88
: Pos:5 Char:'c' Val:24 Sum:120
: Pos:6 Char:'a' Val:26 Sum:156
: Pos:7 Char:'r' Val:9 Sum:63
560
$ ./reverse-degree -v zyx
: Pos:1 Char:'z' Val:1 Sum:1
: Pos:2 Char:'y' Val:2 Sum:4
: Pos:3 Char:'x' Val:3 Sum:9
14
And that's it.