Daily Challenge #110 - Love VS. Friendship

#challenge

dev.to staff Nov 7, 2019 ・1 min read

If　a = 1, b = 2, c = 3 ... z = 26

Then l + o + v + e = 54

and f + r + i + e + n + d + s + h + i + p = 108

So friendship is twice as strong as love :-)

Write a function to find the "strength" of each of these values.

Test cases:
wordsToMarks(attitude) wordsToMarks(friends) wordsToMarks(family) wordsToMarks(selflessness) wordsToMarks(knowledge)

Good luck!

This challenge comes from J or nor J on CodeWars. Thank you to CodeWars, who has licensed redistribution of this challenge under the 2-Clause BSD License!

Want to propose a challenge idea for a future post? Email yo+challenge@dev.to with your suggestions!

Discussion

La blatte • Nov 7 '19

A bit like @nicolas Hervé, using the reduce method:

f=a=>[...a].reduce((t,l)=>t+l.charCodeAt``-96,0)

[...a] transforms the string to an array with every character (["l", "o", "v", "e"])
The reduce methods will loop through every character, get its charCode, subtract 96 to it (the charCode associated with "a" + 1), and sum everything

kespri • Nov 7 '19

Thanks to @ynndvn for showing me reduce method, it exist too in Swift language and it's awesome !!!

This is my solution in Swift :

func wordsToMarks(word: String) -> Int {
    return word.compactMap { $0 }.reduce(0) {
       $0 + Int($1.asciiValue ?? 0) - 96
    }
}

wordsToMarks(word:"attitude")
wordsToMarks(word:"friends")
wordsToMarks(word:"family")
wordsToMarks(word:"selflessness")
wordsToMarks(word:"knowledge")

Michael Kohl • Nov 8 '19

Why the compactMap, you are mapping each letter to itself, works fine without that too.

Also when I tried your solution with the input string "love🐈" it returned -42 since the way you handle the Optional results in non-ASCII characters having a value of -96. The following version simply ignores such characters instead:

func wordsToMarks(word: String) -> Int {
    return word.reduce(0) {
       $0 + Int($1.asciiValue.map { $0 - 96 } ?? 0)
    }
}

kespri • Nov 8 '19

Oh nice! That's better !

I forgot String are a table of Character so you right compactMap is useless.
I have totally ignored emojis because the challenger treat only letter a to z

So thank you for your feedback, your optimisation is truly better 👌

Michael Kohl • Nov 8 '19

I have totally ignored emojis because the challenger treat only letter a to z

True, but you did add some code to deal with the case were a character may not have an asciiValue, I just offered an alternative to that :-)

Donald Feury • Nov 7 '19

I started picking up ruby for rails, so here is my shining gem.

Try it out here

word.rb

class Word
  @@alphabet = ('a'..'z').to_a

  def self.to_mark(word)
    word.chars.sum { |char| @@alphabet.index(char.downcase) + 1 }
  end
end

word_test.rb

require 'minitest/autorun'
require_relative 'word'

class TestWord < Minitest::Test
  def setup
    @cases = [
      { input: 'love', expected: 54 },
      { input: 'friendship', expected: 108 },
      { input: 'attitude', expected: 100 },
      { input: 'friends', expected: 75 },
      { input: 'family', expected: 66 },
      { input: 'selflessness', expected: 154 },
      { input: 'knowledge', expected: 96 }
    ]
  end

  def test_to_mark
    @cases.each { |test| assert_equal test[:expected], Word.to_mark(test[:input]) }
  end
end

Michael Kohl • Nov 8 '19

Just a friendly tip: alphabet really shouldn't be a class variable @@, they are very seldom what you want in Ruby:

railstips.org/blog/archives/2006/1...

In this specific case a constant would be the most common:

ALPHABET = ('a'..'z').to_a

Donald Feury • Nov 8 '19

Ah okay gotcha, thank ya for the tip!

Yechiel Kalmenson • Nov 7 '19

def wordsToMarks(word)
  alphabet = ("a".."z").to_a
  mark = 0
  word.each_char do |char|
    mark += alphabet.index(char.downcase) + 1
  end
  mark
end

Shannon Crabill • Nov 8 '19

("a".."z").to_a is nifty!

Yechiel Kalmenson • Nov 8 '19

Well, my first attempt had me iterating calling .index on ("a".."z") only to find out that you can't call index on a range, so in trying to work around that I realized I could just convert the range into an array... :)

Michael Kohl • Nov 8 '19

Ruby:

def words_to_marks(word)
  word.chars.sum { |c| c.ord - 96 }
end

For fun a second version that builds a lookup table with an endless range (introduced in Ruby 2.6):

ALPHABET = ('a'..'z').zip(1..).to_h

def words_to_marks(word)
  word.chars.sum { |c| ALPHABET.fetch(c, 0) }
end

peter279k • Jul 13 '20

Here is the simple solution with Python:

def words_to_marks(s):
    alphabet = {
        'a': 1,
        'b': 2,
        'c': 3,
        'd': 4,
        'e': 5,
        'f': 6,
        'g': 7,
        'h': 8,
        'i': 9,
        'j': 10,
        'k': 11,
        'l': 12,
        'm': 13,
        'n': 14,
        'o': 15,
        'p': 16,
        'q': 17,
        'r': 18,
        's': 19,
        't': 20,
        'u': 21,
        'v': 22,
        'w': 23,
        'x': 24,
        'y': 25,
        'z': 26,
    }

    sum_length = 0
    for char in s:
        sum_length += alphabet[char]

    return sum_length

Nicolas Hervé • Nov 7 '19

const wordsToMarks = (word) => word.split("").reduce((acc, curr) => acc + curr.charCodeAt(0) - "a".charCodeAt(0) + 1, 0);

Karin • Nov 9 '19

A little solution for Ruby :) this was fun!

def wordsToMarks(word)
  strength = 0
  alphabet = ('a'..'z').to_a
  word.split('').each do |i|
    strength = strength + (alphabet.index(i) + 1)
  end
  return strength
end

Jon Bristow • Nov 7 '19

words_to_marks("",0).
words_to_marks(S,Mark) :-
    string_codes(S,SC),
    length(SC, SCL),
    sum_list(SC,SCS),
    Mark is SCS - SCL * 96.

?- words_to_marks(“love”, V).
% V = 54

?- words_to_marks(“declarative”, V).
% V = 100

Michael Kohl • Nov 8 '19

Nice to see some Prolog :-)

Ed Reeseg • Nov 7 '19

#include <stdio.h>

int main(int argc, char *argv[]) 
{
    if (argc != 2)
    {
        printf("Usage: ./strength string\n");
        return 1;
    }
    int sum = 0;
    char *c = argv[1];
    while(*c != '\0')
    {
        if (*c < 'a' || *c > 'z')
        {
            printf("Invalid argument\n");
            return 1;
        }
        sum += *c - 'a' + 1;
        c++;
    }
    printf("%i\n", sum);
    return 0;
}

Lloric Mayuga Garcia • Nov 11 '19

In PHP


$alpha = function (string $string) {
    $al = array_flip(range('a', 'z'));

    $points = 0;
    foreach (str_split($string) as $s) {
        $points += $al[$s] + 1;
    }
    return $points;
};

echo $alpha('friendship');

EDIT:

array_map(function (string $string) {
    $al = array_flip(range('a', 'z'));

    $points = 0;
    foreach (str_split($string) as $s) {
        $points += $al[$s] + 1;
    }
    echo "$string: $points<br>";
}, ['attitude', 'friends', 'family', 'selflessness', 'knowledge']);

attitude: 100
friends: 75
family: 66
selflessness: 154
knowledge: 96

Amin • Nov 7 '19

Haskell

import Data.Char (ord)

wordsToMarks :: String -> Int
wordsToMarks = sum . map ((flip (-) 96) . ord)

Playground

Here.

Michael Kohl • Nov 8 '19

The prelude defines subtract, so you don't need flip here:

wordsToMarks = sum . map (subtract 96 . ord)

Another relatively common workaround is to use a right-section that adds a negative number, i.e.

wordsToMarks = sum . map ((+(-96)) . ord)

Amin • Nov 8 '19

Thanks for your insight. I didn't know about subtract and when the second solution is kind of obscure to me it seems very cool and short.

Michael Kohl • Nov 8 '19

Since - is both subtraction and unary minus one can’t use a right-section like (-1), but considering that subtracting is the same as adding a negative number the slightly obscure form makes sense. I’d still use subtract though.

Tallie • Nov 12 '19

My solution to the challenge using map and reduce in Python. If anyone has any suggestions on how to improve it, I would love to hear! :)

from functools import reduce 
def wordsToMarks(word):
    print(reduce((lambda x,y: x+y), map((lambda x: ord(x) - 96), list(word))))

Vitaliy Kononenko • Nov 9 '19

Here is my fun solution using Kotlin:

fun wordsToMarks(word: String) = word.toCharArray()
        .map { it + 1 - 'a' }
        .sum()

Zohar Peled • Nov 8 '19

c#, one liner using linq, ignoring non-letters chars:

int WordsToMarks(string word)
{
    return word.Where(c => char.IsLetter(c)).Sum(c => char.ToUpperInvariant(c) - 64);
}

David Kanekanian • Nov 8 '19

My Python Algorithm:

words_to_marks = lambda words: sum(
    map(ord, words.lower()),
    -96 * len(words))

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Daily Challenge #110 - Love VS. Friendship

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