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Posted on Oct 1, 2019

Daily Challenge #80 - Longest Vowel Change

#challenge

The vowel substrings in the word codewarriors are o,e,a,io. The longest of these has a length of 2. Given a lowercase string that has alphabetic characters only and no spaces, return the length of the longest vowel substring. Vowels are any of aeiou.

Good luck!

This challenge comes from KenKamau at 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!

Top comments (19)

willsmart • Oct 1 '19 • Edited

The new JS function matchAll is going to be really useful.
Here's a JS quickie

longestVowelRunLength = string => [...string.matchAll(/[aeiou]+/g)].reduce((acc, [match]) => Math.max(acc, match.length), 0)

SavagePixie • Oct 1 '19 • Edited

Gotta love JavaScript one-liners

const longest = str => str.split(/[^aeiou]+/).reduce((a, b) => b.length > a ? b.length : a, 0)

Bhuvan Ganesan • Oct 1 '19 • Edited

Javascript version

longestVowelList = (str) => {
  const vowels = ["a", "e", "i", "o", "u"];
  const result = vowels.map(vowel => {
    const regex = new RegExp(`[^${vowel}]+`);
    return str.split(regex)
       .filter(r => r !== "")
       .sort((a, b) => b.length - a.length);
  });
  return result.flat(1).join("");
}
console.log('---->',longestVowelList("codewarriors"));
//Output ----> aeioo

Kerri Shotts • Oct 1 '19

Here's mine... would have liked to use String.prototype.matchAll, but my JS engine didn't support it. So used a generator and Array.from instead. :-)

function *matchAll(str, re) {
    let match;
    do {
        match = re.exec(str);       // get the first match of the regular expession
        if (match) yield match;     // if we have one, yield it
    } while (match != null);        // keep going until no more matches
}

const getLongestSequence = (str, re) =>
    Array.from(matchAll(str, re))   // generators are iterable
         .map(([seq]) => seq)       // extract the string from the regex match result
         .reduce((longest, cur) => 
             cur.length > longest.length ? cur : longest    // keep track of longest
         , "");

const getLongestVowelSequence = str => 
    getLongestSequence(str, /[aeiou]+/gi);  // use vowels -- g = global; i = case insensitive

Full Gist: gist.github.com/kerrishotts/a3ec30...

Kerri Shotts • Oct 1 '19

Just noticed that the challenge is to return the length ;-) Oh well -- getLongestVowelSequence("codewarriors").length will do ;-)

Avalander • Oct 2 '19 • Edited

Haskell

import Data.List (groupBy)

longest :: String -> Int
longest = maximum . map length . filter (isVowel . head) . groupBy bothVowels
  where
    bothVowels a b = (isVowel a) && (isVowel b)

isVowel :: Char -> Bool
isVowel = (flip elem) "aeiou"

longest "codewarriors" -- 2

I could have done it with a single fold, but I decided to group the characters by groups of vowels and non-vowels, filter out the groups that are not vowels, map the length of each group and pick the largest number.

Also, almost entirely point-free, a pity I couldn't figure out how to make bothVowels point-free.

heidrichj • Oct 1 '19 • Edited

Hi,
this is my first time sharing my code here, so please don't judge :)
But I really like these little challenges, thank you!

Here we go (python ftw!):

def vowel(input):
        vowels = ['a','e','i','o','u']
        lengthes = []
        lastVowel = False
        for c in input:
                if c in vowels:
                        if not lastVowel:
                                lengthes +=[1]
                        else:
                                lengthes[len(lengthes)-1] += 1
                        lastVowel = True
                else:
                        lastVowel = False
        return max(lengthes)

print(vowel("codewarriors"))

There is probably some room for improvement...

Have a nice day!

Rafael Acioly • Oct 1 '19 • Edited

Hi @heidrichj you could also use the class Counter from collections package :)

from typing import Union
from collections import Counter

def vowel(word: str) -> Union[int, None]:
    letters = ('a', 'e', 'i', 'o', 'u')
    letters_on_word = Counter(word)

    for letter, quantity in letters_on_word.most_common():
        if letter in letters:
            return quantity

Tip:

Always use tuple in a list that is "fixed" to ensure that this list will be immutable

Peter • Oct 1 '19 • Edited

In golang! Could be simpler but lots of loops!

func vowel(input string) string {
    var ls string
    var ret string
    for _, c := range input {
        switch c {
        case 'a', 'e', 'i', 'o', 'u':
            ls += string(c)
        default:
            if len(ls) > len(ret) {
                ret = ls
            }
            ls = ""
        }
    }
    if len(ls) > len(ret) {
        ret = ls
    }

    return ret
}

Go Playground example original
EDIT:
Add new example with changes from comments
Go Playground example with new switch

EDIT 2:
Realized it would not handle vowels at the end of the string
Go Playground

Rafael Acioly • Oct 1 '19 • Edited

Hey @peter , you can also use multiple values in a single case statement


switch c {
case 'a', 'e', 'i', 'o', 'u':
    ls += string(c)
default:
    arr = append(arr, ls)
    ls = ""
}

github.com/golang/go/wiki/Switch#m...

Peter • Oct 1 '19

Hey good to know thanks! Just learning go so didn't realize I could do that. Thanks!

Rafael Acioly • Oct 1 '19 • Edited

there's also strings.Count built-in method ;)

golang.org/pkg/strings/#Count

strings.Count("codewarriors", "a") // 1

edh_developer • Oct 1 '19 • Edited

import re

def longTokenLength(s):
    return max([len(x) for x in re.split("[^aeiou]",s)])

print longTokenLength("codewarriors")

Edit: It being python, I'm condensing my previous code down into a list comprehension.

Dalmo Mendonça • Oct 1 '19 • Edited

const longestVowelSequence = (str) => Math.max(...str.split(/[^aeiou]/).map(s => s.length))

SavagePixie • Oct 2 '19

I like your approach!

Dimitri Marion • Oct 1 '19

Javascript with reduce

const word = "codewarriors"

function findLongestVowels(word) {
  const regex = /[aeiou]/;

  const vowels = Array.prototype.reduce.call(word, 
      (currentLetters, nextLetter, index) => {
       if (regex.test(nextLetter)) {
           return regex.test(word[index-1]) ? 
                    currentLetters.concat('', nextLetter) : 
                    currentLetters.concat(' ', nextLetter);
       } else {
           return currentLetters;
       }
   }, "");

  const longestVowels = vowels.split(' ').reduce( (longest, current) => {
       return current.length > longest.length ? current : longest;
  }, "");

  return longestVowels.length;
}

console.log("Length of the longest vowel: " + findLongestVowels(word));

View full discussion (19 comments)

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Daily Challenge #80 - Longest Vowel Change

Top comments (19)

Javascript version

Haskell

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