Same thing, but using a reducer instead of sorting.

const getLongestWordOf = (sentence = '') => {
  return sentence
    .split(' ')
    .reduce((longest, current) => {
      return current.length > longest.length ? current : longest;


getLongestWordOf('I am just another solution to the same problem');

EDIT: This approach seems to be faster than the sorting one!


This is what I thought to do as well ๐Ÿ˜„


Good one, didn't think of reduce ๐Ÿ‘

DECLARE longest_word TEXT;
  SELECT strs INTO longest_word
  FROM regexp_split_to_table(val, '\s+') AS strs
  ORDER BY char_length(strs) DESC
  LIMIT 1;

  RETURN longest_word;
$$ LANGUAGE plpgsql;

SELECT get_longest_word('a bc def ghij klm no p');

I know one of the tags is 'Javascript' but here is how I would do it in Python:

def get_longest_word(sentence):
    return max(sentence.split(' '), key=lambda x: len(x))

get_longest_word('Iam a verrrrry loooongggg sentence')
# result: loooongggg


Can I chime in with a quick C# and LINQ one liner?

private Longest(string sentance) 
     return sentance.Split(' ').OrderByDescending(l => l.Length).FirstOrDefault();

Well, the return statement is a one liner :)


Recently started digging my way through GHC libraries

import Data.List

cmpLength :: String -> String -> Ordering
cmpLength l r = compare (length r) (length l)

findLongest :: String -> String
findLongest s = head . sortBy (cmpLength) . words $ s

main :: IO ()
main = do
  putStrLn (findLongest "Iam a verrrrry loooongggg sentence")
import Data.List

longerString :: String -> String -> String
longerString l r = if length l > length r then l else r

findLongest :: String -> String
findLongest s = foldr (\longest w -> longerString longest w) "" (words s)

main :: IO ()
main = do
  putStrLn (findLongest "Iam a verrrrry loooongggg sentence")


$_=qq|Iam a verrrrryyy longggggggg sentence|;
print( (sort{length($b) <=> length($a)} split(qq|\s|))[0])


package main

import (

func main() {
    s := "Iam a verrrrryyy longggggggg sentence"
    b := ""
    for _, w := range strings.Split(s, " ") {
        if len(w) > len(b) {
            b = w



It may not be the most eye perfect code however the perf are close to the reducer solution


I would still go to with the reducer solution, but sometimes doing 'old' things works very nice

I also try with for loop / for of, I expected it to be way more faster than everything else but it didn't.



Learned something new regarding jsperf, thank you. ๐Ÿ™


When used according to the specification, this function has O(1) complexity*.

/** @function getLongestWord
 * @param {String} s a string of words with the longest word in the first position.
 * @returns the longest word in a string, if it is the first word in the string, or a random word that occupies the first position in the string otherwise.
const getLongestWord => (s = '') => s.split(' ')[0]

getLongestWord('loooongggg sentence very Iam') // 'loooongggg'

* It doesn't because split probably has complexity O(n) or something.


Ignoring punctuation and only considering spaces... ruby solution:

def longest_word(str)
Classic DEV Post from Feb 13

How Do I Start Giving Talks on Coding?

I don't consider myself an expert in just about any aspect of computer science ...

๐Ÿ‘‹ Hey reader.

Do you prefer sans serif over serif?

You can change your font preferences in the "misc" section of your settings. โค๏ธ