Find the longest word in an array

Question

Your challenge is to take an array of strings and output the longest string in the array. So for

["tiny", "small", "bigger", "biggest"]

the output would be biggest.

If two elements of the array have the same length, you should choose the one that appears first in the array. That means if the array looks like this:

["one", "two", "no"]

the output is one, but if the array looked like this:

["two", "one", "no"]

the output is two.

---

As this is code-golf, the shortest code in bytes wins.

Answer 1

Imperative Tampio, 168 bytes

Listan x on riippuen siitä,onko sen ensimmäisen alkion pituus suurempi tai yhtä suuri kuin sen jokaisen alkion pituus,joko sen ensimmäinen alkio tai sen hännän x.

Online version

Ungolfed:

Listan pisin alkio on riippuen siitä, onko sen ensimmäisen alkion pituus suurempi tai yhtä suuri kuin sen jokaisen alkion pituus, joko

• sen ensimmäinen alkio tai
• sen hännän pisin alkio.

Online version

The only golfing opportunity this has is to replace pisin alkio (meaning "the longest element") with x.

Translation:

The longest item in a list is, depending on whether the length of the first item is greater or equal to the length of each element in the list, either

• the first item in the list, or
• the longest item in the tail of the list.

Answer 2

Python, 23 bytes

lambda a:max(a,key=len)

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Answer 3

Haskell, 35 bytes

-3 bytes thanks to Zgarb.

foldl1(!)
a!b|(a<$a)<(a<$b)=b|1<2=a

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I like this code. You know why? Because Haskell supports much more elegant solutions with functions from random libraries.

maximumBy(compare`on`length).reverse

That's friggin' readable! Except, it's not valid.

import Data.List
import Data.Function
maximumBy(compare`on`length).reverse

If it weren't for the imports, this would've been a perfect submission to get all the upvotes. :P

(Also, this uses one golfing tip and it uses a fold.)

Answer 4

R + pryr, 31 bytes

[-2 bytes thanks to Scrooble]

pryr::f(x[order(-nchar(x))][1])

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---

R, 33 bytes

function(x)x[order(-nchar(x))][1]

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Answer 5

EXCEL, 36 42 bytes

=INDEX(A:A,MATCH(MAX(LEN(A:A)),LEN(A:A),))

Entered as an array formula (ctrl-shift-enter). The input array should be entered in column A.

The formula returns the first match with maximum length.

Depending on your region settings, substitute , with ;; the code length remains unchanged. Of the 16 languages listed here, English function names are the shortest for this formula.

Explanation:

=                                          - return
 INDEX(                                  ) - the item of
       A:A                                 - the input
          ,                                - at
           MATCH(                       )  - the position of
                                       ,   - the first exact match of
                 MAX(        )             - the maximum of
                     LEN(   )              - the array of lengths of
                         A:A               - the input
                              ,            - in
                               LEN(   )    - the array of lengths of
                                   A:A     - the input

Answer 6

Prolog (SWI), 98 92 72 69 bytes

The top level predicate is *.

X/Y:-atom_length(X,Y).
[A]*A.
[A,B|L]*Z:-A/X,B/Y,Y>X,[B|L]*Z;[A|L]*Z.

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Explanation

The first row defines the dyadic predicate / to be a short for atom_length/2which is true if the first argument's length is the second argument. This saves us 3 bytes over using atom_length twice.

Our main predicate is defined as the dyadic * where the first argument is a list and the second argument the longest element of that list.

The second row is our base case which states that the longest element of a one element list is that element.

The third row states that for a list with at least 2 elements, the longest element is:

If the length of the second element is longer than the first element, the longest element is in the list without the first element.

Otherwise the longest element is in the list without the second element.

Answer 7

APL (Dyalog Unicode), 9 bytes^SBCS

⊢⊃⍨∘⊃∘⍒≢¨

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⊢ from the argument,

⊃⍨ pick the element with the index which is the

⊃ first of the

⍒ indices in descending order of the

≢¨ lengths of each

Answer 8

Pyth, 4 bytes

h.Ml

Test suite.

Explanation

h.Ml   | Program
h.MlZQ | With implicit variables filled in
-------+--------------------------------------------------------------------
h      | First element of
 .M  Q | The list of elements from the input list with the maximal value for
   lZ  | The length of the element

Answer 9

JavaScript (Node.js), 38 bytes

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a=>a.sort((a,b)=>a.length<b.length)[0]

Answer 10

PowerShell, 24 bytes

($args[0]|sort l* -d)[0]

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Takes input $args[0], pipes that to Sort-Object based on length in -descending order. Then takes the [0]th one thereof. Since sort is stable, this takes the first element in case of a tie.

Answer 11

Octave, 33 bytes

@(x)x{[~,p]=max(cellfun(@nnz,x))}

Input is a cell array of strings.

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Explanation

cellfun(@nnz,x) applies the nnz function (number of nonzeros) to each string in the input array x. For ASCII strings, nnz is equivalent to numel (number of elements), but shorter. The result is a numeric array with the string lengths.

Then, [~,]=max(...) gives the index of the first maximum in the array of string lengths. The result is used as a curly-brace index into x to obtain the corresponding string.

Answer 12

J, 19, 11, 10 8 bytes

0{>\:#@>

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Thanks to streetster for the hint!

-1 byte thanks to FrownyFrog!

-2 bytes thanks to Conor O'Brien

How it works:

    (  #@>) - unbox each string and find its length
     \:     - sort down the list of strings according to the lengths
0{::        - take and unbox the first string

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Answer 13

C#, 43 + 18 = 61 bytes

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a=>a.OrderByDescending(x=>x.Length).First()

Answer 14

Japt -h, 5 3 bytes

ÔñÊ

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Reverse, sort by length and output the last element.

Answer 15

Perl 6,  14  13 bytes

*.max(*.chars)

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*.max(&chars)

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Answer 16

PHP, 72 bytes

array_reduce($a,function($c,$i){return (strlen($i)>strlen($c))?$i:$c;});

Answer 17

Swift, 54 bytes

{($0 as[String]).reduce(""){$1.count>$0.count ?$1:$0}}

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Answer 18

K (oK), 9 bytes

*x@>#:'x:

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Example:

*x@>#:'x:("edur";"oot";"taht")
"edur"

Explanation

*x@>#:'x: / solution
       x: / store input in variable x
    #:'   / count (#:) each (')
   >      / sort descending
 x@       / apply indices to x
*         / take the first one

Notes:

Undeleted as this is classed as non-trivial, despite being basically 5 steps (would be if written as the function {*x@>#:'x}).

Answer 19

Röda, 30 bytes

{enum|[[#_,-_,_1]]|max|_|tail}

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Explanation:

{
 enum|         /* For each element, push its index to the stream */
 [[#_,-_,_1]]| /* For each element and index, push [length, -index, element] */
 max|          /* Find the greatest element */
 _|            /* Flat the list in the stream */
 tail          /* Return the last item in the stream */
}

Alternative 30 bytes:

{enum|[[#_,-_,_1]]|max|[_[2]]}

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Answer 20

Java (OpenJDK 8), 67 bytes

Another submission in my favourite language! (read: the only one I know).
This doesn't work with an empty array, but that's fine.

Golfed

w->{for(String y:w)if(y.length()>w[0].length())w[0]=y;return w[0];}

Ungolfed

for(String y:w)                           // Loops through all Strings
    if(y.length()>w[0].length())          // If the String is longer than the first String 
                                w[0]=y;   // Store it as the first string.
return w[0];                              // Return the first String.

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Answer 21

Racket, 160 bytes 110 bytes

Try it online! First time contributing, advice appreciated!

(define(m a)(if(>(length a)1)(if(>=(string-length(car a))(string-length(m(cdr a))))(car a)(m(cdr a)))(car a)))

Ungolfed

(define (m a)
    (if (> (length a) 1)
        (if (>= (string-length (car a)) (string-length (m (cdr a))))
            (car a)
            (m (cdr a))
        )
        (car a)
    )
)

Updated solution based on feedback

Answer 22

APL -- 23 16 bytes

a←{((⍴¨⍵)⍳(⌈/(⍴¨⍵)))⌷⍵}

Thanks to everyone for all of your great suggestions and encouragement!

a←{⍵⌷⍨(⍴¨⍵)⍳⌈/⍴¨⍵}

Usage:

a 'duck' 'duck' 'goose'
  'goose'

Explanation:

gets length of each vector of characters (string) then uses maximum as an index. I just started APL 20 min ago so I am sorry if this is a stupid way to do it.

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(edited for clarity)

Answer 23

Bash, 45 bytes

a=;for b;do((${#b}>${#a}))&&a=$b;done;echo $a

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Answer 24

Scratch 27 17 170 160

It expects a global (attached to all sprites, to be more precise) list of strings called mylist. After clicking the green flag, the longest word will be left in the variable w.

I think this is the link

when gf clicked
set[w]to(item[1]of[mylist
set[i]to[0
repeat(length of[mylist
change[i]by(1
set[m]to(item(i)of[mylist
if<(m)>(w)>then
set[w]to(m
end
end
stop[all

Counting as per this meta.

Answer 25

Standard ML (MLton), 55 bytes

fun&(s::r)=foldl(fn(%,$)=>if size% >size$then%else$)s r

Try it online! Example usage: & ["abc","de","fgh"] yields "abc".

Ungolfed:

fun step (current, longest) = 
    if size current > size longest 
    then current 
    else longest

fun longestString (start :: list) = foldl step start list
  | longestString nil = raise Empty

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Answer 26

dc, 39 38 36 bytes

0sl[slssdd]sG[dZdll!>Gooz0<M]dsMxlsp

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Edit: -1 byte, filling up empty space by 'd' instead

Edit: -2 bytes, thanks @brhfl for suggestion!

Answer 27

Julia 0.6, 24 bytes

!s=s[indmax(length.(s))]

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Answer 28

Funky, 38 bytes

a=>a[(v=a::map@#)::find(math.max...v)]

Explained

a=>a[(v=a::map@#)::find(math.max...v)]
        a::map@#                        $ Create a list of the lengths of the input's strings.
      v=                                $ And assign it to v.
     (          )::find(            )   $ Find the first index in this list that equals...
                        math.max...v    $ The largest value of v, eg. the length of the longest string.
   a[                                ]  $ Get the value at that position.

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Answer 29

Bash, 44 bytes

IFS=$'\n';egrep -m1 .{`wc -L<<<"$*"`}<<<"$*"

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Answer 30

SNOBOL4 (CSNOBOL4), 63 57 bytes

I	M =LT(SIZE(M),SIZE(X)) X
	X =INPUT	:S(I)
	OUTPUT =M
END

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Input is on stdin and output on stdout.

Roughly translates to the following pseudocode:

while input exists
 x = input
 if length(m) < length(x)
  m = x
end
return m

Answer 31

Ruby, 21 20 bytes

->s{s.max_by &:size}

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Trivial solution, thanks Snack for -1 byte

Answer 32

APL+WIN, 11 bytes

s[↑⍒∊⍴¨s←⎕]

Prompts for screen input in the form 'abc' 'defg' hijklm'

Answer 33

Perl 5, 38 bytes

say((sort{$a=~y///c<=$b=~y///c}<>)[0])

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Answer 34

C++, 125 bytes

Accepts input as a pair of iterators; returns an iterator to the longest string. The array must not be empty (i.e. the iterators must be different).

#include<algorithm>
template<class I>I f(I a,I b){return std::max_element(a,b,[](auto&a,auto&b){return a.size()<b.size();});}

Demo

#include <string>
#include <iostream>
#include <iterator>
int main()
{
    std::string strings[] = { "Programming", "Puzzles", "and", "Code", "Golf" };

    std::cout << *f(std::begin(strings), std::end(strings)) << std::endl;
}

The definition of std::max_element() guarantees that

If several elements in the range are equivalent to the greatest element, returns the iterator to the first such element.

Answer 35

Javascript ES5, 41 bytes

Since there was already a solution using sort...

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a=>a.reduce((x,y)=>x.length>=y.length?x:y)

Answer 36

Jelly, 4 bytes

LÐṀḢ

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Maximum (ÐṀ) by length (L). Take Head (Ḣ).

---

Jelly, 4 bytes

ṚLÞṪ

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Reverse. Sort by length. Take tail (last element).

Answer 37

Coconut, 15 13 bytes

max$(key=len)

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Answer 38

Perl, 24 bytes

Includes +1 for p

Give words as lines on STDIN

perl -pe '$;[y///c]//=$_}{$_=pop@'
one
two 
no
^D

Answer 39

Add++, 7 bytes

L,bU«bL

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More interesting, 32 byte version:

L,vbU§bLdbLBkÞ{g}@
D,g,@,bLBK=

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How they work

L,	; Create a lambda function
	; Example argument:         		[["two" "one" "abc" "no"]]
    bU	; Evaluate as list; 			STACK = ['two' 'one' 'abc' 'no']
    «bL	; Take the value with the max length;	STACK = ['two']

And the 32 byte version:

L,		; Create a lambda function
		; Example argument: 		['["two" "one" "no"]']
	vbU	; Evaluate;		STACK = ['two' 'one' 'no']
	§bL	; Sort by length;	STACK = ['no' 'two' 'one']
	dbLBk	; Save length;		STACK = ['no' 'two' 'one']	REGISTER = 3
	Þ{g}	; Filter by 'g';	STACK = ['two' 'one']
	@	; Reverse;		STACK = ['one' 'two']
		; Implicitly return the top element:   'two'

D,g,@,		; Create a function 'g'
		; Example argument:		['two']
	bL	; Length;		STACK =	[3]
	BK=	; Equal to register;	STACK = [1]

Answer 40

jq -r, 19 15 bytes

Golfed 4 bytes after learning min_by is a builtin.

min_by(-length)

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(Without -r the output string is surrounded with "s.)

Answer 41

Lua, 70 64 bytes

function f(a,...)x=...and f(...)return x and#x>#a and x or a end

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x is set to the second argument if that is falsy, otherwise the longest of the second argument and up. The function returns x if it is truthy and longer than the first argument, else the first argument.

No type-checking, so works not only for a sequence of strings, but for sequences containing tables, or anything with a __len metamethod that returns a number. For instance f("tiny", {1, 2, 3, 4, 5}, debug.setmetatable(6, { __len = function (self) return self end })) returns 6.

This technically breaks the rule of operating on an array (in Lua, a table) because it operates on an argument list, but operating on a table seemed more complicated (table.sort is not stable). A table can be unpacked into the function: f(table.unpack { 'oh', 'one', 'two' }).

Answer 42

Mathematica, 29 bytes

#&@@#~MaximalBy~StringLength&

Pure function. Takes a list of strings as input and returns a string as output. #~MaximalBy~StringLength finds the longest strings in the list and #&@@ selects the first instance.

Answer 43

Rust, 73 bytes

|mut x:Vec<String>|{x.sort_by(|a,b|b.len().cmp(&a.len()));x[0].clone()}

Very simple code, that can probably be golfed at least a little more. The fact that i had to use String instead of &str, just for the return value, bugs me alot. (I could have used a .to_string() method, but that wouldve added 2 bytes.)

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Answer 44

MY, 24 bytes

1ω74ǵ'ƒ⇹(ω74ǵ'ƒ⇹(⍐=⍸@ω@←

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How?

1ω74ǵ'ƒ⇹(ω74ǵ'ƒ⇹(⍐=⍸@ω@←
                       ← = Output ...
                      ω  = ... the input ...
                       @ = ... at ...
1                    @   = ... the first ...
                   ⍸     = ... truthy index of ...
 ω74ǵ'ƒ⇹(                = ... the length of each element of the input's ...
                  =      = ... equality with ...
         ω74ǵ'ƒ⇹(⍐       = ... the maximum length.

Answer 45

Java 8, 57 bytes (70 bytes)

By the power of streams, my java answer shall be the shortest

Arrays.stream(a).max(Comparator.comparing(String::length))

Honestly I don't know if I cheated with that answer because I don't return anything but you could use that line of code. If we really want to return stuff then it's a bit longer.

Thanks to LukeStevens for pointing out to use lambdas to be more golfy.

w->Arrays.stream(w).max(Comparator.comparing(String::length)‌​).get()

Answer 46

F# (.NET Core), 29 bytes

 Seq.maxBy <| fun s->s.Length

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Explanation

Through partial application of Seq.maxBy, returns a function (string seq -> string)

Answer 47

Stacked, 19 bytes

[:$#'"!:MAX index#]

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Explanation

[:$#'"!:MAX index#]
[                 ]   anonymous function, takes input as list of strings
 :$  "!               on each member:
   #'                   get the length
            index     get the index of...
       :MAX             the maximum length
                 #    and obtain the respective member in the array.

Alternatives

19 bytes: [[#'\#'-]sortby 0#]

24 bytes: [$#'"!:sorted index#_1#]

Answer 48

PHP, 69 bytes

usort($x,function($a,$b){return strlen($b)<=>strlen($a);});echo$x[0];

Not a trick or nice golfing, just sorting array by string length and then outputting the first element.

Answer 49

V, 27 23 21 20 bytes

Í./_
ÚGYu/"
uYHVGp

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Explanation:

Í./_         In every line, replace every character by _
Ú            Sort alphabetically. Cursor ends up in first line
 GY          Go to last line, copy it
   u         Undo sort
    /"       Search for string in register " (the one that was just copied). Non-printable character is <C-r>
u            Undo replace, go to last cursor position. Non-printable character is <C-o> (thanks, Cows quack)
  Y          Copy line
   HVG       Go to first line, select all lines until last line
      p      Paste copied line into selection

Hexdump:

00000000: cd2e 2f5f 0ada 4759 752f 1222 0a75 0f59  ../_..GYu/.".u.Y
00000010: 4856 4770                                HVGp

Answer 50

Common Lisp, 32 bytes

(elt(sort(read)'> :key'length)0)

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Answer 51

Husk, 3 bytes

►L↔

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Explanation

  ↔  -- reverse the list (such that the first gets picked in case of a tie)
►L   -- find maximum by length

Answer 52

Scratch 2.0, 22 blocks

Accepts input by pressing see inside on the project page and manually adding and removing items from list.

Project here. Flash required to run and view, apologies.

All blocks were counted including nested ones. Not sure how I'd count bytes with this language unless you want the length of the JSON.

Answer 53

05AB1E, 3 bytes

Réθ

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Explanation

R     # reverse
 é    # sort by length
  θ   # get the last element

Answer 54

PHP, 59 bytes

while($n=strlen($s=$argv[++$i]))$n>$e&&$e=$n+!$r=$s;echo$r;

takes words as separate command line arguments. Run with -nr or try it online.

Answer 55

Google Sheets, 40 Bytes

Anonymous worksheet function that takes input from range A:A and outputs to the calling cell

=Index(A:A,Match(Max(Len(A:A)),Len(A:A),

Answer 56

APL NARS, 10 chars, 24 bytes

{⍵[↑⍒⍴¨⍵]}

note in the test, that: 'uno' it is not one array of array, so the array of correct deep is ,⊂'uno' instead

⎕fmt  ,⊂'uno'    
  ┌1─────┐
  │┌3───┐│
  ││ uno││
  │└────┘2
  └∊─────┘

so in the test g fail for 'uno'

g←{⍵[↑⍒⍴¨⍵]}
g 'uno'
  u

but not fail for the correct type of input

g ,⊂'uno'
  uno
g 'uno' 'quattro' 'cinque'
  quattro
⎕fmt  g ,⊂''
  ┌───┐
  │┌0┐│
  ││ ││
  │└¯┘2
  └∊──┘

if the community think this is one too much big limitation: than there is the 18 chars solution

{1=≡⍵:⍵⋄∊⍵[↑⍒⍴¨⍵]} 

that evade the problem in the first test calculate the deep of input

f←{1=≡⍵:⍵⋄∊⍵[↑⍒⍴¨⍵]}     
f 'uno'
  uno
f 'uno' 'quattro' 'cinque'
  quattro

The problem of the input type of one Char in Apl as 'a' (it is consider from interpreter one char). We can transform that in one array of array of char, or one type list of arrays of deep 2 write ,⊂,'a'; so g ,⊂,'a' should return the correct answer of type array of array length 1

Answer 57

sed 4.2.2 -rz, 71 72 83 bytes

h
:
s/^.//gm
/[^\n]{2}/t
:b
/^\n/{s/.//
x
s/^[^\n]*.//
x}
tb
x
s/\n.*//

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• -r uses extended regular expressions, in other words, allows golifer syntax as we are able to save several \s

• -z puts all lines of input into one "line", so that they would be processed at the same time rather than the normal cyclic processing.

This was a fun challenge seeing as sed did not have any builtin sort functions. The program works as follows:

• copy the pattern space into the hold space
• find the position of the longest line by repeatedly removing the first character from every line until we are remaining with line(s) with only a single character, this first of these is at the position of the longest line

• to find which line this position corresponds to, we repeatedly

  • remove the first line of the pattern space and hold space until we encounter a non-newline character on the first line of the pattern space
• now that the corresponding line is at the top of the hold space, we remove the other lines
• the contents of the buffer is printed at the end of the program

Answer 58

C++ (gcc), 65 63 bytes

Accepts any container of std::string and returns via reference.

[](auto&c,auto&r){r=c[0];for(auto x:c)r=x.size()>r.size()?x:r;}

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If you are tolerant to output, there is a 58 byte solution which alters the input list and copies the longest string to the first element:

[](auto&c){for(auto x:c)c[0]=x.size()>c[0].size()?x:c[0];}

Answer 59

Attache, 9 bytes

MaxBy&:`#

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This is simply bonding the size operator (`#) to the left of MaxBy. This is equivalent to:

f[x] := MaxBy[Size, x]

...which selects the first element with the largest size in x.

Answer 60

Kotlin, 21 bytes

{it.maxBy{it.length}}

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Answer 61

Tcl, 93 bytes

proc c a\ b {expr [string le $a]<[string le $b]}
proc L l {return [lindex [lsort -c c $l] 0]}

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Quite disappointed by the score, should/could be better. Thought about this one-proc version:

proc L l {
 set m  [lmap e $l {regsub -all . $e 1}]
 return [lindex $l [lsearch $m [expr max([join $m ,])]]]
}

but it's longer.