This is an easy one.

Anagrams are words which have the same characters but in different order. All you have to do is to figure out if two strings are anagram or not.


  • Take two strings as a function argument, command-line argument or from standard input.
  • Write a function or program which returns truthy value if strings are anagram, falsy value otherwise
  • Anagrams may have different number of spaces () characters in them.
  • Anagrams are case insensitive matches.
  • Anagrams can only have printable ASCII characters for this code-golf.

For example(taken from the Wikipedia page itself): Doctor Who is an anagram of Torchwood . Note missing space and lack of capital D and W in Torchwood

This is code-golf so minimum bytes code win!

Happy golfing.

  • \$\begingroup\$ Done - Edited to clarify \$\endgroup\$ – Optimizer Sep 12 '14 at 7:40
  • \$\begingroup\$ If that is considered as standard input (STDIN) then yes. \$\endgroup\$ – Optimizer Sep 12 '14 at 7:54
  • \$\begingroup\$ printable ascii or alphanumeric? what about ,:;...? \$\endgroup\$ – edc65 Sep 12 '14 at 8:49
  • \$\begingroup\$ @MartinBüttner No objection against ARGV from my side. Feel free to edit the question to point it out in clear words as I might not be able to make it clear. \$\endgroup\$ – Optimizer Sep 12 '14 at 9:43
  • \$\begingroup\$ @edc65 Well, these are just words, not sentences, so ideally those characters should not appear. But if they do, they are to be considered. Same for numbers \$\endgroup\$ – Optimizer Sep 12 '14 at 9:44

12 Answers 12


CJam, 15 11 bytes

Just thought I'd practice CJam a bit more. This is basically the same as my Ruby answer (or most answers):


Quick explanation

ea          "Push command line arguments";
ea{     }/  "For each element in that array";
ea{S-   }/  "Push a space character onto the stack and remove spaces from string";
ea{S-el }/  "Convert string to lower case";
ea{S-el$}/  "Sort the characters in the string";
ea{S-el$}/= "Check for equality";

Prints 1 for anagrams and 0 otherwise.

  • \$\begingroup\$ 1. Brackets can be unmatched in CJam, so you don't need the [ in [ll]. ea would be even shorter; it pushes all command-line arguments as an array. Doesn't work with the online interpreter though... 2. S is initialized to " ", so you can use it instead of ' . 3. You can replace %~ with /. \$\endgroup\$ – Dennis Sep 14 '14 at 5:12

Java, 146 131 144 134 131 128

Not a very exciting answer. Returns true for anagrams, false for others. Actually works now. Cheapass abuse of generics in class declaration.

boolean b(T a,T b){return c(a).equals(c(b));}T c(T a){char[]c=a.toLowerCase().toCharArray();sort(c);return(T)valueOf(c).trim();}


boolean b(T a,T b) {
    return c(a).equals(c(b));

T c(T a) {
    char[] c = a.toLowerCase().toCharArray();
    return (T) valueOf(c).trim();
  • 1
    \$\begingroup\$ mmm... cat and faq? \$\endgroup\$ – edc65 Sep 12 '14 at 9:52
  • \$\begingroup\$ Two different strings can have the same ASCII sum. Take, for e.g., the strings "bb"(98 + 98) and "ac"(97 + 99). \$\endgroup\$ – c.P.u1 Sep 12 '14 at 10:10
  • \$\begingroup\$ Is valueOf a static import? \$\endgroup\$ – nanofarad Sep 12 '14 at 10:32
  • \$\begingroup\$ yeah, same with sort \$\endgroup\$ – carnifex Sep 12 '14 at 10:34

Rebol   44   40

(Rebol 2)

Define a function m which takes two values a and b. Sort each string, trim the whitespace and use = to check if the strings are the same.

m: func[a b][(trim sort a)= trim sort b]

m: func[a b][equal? trim sort a trim sort b]


>> m: func[a b][(trim sort a)= trim sort b] 
>> m "Doctor Who" "Torchwood"             
== true

It is useful handy that sort is case insensitive by default. Note: This function does have side effects if you use this code elsewhere as sort modifies a and b directly.

(sort should work in Rebol 3 but doesn't as there is a known bug)

  • 1
    \$\begingroup\$ Yes ... :( <backs slowly away from keyboard and slaps forehead> \$\endgroup\$ – johnk Sep 12 '14 at 11:58
  • \$\begingroup\$ That should work. Just a few chars longer \$\endgroup\$ – johnk Sep 12 '14 at 12:05
  • \$\begingroup\$ Wait, I don't understand. Just because sort sorts case insensitively, how does that make equal? work on strings of different cases? \$\endgroup\$ – Martin Ender Sep 12 '14 at 12:06
  • \$\begingroup\$ Oh equal? itself is case insensitive, too. That's... handy. \$\endgroup\$ – Martin Ender Sep 12 '14 at 12:08
  • \$\begingroup\$ and same? "ABC" "abc" is false :) Also has the shorthand "ABC" == "abc" - double equal signs means really equal! \$\endgroup\$ – johnk Sep 12 '14 at 12:11

Ruby, 65 60 58 57 51 48 46 bytes

s,t=$*.map{|u|u.upcase.scan(/\S/).sort};p s==t

Edit: Switched to STDIN to save a few bytes.

Edit: Switched to ARGV to save three more bytes.

Quite long for such a simple problem, hmmmm.

  • For each string:
    • Turn to upper case (to make it case in-sensitive).
    • Split into non-whitespace characters.
    • Sort (by character code).
  • Then check that the results are equal.
  • \$\begingroup\$ Wow everything i would like to have in C# - nice one +1 \$\endgroup\$ – Stephan Schinkel Sep 12 '14 at 8:27

JavaScript (ES6) 68 99

Generate a key for each word and compare them. If only JS had some shorter ucase method...

  • \$\begingroup\$ You always bring in unique techniques to the table :) \$\endgroup\$ – Optimizer Sep 12 '14 at 10:04
  • \$\begingroup\$ @Optimizer you're kind, but nothing new here, this is really a standard way to handle anagrams \$\endgroup\$ – edc65 Sep 12 '14 at 10:07
  • \$\begingroup\$ No, I was talking about the declare a second method in the third argument and use it to save space trick. \$\endgroup\$ – Optimizer Sep 12 '14 at 10:09

Python 3 -- 73 67 bytes


f=lambda:sorted(i for i in input().upper()if" "!=i);print(f()==f())

As usual, nice looking, but not very short. Typical python...

  • \$\begingroup\$ Does a#b count as an word? Fixed. \$\endgroup\$ – matsjoyce Sep 12 '14 at 10:26
  • \$\begingroup\$ The challenge asks for printable ASCII, so all characters except spaces should be treated as "letters" for the purposes of the challenge. \$\endgroup\$ – Martin Ender Sep 12 '14 at 10:26

Cobra - 130

    print (f=Func<of String,String>(do(s='')=s.toLower.toCharArray.toList.sorted.join('').trim)).invoke(a)==f.invoke(b)
  • 2
    \$\begingroup\$ Sorted first, then tolower: couldn't this create problems? \$\endgroup\$ – edc65 Sep 12 '14 at 10:37

JavaScript (ES6) 82

Returns true if the two strings are anagrams, otherwise returns false or throws an exception

  • \$\begingroup\$ "or throws an exception" violates the rules, though. That's not a falsy value as required. \$\endgroup\$ – Ingo Bürk Sep 12 '14 at 19:58
  • \$\begingroup\$ Throwing makes this an invalid entry :( \$\endgroup\$ – Optimizer Sep 12 '14 at 21:41

Perl - 84

Always need Perl in these...

sub c{map{s/\s//}@a=@_;(join'',sort split//,lc$a[0])eq join'',sort split//,lc$a[1]}

Wanted to be able to modify @_ directly but perl wont allow that

  • \$\begingroup\$ How should this be called? It doesn't seem to return a result. \$\endgroup\$ – i alarmed alien Sep 13 '14 at 16:34
  • \$\begingroup\$ It returns a 1 or a 0 ---> print c('Doctor Who','Torchwood') ? 'Yes' : 'No'; \$\endgroup\$ – mcreenan Sep 16 '14 at 23:37

Bash+coreutils, 62 bytes

s()(tr -d '     '<<<"${@^^}"|fold -1|sort)
cmp <(s $1) <(s $2)

Takes arguments from the command line. Return code 0 for TRUE and 1 for FALSE. Also something like /dev/fd/63 /dev/fd/62 differ: byte 1, line 1 is output in the FALSE case.


$ ./anagram.sh "Doctor Who" Torchwood; echo $?
$ ./anagram.sh "Doctor Watson" Torchwood; echo $?
/dev/fd/63 /dev/fd/62 differ: byte 1, line 1

Clojure : 84 chars


(defn g[s](sort(remove #(= \  %)(seq(.toLowerCase s)))))
(defn f[a b](=(g a)(g b)))


(defn g[s] (sort (remove #(= \  %) (seq (.toLowerCase s)))))
(defn f[a b] (= (g a) (g b)) )

Driver program with execution:

(def a "Doctor   Who")
(def b "Torch woo d")
(println (f a b))

bash$ java -jar clojure-1.6.0.jar anagram.clj 

Groovy : 96 77 chars


g={it.findAll{it!=" "}.collect{it.toLowerCase()}.sort()}


g = { it.findAll{ it!=" " }.collect{ it.toLowerCase() }.sort() }
f = { a,b -> g(a)==g(b) }

Example main program and execution:

a="Doctor Who x x"
b="Tor chwoo dxx"
println f(a,b)

$ groovy Anagram.groovy
$ true

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