Determine whether strings are anagrams

Question

Challenge

Given two strings, work out if they both have exactly the same characters in them.

Example

Input

word, wrdo

This returns true because they are the same but just scrambled.

Input

word, wwro

This returns false.

Input

boat, toba

This returns true

Rules

Here are the rules!

• Assume input will be at least 1 char long, and no longer than 8 chars.
• No special characters, only a–z
• All inputs can be assumed to be lowercase

Test Cases

boat, boat = true
toab, boat = true
oabt, toab = true
a, aa = false
zzz, zzzzzzzz = false
zyyyzzzz, yyzzzzzy = true
sleepy, pyels = false
p,p = true

Answer 1

Python, 32 bytes

f=lambda a,b,S=sorted:S(a)==S(b)

Answer 2

Golfscript, 3 chars?

$$=

usage:

'boat'$'baot'$=
1

'toab'$'boat'$=
1

'oabt'$'toab'$=
1

'a'$'aa'$=
0

'zzz'$'zzzzzzzz'$=
0

'zyyyzzzz'$'yyzzzzzy'$=
1

'sleepy'$'pyels'$=
0

'p'$'p'$=
1

Answer 3

J, 8

-:&(/:~)

Literaly, match (-:) on (&) sort up (/:~)

Sample use:

   'boat' -:&(/:~) 'boat'
1
   'toab' -:&(/:~) 'boat'
1
   'oabt' -:&(/:~) 'toab'
1
   'a' -:&(/:~) 'aa'
0
   'zzz' -:&(/:~) 'zzzzzzzz'
0
   'zyyyzzzz' -:&(/:~) 'yyzzzzzy'
1
   'sleepy' -:&(/:~) 'pyels'
0
   'p' -:&(/:~) 'p'
1

Where do the 64-bit integers come into play?

Answer 4

Javascript, 192 157 152 147 125 bytes

Ok some of these languages are a lot more flexibile than I thought! Anyway this is the longer way I guess, but a different technique at least.

Compressed

Thanks to Peter and David for squeezing more chars out!

for(a=[j=p=2];j<123;)a[j]?p%a[++j]<1&&p++&&(j=0):(a[j]=p,j=0);function b(c,i){return c[i=i||0]?a[c.charCodeAt(i)]*b(c,++i):1}

Then do:

alert(b("hello")==b("elloh"));

Expanded Code

The compressed has had lots of changes, but this is the basic theory:

// Define dictionary of primes
a = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101];

// Returns the unique ID of the word (order irrelevant)
function b(c) {
    r = 1;
    for (i = 0; i < c.length; i++)
        r *= a[c[i].charCodeAt(0) - 97];
    return r
}

alert(b("hello") == b("hlleo"));

Answer 5

Golfscript, 8 bytes

This defines a function called A

{$\$=}:A

Test cases

;
'boat' 'boat' A
'toab' 'boat' A
'oabt' 'toab' A
'a' 'aa' A
'zzz' 'zzzzzzzz' A
'zyyyzzzz' 'yyzzzzzy' A
'sleepy' 'pyels' A
'p' 'p' A

Answer 6

Haskell, 31 bytes

function - 31

import List
f=(.sort).(==).sort

program - 81 58 55

import List
g=sort`fmap`getLine
main=(`fmap`g).(==)=<<g

Usage:

$ runghc anagram.hs
boat
boat
True
$ runghc anagram.hs
toab
boat
True
$ runghc anagram.hs
a
aa
False

Kudos to lambdabot and its pointfree refactoring.

Answer 7

C#, 129 chars

namespace System.Linq{class X{static void Main(string[]a){Console.Write(a[0].OrderBy(_=>_).SequenceEqual(a[1].OrderBy(_=>_)));}}}

Readable:

namespace System.Linq
{
    class X
    {
        static void Main(string[] a)
        {
            Console.Write(a[0].OrderBy(_ => _)
                  .SequenceEqual(a[1].OrderBy(_ => _)));
        }
    }
}

Answer 8

Ruby, 34 bytes

Using the IO scheme of Peter Taylors Perl solution:

p gets.chars.sort==gets.chars.sort

Answer 9

C program, 118

t[52],i;main(c){for(;i<52;)(c=getchar())<11?i+=26:t[i+c-97]++;
for(i=27;--i&&t[i-1]==t[i+25];);puts(i?"false":"true");}

Answer 10

Perl, 58 bytes

(complete program, unlike the other Perl answer which is only a function)

($c,$d)=map{[sort split//]}<>;print"@$c"eq"@$d"?true:false

49 as a function

sub f{($c,$d)=map{[sort split//]}<>;"@$c"eq"@$d"}

Answer 11

C++ (104 non-ws chars)

Based on counting sort. Note: Assumes strings of the same length, which seems to be implied (though not stated) by the question.

int u[123], i;

int a(char **c) {
    for(; c[0][i]; ) {
        u[c[0][i]]++;
        u[c[1][i++]]--;
    }

    i=123;
    while(i && !u[--i]);
    return !i;
}

Answer 12

Clojure - 23 chars

As an anonymous function:

#(apply = (map sort %))

Test case example:

(#(apply = (map sort %)) ["boat" "boat"])
=> true

Answer 13

JavaScript

Based on @zzzzBov's solution.

Comparison, 65 chars (40 without function)

function f(a,b){return a.split('').sort()==b.split('').sort()+""}

Comparator, 43 chars

function f(a){return a.split('').sort()+""}

Answer 14

Perl, 62 bytes

This function takes the strings as arguments and returns true or false.

sub f{my@a;for$.(1,-1){$a[ord]+=$.for split//,pop}!grep{$_}@a}

Stores the ASCII values in an array and checks if it evens out. Increments for the first word and decrements for the second word.

Answer 15

Perl, 77 75 chars

The I/O of the problem aren't well specified; this reads two lines from stdin and outputs true or false to stdout.

sub p{join'',sort split'',$a}$a=<>;$x=p;$a=<>;print$x eq p()?"true":"false"

(Thanks to Tim for 77 -> 75)

Answer 16

PHP (command line, 87 characters)

function d($s){
    return array_count_values(str_split($s));
}

echo d($argv[1]) == d($argv[2]);

Answer 17

Python 3, 107 97 76 64

s=sorted;a,b=input().split(', ')
print(str(s(a)==s(b)).lower())

Obviously this can be shortened if we don't take the OP's wording literally and lowercase "true" and "false"...

Answer 18

Python, 32 bytes

p=sorted
f=lambda a,b:p(a)==p(b)

Answer 19

Bash, 88 characters

diff <(grep -o .<<<$1|sort) <(grep -o .<<<$2|sort)>/dev/null && echo true || echo false

Answer 20

Javascript

A (very) slightly shorter version of @zzzzBov's solution, that uses .join() instead of String boxing:

function a(b,c){return b.split('').sort().join()==c.split('').sort().join()}
alert(a('abc','bca')); //true

Similarly:

function a(b){return b.split('').sort().join()}
alert(a('abc')==a('bca')); //true

Answer 21

Clojure REPL 41 chars

(= (sort (read-line)) (sort (read-line)))

Answer 22

Java

(everyone's favorite language apparently!)

173 chars:

import java.util.*;class g{public static void main(String[]p){String[]a=p[0].split(""),b=p[1].split("");Arrays.sort(a);Arrays.sort(b);System.out.print(Arrays.equals(a,b));}}

(Doesn't print newline char to save 2 chars from println)

Compile and run:

javac g.java
java -cp . g abcdef fedcba
true

Love to see a shorter one...

Answer 23

R, 54 bytes

function(x,y,`~`=sort,`+`=utf8ToInt)identical(~+x,~+y)

Try it online!

Answer 24

sed, 45 chars

It's even possible in my favourite - sed! Just one regular expression to solve the anagram! Just keep removing the corresponding letters:

:
s/(.)(.*,.*)\1/\2/
t
/\w/{i\false
d}
i\true

(to be invoked with -nE)

Perl, 48

1while s/(.)(.*,.*)\1/\2/;$_=/\w/?"false":"true"

To be invoked with -p.

Perl function, 39

sub f{$,while s/(.)(.*,.*)\1/\2/;!/\w/}

Answer 25

APL, 2 chars

≡⍦

This is the Multiset Match function from Nars2000, one of the leading-edge APL implementations. When applied to strings, it computes exactly the function required:

      'elvis' ≡⍦ 'lives'
1
      'alec guinness' ≡⍦ 'genuine class'
1

Answer 26

05AB1E, 6 4 bytes (non-competing)

{I{Q

Try it online!

This took a while because of input difficulties. Golfed down due to pop.

Explanation:

{I{Q    Original code

{       Takes first input e.g. word and sorts -> 'dorw'
 I      Takes second input e.g. 'wrdo'
  {     Sorts second input -> 'dorw'
   Q    Compare if sorted 1 = sorted 2, then print result. 'dorw' = 'dorw', so prints 1.

Answer 27

JavaScript, 41

Comparison function (41):

a=b=>''+[...b].sort()
b=(c,d)=>a(c)==a(d)

alert(b('abc', 'cba')) // true

Comparator function (21):

a=b=>''+[...b].sort()

alert(a('abc') == a('bca')); //true

Comparator function (48):

function a(b){return String(b.split('').sort())}

alert(a('abc')==a('bca')); //true

Comparison function (78):

function a(b,c){return String(b.split('').sort())==String(c.split('').sort())}

alert(a('abc','bca')); //true

Assumes String has split and Array has sort.

Answer 28

Scala in REPL (32)

readLine.sorted==readLine.sorted

Scala function (43)

def f(a:String,b:String)=a.sorted==b.sorted

Scala program (61)

object A extends App{println(args(0).sorted==args(1).sorted)}

These leverage a neat feature of Scala whereby a String can also be treated as a sequence of characters (Seq), with all the operations on Seq being available.

Answer 29

APL - 13 chars

{(⍺[⍋⍺])≡⍵[⍋⍵]}

Call like this:

      'boat' {(⍺[⍋⍺])≡⍵[⍋⍵]} 'baot'
1
      'baot' {(⍺[⍋⍺])≡⍵[⍋⍵]} 'boat'
1
      (,'a') {(⍺[⍋⍺])≡⍵[⍋⍵]} 'aa'
0

In the last example, 'a' represents a single character, and the prefix , will convert it into a string.

Answer 30

Java (134 bytes)

int[][]c=new int[2][26];
for(int i=0;i<2;i++)for(byte a:args[i].getBytes())c[i][a-97]++;
System.out.print(Arrays.equals(c[0],c[1]));`

This makes an array to count the number of times each letter appears, and then compares the arrays to check if they are equal.

Answer 31

Python, 151, 89 bytes

Handles arbitrarily many inputs from stdin. Comma separated, one pair per line.

import sys
f=sorted
for l in sys.stdin:a,b=l.split(',');print f(a.strip())==f(b.strip())

Answer 32

JavaScript (67)

function(a,b){return ""+a.split("").sort()==""+b.split("").sort()}

Answer 33

VBScript

172 characters

wscript.echo "boat, boat         = " & s("boat, boat")
wscript.echo "toab, boat         = " & s("toab, boat")
wscript.echo "oabt, toab         = " & s("oabt, toab")
wscript.echo "a, aa              = " & s("a, aa")
wscript.echo "zzz, zzzzzzzz      = " & s("zzz, zzzzzzzz")
wscript.echo "zyyyzzzz, yyzzzzzy = " & s("zyyyzzzz, yyzzzzzy")
wscript.echo "sleepy, pyels      = " & s("sleepy, pyels")
wscript.echo "p,p                = " & s("p,p")

function s(a):b=split(replace(a," ",""),","):c=0:for x=1 to len(b(0)):if instr(b(1),mid(b(0),x,1)) then c=c+1 
next:if len(b(1))-c=0 then s=true else s=false
end function

I was kinda suprised I could get it under 200.

Answer 34

Ruby (40)

a.unpack('c*').sort==b.unpack('c*').sort

Answer 35

Another Ruby (46)

(a.size==b.size)&&(a<<b.count(a,b)==a<<b.size)

Answer 36

C function (147 chars), using brute-force

int c(char*x,char*y){int i=0,l=0;for(;y[l]&&x[l];l++);if(y[l]||x[l])return 0;
while(*x&&i!=l){for(i=0;i<l&&y[i]!=*x;i++);y[i]=0,x++;}return(i!=l);}

Answer 37

Ruby 1.9 - 32

x=->{gets.chars.sort}
p x[]==x[]

Answer 38

Ruby (39)

Accepts the input as given in the question. Run with ruby -n.

$ cat t.rb
$_=~/, /;p $'.chars.sort==$`.chars.sort

$ echo -n "word, wrdo" | ruby -n t.rb
true

Answer 39

Brain-Flak, 170 bytes

{({}[(()()()()()){}]<>)<>}<>([]){{}(<>())<>{({}[(((((()()()()()){}){}){})){}])({({})({}[()])()}{}{}())({<({}[()])><>({})<>}{}<><{}>)<>}{}([])}<>({}[{}])((){[()](<{}>)}{})

Try it online!

Takes input on two lines.

This program makes a Gödel encoding of letter frequencies in each string. This means I need to map each letter to a unique prime number. For that, I use the polynomial n^2 + n + 41, which is known to give prime results for 0 <= n <= 40. I subtract 90 from the ascii code for each letter before passing it to the polynomial, to get a number in the proper range. When the primes are multiplied together, the results should be equal only when the strings are permutations of each other.

Answer 40

Brachylog, 2 bytes

pᵈ

Try it online!

The p predicate is essentially a declaration that the variables on either side of it are permutations of each other. The ᵈ meta-predicate modifies the predicate to which it is attached to interpret the input variable as a list of two items, which are effectively placed on either side of the predicate (with the output variable being unified with the second element of the input). So, given a pair of strings as input, the predicate pᵈ will succeed if they are anagrams, and fail if they are not.

If anyone's suspicious of the header on TIO, it just goes through the list of test cases and prints a list of "true"/"false" values based on whether the predicate succeeded or failed for each case. The predicate can be run as a standalone program that prints "true." if it succeeds for the single test case it is given and "false." if it fails.

Answer 41

Matlab: 10 characters without using sort

a*a'==b*b'

Answer 42

Julia - 36

f=s->==(map(sort,map(collect,s))...)

Used as f(["foo", "bar"])

Using a list for calling it kind of feels like cheating though.

43 characters

f=(a,b)->sort(collect(a))==sort(collect(b))

Used as f("foo", "bar")

Both solutions just sort the words and compare the result

Answer 43

Woohoo, my first real CodeGolf submission ^_^

Mathcad, 38 chars including non-character keys

s(a):sort(str2vec(a))
f(a,b):s(a)©=s(b)

© stands for the Ctrl key.

Displayed by Mathcad formatted as:

s(a):=sort(str2vec(a))
f(a,b):=s(a)=s(b)

Converts the strings to vectors (one-dimensional arrays) of symbols' ASCII values, sorts them, then compares the vectors. Input is supplied to function f. Returns 1 on success and 0 on failure.

Example: f("toab","boat") returns 1

Answer 44

R, 77

f=function(x,y)identical(sort(strsplit(x,"")[[1]]),sort(strsplit(y,"")[[1]]))

Sample output:

f("boat","boat")
[1] TRUE
f("toab","boat")
[1] TRUE
f("oabt","toab")
[1] TRUE
f("a","aa")
[1] FALSE
f("zzz","zzzzzzzz")
[1] FALSE
f("zyyyzzzz","yyzzzzzy")
[1] TRUE
f("sleepy","pyels")
[1] FALSE
f("p","p")
[1] TRUE

Answer 45

Pyth 10 7

qFSMczd

Takes input as two space-separated words.

Try it here.

Essentially, it splits the elements into an array, sorts each element, and checks if the elements are unique.

My first attempt at Code Golf. Any advice appreciated :o)

Edit: It should be noted that this is not a valid answer to the code-golf question, seeing as the language Pyth was invented after the question was asked.

Answer 46

PHP, 44 Bytes

<?=($c=count_chars)($argv[1])==$c($argv[2]);

Answer 47

Perl, 41 bytes

40 bytes code + 1 for -p.

Since this uses a slightly different trick to the other Perl answers I thought I'd share it. Input is separated by newlines.

@{$.}=sort/./g}{$_="@1"eq"@2"?true:false

Uses the magic variable $. which tracks the line number to store the words, as character lists, in @1 and @2 which are then compared.

Usage

perl -pe '@{$.}=sort/./g}{$_="@1"eq"@2"?true:false' <<< 'wrdo
word'
true

perl -pe '@{$.}=sort/./g}{$_="@1"eq"@2"?true:false' <<< 'word
wwro'
false

perl -pe '@{$.}=sort/./g}{$_="@1"eq"@2"?true:false' <<< 'boat
toba'
true

Try it online.

Answer 48

C++14, 104 bytes

As generic function returning via reference parameter. Accepts char[] or std::string or any other container that supports range-based for loop.

Returns 0 for anagram, anything else for non-anagram

#define F(X) for(auto x:X)
void f(auto&A,auto&B,int&r){int C[256]={r=0};F(A)C[x]++;F(B)C[x]--;F(C)r|=x;}

Ungolfed and usage:

#include<iostream>

#define F(X) for(auto x:X)

void f(auto&A,auto&B,int&r){
  int C[256]={r=0}; //declare counting array and set return value to 0
  F(A)C[x]++;       //increase first string chars
  F(B)C[x]--;       //decrease second string chars
  F(C)r|=x;         //if any char is not zero
}

int main(){
  int r;
  #define P(a,b) f(a,b,r); std::cout << a << ", " << b << " -> " << r << "\n"
  P("hello","wrong");
  P("hello","olleh");
  P("zz","zzzzzz");
}

Answer 49

Jelly, 3 bytes (probably non-competing)

Ṣ€E

Try it online!

Similar to X88B88's solution, but this takes a list of strings instead of two arguments, like ["String1", "String2"].

Explanation:

Ṣ€      Sort each input
  E     Check equality

Answer 50

Mathematica 28 Bytes

Split into list of characters, sort and test for equality.

Equal@@ Sort/@Characters[#]&

Usage

%@{"BOAT", "TOAB"}

Output

True

Answer 51

Lua, 140 chars

t={io.read():match"(%w+), (%w+)"}T=table for k=1,2 do w={t[k]:byte(1,-1)}T.sort(w)t[k]=T.concat{string.char(unpack(w))}end print(t[1]==t[2])

... like driving a screw with a light weight hammer, at least for code-golfing

Answer 52

Scala, 84 characters

object A{def main(a: Array[String]){println(a(0).sortWith(_<_)==a(1).sortWith(_<_))}}

This one's slightly longer, but doesn't use sorting (92 characters):

object A{def main(a:Array[String]){print((a(0)diff a(1)).isEmpty&&(a(1)diff a(0)).isEmpty)}}

Answer 53

Python 2, 79 bytes

I thought I would add a different Python approach:

import sys
for l in sys.stdin: not reduce(cmp,map(sorted,l.strip().split(',')))

Or as a function:

def f(*v): return not reduce(cmp,map(sorted,v))

Answer 54

Yet another Python answer :). (43 characters including whitespace)

This also includes reading in input and displaying output.

i,s=raw_input,sorted
print s(i())==s(i())

Answer 55

C++ Counting sort, fixed version

int a(char*x,char*y){int i=0,u[256];while(i<256)u[i++]=0;
while(*x&&*y)u[*x++]++,u[*y++]--;if(*x||*y)return 0; 
for(i=255;i&&!u[i--];);return!i;}

Unrolled, so you can see what's going on:

int a(const char *x,const char *y) {
    int i=0,u[256];for(;i<256;u[i++]=0);
    for(i=0;x[i]&&y[i];i++)u[x[i]]++,u[y[i]]--;
    if(x[i]||y[i])return 0;
    for(i=0;i<256 && !u[i];i++);
    return (i==256);
}

(with due credit to Matthew Read's very elegant counting sort strategy)

Answer 56

Matlab (24)

Given two strings a and b.

isequal(sort(a),sort(b))

Answer 57

Common Lisp, 50 bytes

(lambda(a b)(string=(sort a'char<)(sort b'char<)))

Try it online!

Answer 58

Perl 5, 27 + 3 (-lF) = 30 bytes

@{$.}=sort@F}{say"@1"eq"@2"

Try it online!

Answer 59

Perl - 78 characters[1]

@x=map{join("",sort(split("")))}split(",",<>);print$x[0]eq$x[1]?"true":"false"; 

[1] Unlike some other Perl code above, this actually reads the input in "foo,bar" format and prints "true" or "false". Could be shorter otherwise.

Answer 60

Python 3 (64)

This is not a particularly good golf given how short some of the earlier Python solutions that use sorted() are, but here's a version that uses collections.Counter (an unlovable module from a golfing perspective, but with some neat stuff). It reads two lines from input and outputs True or False. Going with Python 3 versus 2.7 saved 4 chars with input() instead of raw_input(), but lost 1 for the parens for print since it is now a function.

from collections import *;c,i=Counter,input;print(c(i())==c(i()))

Answer 61

SmileBASIC, 49 bytes

INPUT A$,B$WHILE""<A$B$[INSTR(B$,POP(A$))]="
WEND

Throws an error when the strings aren't anagrams.

Answer 62

JavaScript, shortest JS answer so far: 57 56 characters, acccepts user input

This prompts the user for a value to compare.

function _(){return prompt().split(0).sort()+''}_()==_()

If no user input is required, this can be trimmed down to 53 52 characters.

Assuming the following variables are set:

var a='test', b='sets';

you can test for it with the following:

function _(a){return a.split(0).sort()+''}_(a)==_(b)

Note: this answer relies on some quirk that allowed using .split(0) instead of .split(""). This behavior no longer exists (at least in Firefox), so to get it running today, you have to replace 0 with "".

Answer 63

Python - 137 chars

def h(s):
    r={}
    for c in s:
        try:r[c]+=1
        except:r[c]=1
    return r
x=raw_input().split(',')
print h(x[0])==h(x[1])

Sample: (I defined a function anagram to do the work of the last 2 lines.)

   anagram('boat','boat')
True
    anagram('toab','boat')
True
    anagram('oabt','toab')
True
    anagram('a','aa')
False
    anagram('zzz','zzzzzzzz')
False
    anagram('zyyyzzzz','yyzzzzzy')
True
    anagram('sleepy','pyels')
False
    anagram('p','p')
True

Answer 64

Python 2, without sorting as that starts to get boring - 104 chars

def f(a,b):
 a,b=list(a),list(b)
 while a:
    try:b.remove(a.pop())
    except:return
 return len(a)==len(b)

Answer 65

Haskell, 48

Not counting imports, Haskell can do it in 10 chars:

import Data.List
import Data.Function
on(==)sort

Called like this:

λ> on(==)sort "balaclava" "aabllcav"
False

λ> on(==)sort "balaclava" "aabllcava"
True

Answer 66

Clojure - 30 chars

Too many people seem to be relying on sorting so I thought of an interesting alternative way to do this via a histogram:

#(apply = (map frequencies %))

Use this as a function, i.e.:

(#(apply = (map frequencies %)) ["boat" "toab"])
=> true

Answer 67

Python 3 (66)

Like jloy's answer, I eschewed using sorted because so many other answers did so. I also didn't want to rehash his use of collections.Counter so I used str.count instead. With these constraints I got within 3 characters of jloy.

i=input;a=i();b=i();print(all(a.count(s)==b.count(s)for s in a+b))

Answer 68

k4 - 11 chars

Given strings a and b:

(a@<a)~b@<b

  a:"word"
  b:"wrdo"
  (a@<a)~b@<b
1b

At the cost of 2 chars this can be made into a function:

{(x@<x)~y@<y}["word";"wrdo"]
1b

Implementation is same as the J implementation; sort the vectors then compare equivalence.

~ is match

< is grade up (indices were the vector to be sorted ascending)

@ is index

Answer 69

Pyth - 8 7 6

MqSGSH

Defines function with two args, the two words.

M       define function g with two args, G and H
 q       equals
  SQ     sorted first arg
  SH     sorted last arg

If that cheating golfscript program counts with hardcoded input, we can do that too with : qSS

And for just two more characters you can have it check an infinite number of words:

ql{mSdQ1

q    1       Equals 1
 l           Length
  {          Set constructor (eliminate all duplicates)
   m  Q      Map on evaluated input
    Sd       Sort each element

Answer 70

Q, 25

{(~). asc each(,/)each x}  

sample output:

q){(~). asc each(,/)each x}("boat";"boat")
1b
q){(~). asc each(,/)each x}("toab";"boat")
1b
q){(~). asc each(,/)each x}("oabt";"toab")
1b
q){(~). asc each(,/)each x}("a";"aa")
0b
q){(~). asc each(,/)each x}("zzz";"zzzzz")
0b
q){(~). asc each(,/)each x}("zyyyzzzzz";"yyzyzzzzz")
1b
q){(~). asc each(,/)each x}("p";"p")
1b

Answer 71

C - 107 chars

Mark off chars in second string as we go. At the end, if we've passed over the entirety of both strings, then we've got a match.

i;main(p,v)char**v,*p;{for(;*v[1]&(p=strchr(v[2],*v[1]++));)*p=1,i++;puts(*(v[2]+i)|*v[1]?"false":"true");}

Answer 72

PowerShell, 78 51 bytes

param([char[]]$a,[char[]]$b)(diff $a $b).Length-eq0

Takes the two string inputs, and re-casts them as char-arrays. The diff function (an alias for Compare-Object) takes the two arrays and returns items that are different between the two. We leverage that by re-casting the return as an array with (), and then checking its length. If the length is zero, that means that all items of both character arrays are exactly the same (because nothing was returned). PowerShell has an implicit write for evaluated statements like this, so will automatically write out True or False as required.

Answer 73

C, (108)

char c[192]={};main(){for(;*a;c[127-*a++]++);for(;*b;c[223-*b++]++);puts(memcmp(c,c+96,95)?"false":"true");}

Answer 74

Q, 15 Bytes

f:{~/{x@<x}'x}

f is the name of the function

Test

f("boat";"boat")            /1b
f("toab";"boat")            /1b
f("oabt";"toab")            /1b
f(,"a";"aa")                /0b
f("zzz";"zzzzzzzz")         /0b
f("zyyyzzzz";"yyzzzzzy")    /1b
f("sleepy";"pyels")         /0b
f(,"p";,"p")                /1b

Explanation

Argument of the function is a sequence with both words. At each word applies {x@<x}, that sorts x (take from x in ascending index ordering). ~/ reads as "match over", and compares both transformed words

Answer 75

awk, 63 65 bytes

It only accepts chars a-z (0141-0172). gsub counts occurrances of each char in alphabet, appends them to a variable (variable looks something like 140213000000000000..., starts with 140 for initialization) and compares frequencies in variables in the end. It returns the value of the comparison on exit:

{for(a=i=140;++i<173;)a=a gsub("\\"i,1);if(p>1)exit(p==a);p=a}

Test it:

$ cat file
aabccc
abcacc
$ awk '{for(a=i=140;++i<173;)a=a gsub("\\"i,"");if(p>1)exit(p==a);p=a}' file
$ echo $?  # will output 0 or 1 where 1=true and 0=false
1

Answer 76

Japt, 10 9 5 4 bytes

á øV

Try it

á øV     :Implicit input of strings U & V
á        :Permutations of U
  øV     :Contains V?

Answer 77

APL (Dyalog), 10 bytes

≡/(⊂∘⍋⌷⊢)¨

Try it online!

The program takes in two strings in a single array as its right argument.

Explanation

¨            For each string
 (⊂∘⍋⌷⊢)      Sort it
≡/           And check if both strings are equal

Answer 78

Perl, 60 bytes

Perl one-liner

perl -lanF -e 'push@a,join"",sort@F;END{exit($a[0]ne$a[1])}'

Takes first two lines of STDIN and compares them. Exits 0 if they're anagrams, 1 otherwise.

$ echo -e "word\nwrdo" | \
> perl -lanF -e 'push@a,join"",sort@F;END{exit($a[0]ne$a[1])}' && echo anagram
anagram
$ echo -e "word\nwrro" | \
> perl -lanF -e 'push@a,join"",sort@F;END{exit($a[0]ne$a[1])}' && echo anagram
$

Answer 79

C#

bool IsJumbledPair(string a, string b) {
   if(a.Length!=b.Length) return false;
   foreach(char c in a.ToCharArray()) {
      int i = b.IndexOf(c);
      if(i<0) return false;
      b = b.Remove(i,1)
   }
   return (b.Length==0);
}

Readable, and does not require sorting.

Another method:

bool IsJumbledPair(string a, string b) {
   string c;
   while {
     if(a.Length!=b.Length) return false;
     if(a.Length==0) return true;
     c = a.Chars(0).ToString();
     b = b.Replace(c, "");
     a = a.Replace(c, "");
   }
}

The above version works better when the strings contain a number of duplicated characters, as each distinct character is only compared once using the relatively fast Replace() method.

VB.NET

Same as the first C# example, slightly simpler using Replace() method, but doesn't short-cut like the C# version:

Function IsJumbledPair(a As String, b As String) As Boolean
   If a.Length <> b.Length Then Return false
   For Each(c As Char In a.ToCharArray())
      b = Replace(b, c.ToString(), "", 1, 1)
   Next
   Return (b.Length=0)
}

Answer 80

PHP (51 chars, compressed)

function d($s){$s=str_split($s);sort($s);return$s;}

(Split string into an array, sort the array and return)

Example:

var_dump(d("word")===d("drow"));

Answer 81

Ruby (35)

a,b=ARGV;a.chars.sort==b.chars.sort

Answer 82

C# 118 chars

using System.Linq;namespace A{class P{static void Main(string[] a){System.Console.Write(!a[0].Except(a[1]).Any());}}}

Readable:

using System.Linq;

namespace A
{
    class P
    {
        static void Main(string[] a)
        {
            System.Console.Write(!a[0].Except(a[1]).Any());
        }
    }
}

Answer 83

Groovy, 54

print args[0].toList().sort()==args[1].toList().sort()

Answer 84

Excel VBA, 87 Bytes

Anonymous VBE immediate window function that takes input from range A1:B1 and outputs whether the two inputs are anagrams of one another to the VBE immediate window. This a destructive process as the value inputted into range B1 is destroyed

For i=1To[Len(A1)]:[B1]=Replace([B1],Mid([A1],i,1),"|",,1):Next:?[B1=Rept("|",len(A1))]

Answer 85

Wolfram Language 26 bytes ( Mathematica )

ContainsAll@@Characters@#&

Usage:

%@{"BOAT","TOAB"}

Output:

True

Answer 86

J, 8 bytes

-:&(/:~)

Match -: after sorting /:~ both args &.

Try it online!

Answer 87

Axiom, 45 bytes

f(a:String,b:String):Boolean==sort(a)=sort(b)

Answer 88

Julia, 41 bytes

g(a,b)=sort(collect(a))==sort(collect(b))

Answer 89

Python, 99 chars

f=lambda w,x,y:w==y if x==""else any([f(w+a,x.replace(a,"",1),y)for a in x]) 
g=lambda x,y:f("",x,y)

Try it Online!

Recursively finds all permutations of x and compares them to y.

Answer 90

Tcl, 82 bytes

proc A x\ y {expr {[string le $x]==[string le $y]&[lsort -u [split $x $y]]=="{}"}}

Try it online!

---

Tcl, 83 bytes

proc A x\ y {expr {[string le $x]==[string le $y]&&[lsort -u [split $x $y]]=="{}"}}

Try it online!

Tcl, 84 bytes

proc S s {lsort [split $s ""]}
proc A x\ y {set x [S $x]
set y [S $y]
expr {$x==$y}}

Try it online!

Previous version was failing for comparing 2 to 20, because after the lsort and join the comparison became 2==02, and == was doing a numerical comparison instead of a numerical one!

---

Tcl, 94 bytes

proc S s {join [lsort [split $s ""]] ""}
proc A x\ y {set x [S $x]
set y [S $y]
expr {$x==$y}}

Try it online!

Answer 91

Pyth, 4 bytes

_ISM

Try it Online

Explanation

_ISM
  SMQ   Sort both (implicit) inputs.
_I      Check if the result is invariant under reversing.

Answer 92

Perl, 26 bytes

Includes +1 for p (using -F and @F ends up with the same score)

Give the input strings as 2 lines on STDIN.

(echo word; echo wrdo) | perl -pE '$\=${join W,sort/./g}++}{'

This prints 0 if not an anagram or 1 if it is an anagram

(echo word; echo wrdo) | perl -pE '$_=${join W,sort/./g}++'

comes in at 24 bytes and prints 00 if not an anagram and 01 if it is. If these values are numbers they are in fact valid falsy and thruthy in perl, but as strings they are both thruthy. It's probably fairer to consider them as strings so this solution is invalid

So this is the same length (but without warnings) as Xcali's answer (after optimizing):

(echo word; echo wrdo) | perl -F -E '@{$.}=sort@F}{say@1~~@2'

Answer 93

MathGolf, 4 bytes

ms~=

Try it online!

Explanation

m      explicit map
 s     sort(array)
  ~    dump array to stack
   =   pop(a, b), push(a==b)

Answer 94

Pepe, 58 bytes

REEREEEEeEERREEeREEEEeEeErRREEEEEeEEreerrEEREEeeReEErEereE

Try it online! Input is in the form a;b. Outputs 0 for truthy and none for falsy.

Answer 95

Pushy, 13 bytes

Assumes the two inputs are separated by a space.

K32-$v;.gFgx#

Try it online!

                 \ Implicit: input on stack as character codes
K32-             \ Subtract 32 from all, making SPACE = 0
    $            \ Until 0 (space) on top of stack:
     v;          \   Send last char to auxiliary stack
       .         \ Space has been reached, pop it.
                 \ Now each stack contains all the letters of one of the words.
        gFg      \ Sort both stacks
           x     \ Check their equality
            #    \ Print result (1 if true, 0 if false)

Answer 96

C# (.NET Core), 108 bytes

Without LINQ. Returns 1 for true, 0 for false.

(a,b)=>{int x=a.Length==b.Length?1:0;foreach(char c in a){try{b.Remove(b.IndexOf(c));}catch{x=0;}}return x;}

Try it online!

Answer 97

Lua 112

function q(a,b)s=table.sort;t=table.concat;a={a:byte(1,#a)}s(a);b={b:byte(1,#b)}s(b)return t(a,".")==t(b,".")end

Answer 98

JavaScript 87

alert((a=prompt().split(","))[0].split("").sort().join()==a[1].split("").sort().join())

Prompt requires comma separated list of two "words"

[Prompt]

btoa,boat

Output: true

Answer 99

import java.util.Scanner;
public class Match 
{
    public static void main(String[] args) 
    {
        Scanner s=new Scanner(System.in);
        String s1=s.nextLine();
        String s2=s1.toLowerCase();
        char []a=s2.toCharArray();
        String s3=s.nextLine();
        String s4=s3.toLowerCase();
        char []e=s4.toCharArray();
        int h,g = 0,l=0,r=0,h1=0,m=0;
        int b=(int)a[0];
        int f=(int)e[0];
        //System.out.println();
        for(int k=0;k<a.length;k++)
        {
            for(h=1+k;h<a.length;h++)
            {
                if(b>(int)a[h])
                {   
                    b=a[h];
                    char temp=a[h];
                    a[h]=a[k];
                    a[k]=temp;
                }

            }
        if(l<a.length-1)
        {
                b=a[++l];
        }
        }
        for(int k1=0;k1<e.length;k1++)
        {
            for(h1=1+k1;h1<e.length;h1++)
            {
                if(f>(int)e[h1])
                {   
                    f=e[h1];
                    char temp=e[h1];
                    e[h1]=e[k1];
                    e[k1]=temp;
                }

            }
        if(m<e.length-1)
        {
            f=e[++m];
        }
        }
        if(a.length==e.length)
        {
            int flag=0;
            for(int j=0;j<a.length;j++)
            {
                if(a[j]==e[j])
                {
                    flag++;
                }
            }
            if(flag==e.length)
            {
                System.out.println("True");
            }
            else
            {
                System.out.println("false");
            }
        }
        else
        {
            System.out.println("False");
        }   
    }
}

Answer 100

C#, 144 chars

namespace System.Linq{class m{static void Main(string[]a){Console.Write(a[0].Select(t=>a[1].Select(y=>t!=y)).Count()*2==(a[0]+a[1]).Length);}}}

Function: 90 chars

bool i(string a,string b){return a.Select(t=>b.Select(y=>t!=y)).Count()*2==(a+b).Length;}

Answer 101

Bash (66 58)

f(){ fold -w1<<<$1|sort;}
g(){ [ "$(f $1)" == "$(f $2)" ];}

Call it with g <word1> <word2>.

Edit: Stupid me, I do not need to unic -c after I sort

Answer 102

VB.net

Module Q
Sub Main(a As String())
 Console.WriteLine(a(0).Count=-a(0).Sum(Function(c)a(1).Sum(function(x)x=c)))                          
End Sub
End Module

Answer 103

C 138

Code finds the sum of the differences of the characters then checks if they are same length and sum == 0.

int s,t,i,j;
void f(char*a,char*b)
{
while (*a&&*b){
s+=*a-*b;
a++;b++;
i++;j++;
}
if (i==j&&s==0)t=1;
puts(t?"true":"false");
}

Answer 104

Groovy 44

def f(a,b){print !(a-b)&&a.size()==b.size()}

Answer 105

Groovy 42

def f(a,b){print ((b as Set)==(a as Set))}

Answer 106

Python, 20 bytes

lose to golfscript again...... :(

sorted(x)==sorted(y)

Answer 107

PHP (71)

function a($a){while($i<$r=strlen($a))$t+=ord($a[$i++]);return"$r|$t";}

Use:

echo (a("abc")==a("cba"));

Answer 108

F# (59 chars)

let f a b=(Seq.sort>>Seq.toArray)a=(Seq.sort>>Seq.toArray)b

Very annyoying that sequences can' be directly compared.

Answer 109

C - 79 90 chars (using strchr & strlen)

r(char*s,char*t){int i=0;for(;*s&&strchr(t,*s++);i++);return i==strlen(t)?1:0;}

Ungolfed...

r( char *s, char *t )
{
    int i=0;
    for (; *s && strchr(t, *s++); i++)
        ;
    return i == strlen(t) ? 1 : 0;
}

C - 115 110 chars (brute with histogram)

int k,u[256];ρ(char*s,char*t){char*v=s;for(;*s;)u[*s++]++;for(;*t;)u[*t++]--;while(*v&&!(k=u[*v++]));return k?0:1;}

Ungolfed...

int k, u[256];
r(char *s, char*t)
{
    char *v = s;

    for (;*s;)
        u[*s++]++;
    for (;*t;)
        u[*t++]--;

    while(*v && !(k=u[*v++]) )
        ;
    return k ? 0 : 1;
}

EDIT: fixed bug in brute version.

EDIT: added brute version (no library functions).

Answer 110

Python 121 Characters

It's not a winner wrt length, but I didn't use sorted!

from sys import argv as s
for i in s[1]+s[2]:
 if not s[1].count(i)==s[2].count(i):
  print 'False'
  break
else:
 print 'True'

Answer 111

Ada 2005 - 314 218

type V is array(Character)of Natural;function A(L,R:String)return Boolean is
C,D:V:=(others=>0);begin
for I in L'Range loop
C(L(I)):=C(L(I))+1;end loop;for I in R'Range loop
D(R(I)):=D(R(I))+1;end loop;return C=D;end;

Answer 112

Coffee Script (52)

a=(b,c)->`b.split('').sort()==c.split('').sort()+''`

usage

console.log a 'god', 'dog'

Answer 113

pure bash 136

There is a method without sorting step!

anagramCmp() {
    x=0 y=0
    for((i=0;i<${#1};i++));do
        ((x+=7**(36#${1:i:1}-10)))
        ((y+=7**(36#${2:i:1}-10)))2>/dev/null
    done
    return $((${#1}$x==${#2}$y?0:1))
}

Then now:

if anagramCmp Blah Halb ; then echo Yo; else echo Uh; fi
Yo

if anagramCmp Blahblah Hhaallab ; then echo Yo; else echo Uh; fi
Uh

Answer 114

Javascript - 57 chars

function o(a,b){return b?o(a)==o(b):1+a.split("").sort()}

Answer 115

Groovy, 46

b={args[it].toList().sort()}
print b(0)==b(1)

Answer 116

Java, 173 chars

import java.util.*;class C{public static void main(String[]a){byte[]d,f;Arrays.sort(d=a[0].getBytes());Arrays.sort(f=a[1].getBytes());System.out.print(Arrays.equals(f,d));}}

It's the same that Guus one but changing a couple of methods.

Answer 117

Pyth - 19

qS@cz", "0S@cz", "1

Try it here. Note that the two words in the input must be separated by a comma and a space.

Python Mapping and explanation

q                     #      equal(              "Check equality"
 S                    #      sorted(             "Sort"
  @                   #      list looku          "Extract element from list"
   c                  #      chop(               "Separate by delimiter"
    z                 #      input(              "Input variable"
     ", "             #      ", "                "This delimiter"
         0            #      0                   "The 0th element"
          S@cz", "1   #      ..                  "The same thing with the 1st element

I'm sure someone can come up with a shorter implementation, but I've only been learning Pyth for a few days. It would also be a lot shorter if I didn't stick strictly to the input format given in the question.

For instance:

qS@Q0S@Q1

Which is only 9 bytes, works if the input is formatted like 'parse','spare'. So which one do you count?

Good practice though!

Answer 118

Ruby, 32

h=->{gets.chars.sort}
p h[]==h[]

Answer 119

Javascript, 143 119 bytes

a=prompt()[s="split"](" ");a[0]=a[0][s]("");a[1]=a[1][s]("");a[0].sort();a[1].sort();alert(a[0][j="join"]()==a[1][j]())

Takes a single space-seperated pair of strings as input.

Answer 120

rs, 35 bytes

+(.)(.*) (.*)\1/\2 \3
\s/
.+/0
^$/1

Yay regexes!!

BTW, rs was created way after this was posted, so this technically doesn't count. Still cool.

Live demo and all test cases.

Answer 121

CoffeeScript 129

Longer than the other CoffeeScript entry, but this one uses recursive string comparison, rather than just comparing sorted strings:

z=(x,y)->d=y.length;e=x.length;return 1if(!d&&!e);b=y.indexOf x[0];return 0if b<0;f=x[1..e];g=y[b+1..d];g=y[0..b-1]+g if b;z(f,g)

Outputs 1 or 0 indicating whether the strings are anagrams or not.

Answer 122

Javascript (ES6) 35

Requires lambdas

Just like the other solutions, but with lambda

f=x=>x.split``.sort()
f(a)==f(b)+""

Usage

// a, b = input
f=x=>x.split``.sort()
console.log(f(a)==f(b)+"")

thanks @comment

Answer 123

R, 63 Bytes

length(unique(lapply(strsplit(c(s,m),''),sort))[[1]])==nchar(m)

It seens it passes the test cases.

(just realized this question is 4 year old, but posting anyway)

Answer 124

Powershell 48 bytes

param([char[]]$a,[char[]]$b)![bool](diff $a $b)

Cast incoming strings as an array of char. Use diff (compare-object) on the two objects and cast as bool. Since a blank result (diff only shows differences not similarities) is False, negating it with ! will result in true for identical strings.

Added benefit: it will work on any arbitrary length strings (if equal whitespace).

Answer 125

Javascript 70 (with primitive GUI)

Here's a Javascript entry that also includes a primitive GUI via two prompts and an alert.

function a(){return prompt('').split('').sort().join()}alert(a()==a())

Have a play – http://jsfiddle.net/liamnewmarch/jGues/

Answer 126

Pyke, 4 bytes (non-competing)

Pyke is FAR newer than this challenge

mSXq

Try it here!

Also 4 bytes

SRSq

Try it here!

Answer 127

Retina, 10 bytes

This answer is non-competing since Retina is much newer than this challenge. Byte count assumes ISO 8859-1 encoding.

%O`.
D`
¶$

Input is linefeed-separated.

Try it online!

Explanation

%O`.

This sorts (O) the individual characters (.) in each line (%), i.e. it sorts each input string separately.

D`

This deduplicates the input on the (implicit) regex .*, which means it removes the characters from the second line if both strings are equal.

¶$

Finally, this tries to match a linefeed followed by the end of the string. Since the input strings are guaranteed to be non-empty, this can only happen if the second string was removed in the previous stage.

Answer 128

Jelly, 5 Bytes (Non-competitive)

Ṣ⁼⁴Ṣ¤

Special thanks to Leaky Nun for helping me fix a problem.

Outputs one if true. Try it Online!

Answer 129

R, 69 bytes

a=function(b,d)identical(table(strsplit(b,"")),table(strsplit(d,"")))

I think this is the shortest R implementation which deals with a("b","bb") being FALSE.

Answer 130

Racket 92 bytes

(λ(a b)(define(g s)(sort(map(λ(x)(char->integer x))(string->list s))>))(equal?(g a)(g b)))

Ungolfed:

(define f
  (λ (a b)
    (define (g s)    ; fn to get sorted list of string chars as numbers.
      (sort
       (map
        (λ(x)(char->integer x))
        (string->list s))
       >))
    (equal? (g a) (g b))))

Testing:

(f "test" "ttse")
(f "boat" "boat")
(f "toab" "boat")
(f "oabt" "toab")
(f "zyyyzzzz" "yyzzzzzy")
(f "a" "aa")
(f "zzz" "zzzzzzzz")
(f "sleepy" "pyels")

Output:

#t
#t
#t
#t
#t
#f
#f
#f

Answer 131

Scala - 64 Bytes

Could be made shorter (the toList could be changed to toSeq, I believe) ('a'to'z').forall(c=>y.toList.count(_==c)==z.toList.count(_==c))

Answer 132

C 110 bytes

char*a,*b;l(i,j,k){y:if(!b[j]||!a[i])return i==k+1;if(a[i]==b[j]){b[j]=1;return l(i+1,0,j>k?j:k);}++j;goto y;}

this is one recursive function that has one not common way to have its input and write in one its argument... but not use library functions....

/*
l(i,j,k)
{y: if(!b[j]||!a[i])return i==k+1;
    if(a[i]==b[j]){b[j]=1;return l(i+1,0,j>k?j:k);}
    ++j;goto y;   
}
//110
r=1 m1=12345  m2=?????
r=0 m3=a  m4=e
*/

main()
{char m1[]="12345", m2[]="54321", m3[]="a", m4[]="e";
 int  r;

 a=m1;b=m2;r=l(0,0,0);printf("r=%d m1=%s  m2=%s\n", r, m1, m2);
 a=m3;b=m4;r=l(0,0,0);printf("r=%d m3=%s  m4=%s\n", r, m3, m4);
}

Answer 133

C 87 bytes

char*a,*b,*c;d(i,k){return!a[i]||!(c=strchr(b,a[i]))?i==k+1:(*c=1,d(i+1,k<c-b?c-b:k));}

this is one recursive function that has one not common way to have its input and write in one its argument...this use recursion + library function...

/*
char*a,*b,*c;
d(i,k)
{return !a[i]||!(c=strchr(b,a[i]))?i==k+1:(*c=1,d(i+1,k<c-b?c-b:k));}
87 
*/

 main()
{char m1[]="12345", m2[]="54321", m3[]="a", m4[]="e";
 int  r;

 a=m1;b=m2;r=d(0,0);printf("r=%d m1=%s  m2=%s\n", r, m1, m2);
 a=m3;b=m4;r=d(0,0);printf("r=%d m3=%s  m4=%s\n", r, m3, m4);
}