89
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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 az
  • 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
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6
  • 10
    \$\begingroup\$ 9 answers in 13 views... wow! \$\endgroup\$
    – Tom Gullen
    Mar 8, 2011 at 16:44
  • \$\begingroup\$ @Tom, because everyone wanted to prove that your comment about using a 64-bit integer was pointing in the wrong direction :P \$\endgroup\$ Mar 8, 2011 at 18:15
  • 5
    \$\begingroup\$ Title request: Cod Elf, Go! \$\endgroup\$
    – user54200
    Jul 9, 2016 at 12:48
  • 7
    \$\begingroup\$ "Falcon Rage, go mad!" \$\endgroup\$
    – Geobits
    Oct 6, 2016 at 17:25
  • 11
    \$\begingroup\$ My name suggestion: "are they anagrams" → "manage the arrays" \$\endgroup\$ Oct 31, 2017 at 4:32

145 Answers 145

1
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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
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1
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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
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1
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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
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2
  • \$\begingroup\$ How would qSS work? I am trying to learn pyth and I am curious. \$\endgroup\$ Jan 27, 2015 at 20:07
  • \$\begingroup\$ @ericmark26 its the same as the golfscript and its cheating because it need hardcoding. qss just means equals, sort, sort. You'll have to put the strings after the S's \$\endgroup\$
    – Maltysen
    Jan 27, 2015 at 20:35
1
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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");}
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1
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C, (108)

char c[192]={};main(){for(;*a;c[127-*a++]++);for(;*b;c[223-*b++]++);puts(memcmp(c,c+96,95)?"false":"true");}
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1
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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.

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1
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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

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1
  • \$\begingroup\$ If you're using k rather than Q you can just do ~/x@'<:'x: for 10 bytes. No need to create the outer function f nor the inner function if you set x to be the input. \$\endgroup\$
    – mkst
    Oct 31, 2017 at 9:19
1
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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 "".

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1
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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");
}
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1
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Mathematica 28 Bytes

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

Equal@@ Sort/@Characters[#]&

Usage

%@{"BOAT", "TOAB"}

Output

True
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1
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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
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0
1
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Wolfram Language 26 bytes ( Mathematica )

ContainsAll@@Characters@#&

Usage:

%@{"BOAT","TOAB"}

Output:

True

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1
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Perl 5, 27 + 3 (-lF) = 30 bytes

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

Try it online!

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1
  • \$\begingroup\$ The -l option is not needed. It would still count as +3 though since you must count the space too (since you cannot bundle the -F with -E. You can however do the final match using @1~~@2 saving 4 bytes (or 3 if you add an extra X option letter to supress the warning). Very neat solution by the way. Have a +1 \$\endgroup\$
    – Ton Hospel
    Feb 4, 2018 at 11:43
1
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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

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1
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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))
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1
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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)}}
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3
  • 1
    \$\begingroup\$ You can use sorted instead of sortWith. I posted another reply with sorted as well as REPL and function versions. \$\endgroup\$
    – ebruchez
    Mar 9, 2011 at 6:15
  • \$\begingroup\$ @ebruchez Thanks, I'm just learning Scala at the moment and these make good exercises. :-) \$\endgroup\$
    – Gareth
    Mar 9, 2011 at 10:19
  • \$\begingroup\$ Still learning myself ;) \$\endgroup\$
    – ebruchez
    Mar 9, 2011 at 16:17
1
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Common Lisp, 50 bytes

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

Try it online!

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1
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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?
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2
  • \$\begingroup\$ As of v2.0a0, ¬n eV¬n works with 7 bytes. \$\endgroup\$
    – Bubbler
    May 21, 2018 at 7:42
  • \$\begingroup\$ øVá \$\endgroup\$
    – Oliver
    Feb 1, 2019 at 2:22
1
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SmileBASIC, 49 bytes

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

Throws an error when the strings aren't anagrams.

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1
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Symja, 26 bytes

f(x_,y_):=Sort(x)==Sort(y)

Try It Online!

Try With Changeable Tests

Assumes strings are given as a list of characters.

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1
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Vyxal, 1 byte

Try it Online!

An empty list signifies truthy and a non-empty list signifies falsiness. is the built-in for set(a) ^ set(b) in Python (I'm still not sure what that actually does).

But if that's bending things a bit too far:

Vyxal, 2 bytes

⊍¬

Try it Online!

Explained

⊍¬
⊍  # set(first) ^ set(second)
   # logical not that and implicitly output
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1
1
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TI-Basic, 83 bytes

Prompt Str1,Str2
"seq(inString(Str1+Str2,sub(Ans,I,1)),I,1,length(Ans→u
Str1
u→A
SortA(ʟA
Str2
u→B
SortA(ʟB
0
If dim(ʟA)=dim(ʟB
min(ʟA=ʟB
Ans

Output is stored in Ans and displayed. Outputs 1 if the inputs are anagrams, otherwise 0.

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0
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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"));
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0
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Ruby (35)

a,b=ARGV;a.chars.sort==b.chars.sort
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1
  • 1
    \$\begingroup\$ (Five years later...) $* is a handy alias for ARGV for 2 bytes. \$\endgroup\$
    – Jordan
    Aug 18, 2016 at 5:32
0
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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());
        }
    }
}
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1
  • \$\begingroup\$ You can remove an extra character: string[]a You also don't need a namespace. \$\endgroup\$
    – ICR
    Dec 7, 2011 at 13:54
0
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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)
}
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3
  • \$\begingroup\$ It is likely that the downvotes are because you have made no effort to golf (see the tag?) your solutions, or because you have mixed several distinct solutions in one answer. \$\endgroup\$ Mar 9, 2011 at 22:50
  • 1
    \$\begingroup\$ Meh. If I play real golf, it's for socializing or beer, not score-keeping. Guess I should have actually been playing to win... :) \$\endgroup\$ Mar 11, 2011 at 6:17
  • \$\begingroup\$ People generally respond well to solutions in wordy languages (I play in Fortran 77 from time to time), as long as you make an effort to write aggressively compacted <wordy language>, and especially if your languages supports a interesting trick (i.e. fortran has computed branches). \$\endgroup\$ Mar 11, 2011 at 16:10
0
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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.

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0
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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;
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0
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Coffee Script (52)

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

usage

console.log a 'god', 'dog'
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0
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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'
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3
  • \$\begingroup\$ You can save 2 characters by changing the import to from sys import* (newline) s=argv. \$\endgroup\$ Sep 13, 2011 at 14:45
  • \$\begingroup\$ Algorithm fails with s[1]='the',s[2]='them'. \$\endgroup\$ Sep 13, 2011 at 14:54
  • \$\begingroup\$ You could avoid the need to dereference s by from sys import* (newline) _,a,b=argv. \$\endgroup\$ Sep 13, 2011 at 17:19

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