# Determine if all decimal digits are unique

Deleted questions on Stack Overflow sometimes make for great golf material.

Write a function that takes a nonnegative integer as input, and returns true if all the digits in the base 10 representation of that number are unique. Example:

48778584 -> false
17308459 -> true


Character count includes only the function.

If you choose to answer in C or C++: no macros, no undefined behaviour; implementation-defined behaviour and compiler warnings are fine.

• I'd still be interested in other C or C++ solutions as per the question that inspired this one. – Thomas May 21 '14 at 19:59
• Why no C or C++ macros or undefined behavior? That's oddly limiting to just two languages. – dfeuer Apr 14 '19 at 2:05

## C, 76

This is no where near winning, but I'll post it anyway just to show an alternative approach.

c;i;a[99];main(){while(~(c=getchar()))a[c]++;for(;i<99;)a[i++]>1&&puts("");}


Prints a new line if false, prints nothing if true.

• This program has an undefined behavior. The correct signatures for main are int main(int, char **) or int main(void). int main(int) is not valid. – Konrad Borowski May 22 '14 at 13:11
• @xfix I assume main() is ok then? – user12205 May 22 '14 at 13:22
• Yes, it's fine. It means the same thing as main(void) (when used in definition, in declaration it declares a function with unknown parameter list). – Konrad Borowski May 22 '14 at 13:22

# k4 (8)

  {x=.?$x}48778584 0b {x=.?$x}17308459
1b


inspired by a combination of the J and Golfscript answers

# Python 2.7 114 106 bytes

First golf, and this is no way going to win :P

def d(a):
f=[];
for i in str(a):
if i in f:return False
f.append(i)
return True


Kinda want to see how other people would improve this :)

def d(a):
f=[];
for i in str(a):
if i in f:return 1<0
f+=[i];
return 0<1

• return True can be shortened to something like 0<1. And f.append(i) can be shortened to f+=[i]. – ProgramFOX Dec 21 '14 at 15:40
• You're welcome! Also, you can save one more character by dropping the ; after f+=[i]. It is not necessary there. – ProgramFOX Dec 21 '14 at 19:03
• Ugh... Too much php :) – JamJar00 Dec 21 '14 at 19:04

# Python, 40

f=lambda x:len(str(x))==len(set(str(x)))


The built-in set removes the duplicates, so if the length of a thing and the length of his set are equal, the thing does not have repeated elements.

## Burlesque, 2 bytes

Using the Unique built-in:

blsq ) 17308459U_
1
blsq ) 48778584U_
0


# Go, 138 bytes

run it like uniqchars 112345678

package main
import(."fmt"
."os"
."strings")
func main(){s:=Args[1]
f:=true
for _,e:=range s{if Count(s,string(e))>1{f=false}}
Println(f)}
• This is post 65536 on the site! That's 2^16. – wizzwizz4 Jan 11 '16 at 21:11

# PARI/GP, 26 bytes

n->d=digits(n);#d==#Set(d)


Try it online!

L,BDdq=


Try it online!

# Perl 6, 12 bytes

false
$echo 17308459| perl -F// -alpe '$"="";$_= 0<"@{[map {$a{$_}++} @F]}"?"false":"true"' true  C# 72 69 67 characters (no libraries needed) for(;d>0;d/=10)for(int f=d/10;f>0;f/=10)if(d%10==f%10) return true;  Ungolfed: for (; d > 0; d /= 10) for (int f = d / 10; f > 0; f /= 10) if (d % 10 == f % 10) return true;  I'm just using simple maths here.(i.e. number 1231): • Take the last digit (1) • Iterate through the quotient (123) • If the number is equal to our digit (1), then return true • 3 == 1, 2 == 1, 1 == 1 - found it! Ruby(47 characters) ->(x){x.to_s.chars.sort.uniq.size==x.to_s.size}  Example usage ->(x){x.to_s.chars.sort.uniq.size==x.to_s.size}[10] => true  C# (44 - 64) Func As a lambda (44): y=>!(y+"").GroupBy(x=>x).Any(x=>x.Count()>1)  alt lambda (also 44): y=>(y+"").GroupBy(x=>x).All(x=>x.Count()==1)  As a Func (63): Func<int, bool> f=y=>(y+"").GroupBy(x=>x).All(x=>x.Count()==1);  As a method (64): bool f(int y){return (y+"").GroupBy(x=>x).All(x=>x.Count()==1);}  # Extended BrainFuck: 100 not included the unnecessary linefeed. {a>[>>]}>>,10-[38-[-&a+[<<]>]&a>[[-]3<[-<<]<+2>&a>] +3<[-<<]>,10-]<+<([-]>-|"false"[-])>(-|"true"[-])  Usage: %> beef ebf.bf < unqiue.ebf > unqiue.bf %> echo 48778584 | beef unique.bf -> false %> echo 17308459 | beef unique.bf -> true  ## VBA (145) Function jkl(f As String) As Boolean For i = 1 To Len(f) jkl = InStr(1, f, Left(Mid(f, i), 1)) <> i If jkl = True Then Exit Function Next End Function  Calling the function and output with opposite of the result: Sub jj() Dim f As String f = "1234256789" MsgBox Not (jkl(f)) End Sub  # PHP 74 Chars <?php$a=str_split($argv[1]);var_dump(count($a)==count(array_unique($a)));  ## VB.NET - 189 184 Function U(I As Integer) As Boolean U = True Dim D(9) Dim S = I.ToString For Each E As Char In S If IsNumeric(E) Then If D(Val(E)) + 1 > 1 Then U = False Else D(Val(E)) += 1 Next End Function  First code-golf attempt. I'm aware it's an unsuitable language and poor result, but I wanted to try. • I counted 189 characters even without necessary whitespace. Please be aware that, unless explicitly stated in the question, we usually count necessary whitespaces in answers. However, you can reduce your source size by removing unnecessary whitespaces (spaces around assignment operators, indentations etc) and using one-character identifier names, e.g. U instead of Unique, i instead of IA etc. Also, I cannot test it now but I think you can skip the type definitions of your variables/function (i.e. using only Dim D(9),i()=I.ToString.ToCharArray) – user12205 May 24 '14 at 15:17 • So your first line may become something like Function U(I). Your function can then return integers 0 and 1 instead of false/true. Then your second line can be changed to U=1 and your Return False can be changed to U=0. Finally, I'm not sure whether it works but perhaps making IA a string instead of a char array can save you a lot of bytes. – user12205 May 24 '14 at 15:22 • I used an online tool out of laziness, I may have to write a character-counting function instead. Do you count line returns and tabs as characters? – Lou May 24 '14 at 15:23 • Visual Studio 2012 religiously enforces spacing and indentation, even correcting it before beginning a debug - I don't see a way I can cut them out. – Lou May 24 '14 at 15:24 • Also, how do I count bytes in code? Sorry, I looked in Meta and the Help Centre but I couldn't find an answer. – Lou May 24 '14 at 15:27 ## Clojure - (42 - 52) As a named function (52): (defn f[i](let[s(str i)](=(count s)(count(set s)))))  As an anonymous function (42): #(let[s(str %)](=(count s)(count(set s))))  # Scala Cheating with library functions (29/12 characters): def t(i:String)=i==i.distinct  Scala version of the GolfScript version (36/19 characters): def t(i:String)=i.intersect(i).size>0  Doing the counting work manually (91/74 characters): def t(i:String)=(Map.empty[Char,Int]/:i)((m,c)=>m+(c->(m.getOrElse(c,0)+1))).forall(_._2<2)  ## PHP - 44 33 var_dump(max(count_chars($n))<2);


Or Proper function 42 Characters

function(){return max(count_chars($n))<2;}  UPDATE : Thanks for @scrblnrd3 pointed out for shorter solution. • You could cut out a few chars by doing <2 instead of the ternary operator – scrblnrd3 May 23 '14 at 19:49 • <2 in PHP, DO you have some reference page, how it used in php as I never seen this. – kuldeep.kamboj May 26 '14 at 5:20 • I mean you could use var_dump(max(count_chars($n))<2); instead – scrblnrd3 May 26 '14 at 13:19
• <?=max(count_chars($argn))<2); 30+1 bytes: run as pipe with -F, prints 1 for true, nothing for false – Titus Nov 2 '17 at 20:58 # Haskell (41) import Data.List f x=show x==nub(show x)  About 23 characters shorter than the other Haskell answer and a fair bit more intuitive I think. Basically the same process but using Haskell's built-in list functions is much easier and shorter. • I don't think you're required to count the last newline. – seequ Jun 3 '14 at 17:42 ## C - 62 int m[10];f(int a){while(m[a%10]++?0:a/=10);return m[a%10]<2;}  I've included the count of the globally declared array (which ensures initialisation to 0). As a bonus, here is a much shorter technically correct answer: f(int a){return 1}  The question does not state that non-unique digits should return false. Therefore I claim the current lead for C code with 18 characters, although I expect to be beaten by a scripting language with an answer like: 1  • Welcome to Programming Puzzles and Code Golf. I think the OP states quite clearly that a number with duplicate digits should return false. Perhaps you could explain why you think differently? And explain how an undefined function performs the requested task, and how a singly character can do so in any scripting language. – David Wilkins May 23 '14 at 17:24 • @David Wilkins One of the examples returns false; the actual OP just states that unique digits should return true, which the C program I added in jest certainly does. – Alchymist May 27 '14 at 7:40 • Also, if I knew golfscript, for example, I would have written a function returning true in that rather than describing an unspecified "scripting language" – Alchymist May 27 '14 at 7:59 # Julia - 29 28 f(n)=unique("$n")=="$n".data  Old version: f(n)=join(unique("$n"))=="\$n"


F# - 82

let f i =
let s = i.ToString()
s |> Set.ofSeq|> Seq.length = s.Length;;