Determine if all decimal digits are unique

Question

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.

Answer 1

Golfscript, 8 7 characters:

{`..&=}

• ` - stringify the argument
• .. - clone twice
• & - intersect with itself (remove duplicates)
• = - check for equality.

if the function needs to be named (10 9 characters):

{`..&=}:a

if a program suffices (5 4 characters):

..&=

Answer 2

Python 2 (28) (32)

lambda n:10**len(set(`n`))>n

The backticks take the string representation. Converting to a set removes duplicates, and we check whether this decreases the length by comparing to 10^d, which is bigger than all d-digit number but no (d+1)-digit numbers.

Old code:

lambda n:len(set(`n`))==len(`n`)

Answer 3

APL (6)

≡∘∪⍨∘⍕

One of the few times where tacit style is shorter in APL too.

It's 8 characters to give it a name,

f←≡∘∪⍨∘⍕

but that's not necessary to use it:

      ≡∘∪⍨∘⍕ 199
0
      ≡∘∪⍨∘⍕ 198
1
      f←≡∘∪⍨∘⍕
      f¨ 198 199 200 201
1 0 0 1
      ≡∘∪⍨∘⍕¨ 198 199 200 201
1 0 0 1

Answer 4

Perl, 19 characters

print<>!~/(\d).*\1/

Answer 5

Rebmμ (10 characters)

e? AtsAuqA

Rebmu's "mushing" trick is that it's case-insensitive, so characters are run together. Whenever a case transition is hit, that splits to the next token. By using transitions instead of a CamelCase kind of thing, the unique choice to start with a capital run means a "set-word" is made. (While set-words can be used for other purposes in symbolic programming, they are evaluated as assignments by default).

So this "unmushes" to:

e? a: ts a uq a

The space is needed because once you've begun a series of runs of alternating cases, you can't use that trick to get a set-word after the first unless you begin a new run. So e?AtsAuqA would have gotten you e? a ts a uq a...no assignment.

(Note: For what may be no particularly good reason, I tend to prefer rethinking solutions so that there are no spaces, if character counts are equal. Since brackets, parentheses, and strings implicitly end a symbol...there are often a fair number of opportunities for this.)

In any case, when mapped to the Rebol that it abbreviates:

equal? a: to-string a unique a

Throwing in some parentheses to help get the gist of the evaluation order:

equal? (a: (to-string a)) (unique a)

So the prefix equality operator is applied to two arguments--the first the result of assigning to a of the string version of itself, and the second the result of unique being run against that string. It so happens that unique will give you back the elements in the same order you passed them...so unique of "31214" is "3124" for instance.

Run it with:

>> rebmu/args "e? AtsAuqA" 17308459             
== true

There's also some stats and debug information:

>> rebmu/args/stats/debug "e? AtsAuqA" 48778584 
Original Rebmu string was: 10 characters.
Rebmu as mushed Rebol block molds to: 10 characters.
Unmushed Rebmu molds to: 15 characters.
Executing: [e? a: ts a uq a]
== false

If the requirement is that one must define a named/reusable function you can make an "A-function" which implicitly takes a parameter named a with a|. (A B-function would be created with b| and take a parameter named A then one named B). So that would add five more characters...let's say you call the function "f"

Fa|[e? AtsAugA]

"You laugh! They laughed at Einstein! Or wait...did they? I...don't know."

Answer 6

FRACTRAN - 53 38 fractions

47/10 3/5 106/47 3599/54272 53/61 2881/27136 2479/13568 2077/6784 1943/3392 1541/1696 1273/848 1139/424 871/212 737/106 469/53 142/3953 67/71 5/67 1/147 1/363 1/507 1/867 1/1083 1/1587 1/2523 1/2883 1/4107 1/5547 1/7 1/11 1/13 1/17 1/19 1/23 1/29 1/31 1/37 1/43

Uses division to count the number of occurrences of each digit. Call by putting n in register 2 and setting register 5 to 1, gives output in register 3 (0 if false, 1 if true). Also, make sure the rest of your program uses only registers > 71.

---

Edit 25/12/14: It's been 7 months and we've since gotten Stack Snippets, so here's one to test the code (using my could-be-better interpreter here).

var ITERS_PER_SEC=1E5;var TIMEOUT_MILLISECS=5E3;var ERROR_INPUT="Invalid input";var ERROR_PARSE="Parse error: ";var ERROR_TIMEOUT="Timeout";var ERROR_INTERRUPT="Interrupted by user";var running,instructions,registers,timeout,start_time,iterations;function clear_output(){document.getElementById("output").value="";document.getElementById("stderr").innerHTML=""};function stop(){running=false;document.getElementById("run").disabled=false;document.getElementById("stop").disabled=true;document.getElementById("clear").disabled=false}function interrupt(){error(ERROR_INTERRUPT)}function error(msg){document.getElementById("stderr").innerHTML=msg;stop()}function factorise(n){var factorisation={};var divisor=2;while(n>1){if(n%divisor==0){var power=0;while(n%divisor==0){n/=divisor;power+=1}if(power!=0)factorisation[divisor]=power}divisor+=1}return factorisation};function fact_accumulate(fact1,fact2){for(var reg in fact2)if(reg in fact1)fact1[reg]+=fact2[reg];else fact1[reg]=fact2[reg];return fact1};function exp_to_fact(expression){expression=expression.trim().split(/\s*\*\s*/);var factorisation={};for(var i=0;i<expression.length;++i){var term=expression[i].trim().split(/\s*\^\s*/);if(term.length>2)throw"error";term[0]=parseInt(term[0]);if(isNaN(term[0]))throw"error";if(term.length==2){term[1]=parseInt(term[1]);if(isNaN(term[1]))throw"error";}if(term[0]<=1)continue;var fact_term=factorise(term[0]);if(term.length==2)for(var reg in fact_term)fact_term[reg]*=term[1];factorisation=fact_accumulate(factorisation,fact_term)}return factorisation}function to_instruction(n,d){instruction=[];divisor=2;while(n>1||d>1){if(n%divisor==0||d%divisor==0){reg_offset=0;while(n%divisor==0){reg_offset+=1;n/=divisor}while(d%divisor==0){reg_offset-=1;d/=divisor}if(reg_offset!=0)instruction.push(Array(divisor,reg_offset))}divisor+=1}return instruction};function run(){clear_output();document.getElementById("run").disabled=true;document.getElementById("stop").disabled=false;document.getElementById("clear").disabled=true;timeout=document.getElementById("timeout").checked;var code=document.getElementById("code").value;var input=document.getElementById("input").value;instructions=[];code=code.trim().split(/[\s,]+/);for(i=0;i<code.length;++i){fraction=code[i];split_fraction=fraction.split("/");if(split_fraction.length!=2){error(ERROR_PARSE+fraction);return}numerator=parseInt(split_fraction[0]);denominator=parseInt(split_fraction[1]);if(isNaN(numerator)||isNaN(denominator)){error(ERROR_PARSE+fraction);return}instructions.push(to_instruction(numerator,denominator))}try{registers=exp_to_fact(input)}catch(err){error(ERROR_INPUT);return}running=true;iterations=0;start_time=Date.now();fractran_iter(1)};function regs_to_string(regs){reg_list=Object.keys(regs);reg_list.sort(function(a,b){return a-b});out_str=[];for(var i=0;i<reg_list.length;++i)if(regs[reg_list[i]]!=0)out_str.push(reg_list[i]+"^"+regs[reg_list[i]]);out_str=out_str.join(" * ");if(out_str=="")out_str="1";return out_str};function fractran_iter(niters){if(!running){stop();return}var iter_start_time=Date.now();for(var i=0;i<niters;++i){program_complete=true;for(var instr_ptr=0;instr_ptr<instructions.length;++instr_ptr){instruction=instructions[instr_ptr];perform_instr=true;for(var j=0;j<instruction.length;++j){var reg=instruction[j][0];var offset=instruction[j][1];if(registers[reg]==undefined)registers[reg]=0;if(offset<0&&registers[reg]<-offset){perform_instr=false;break}}if(perform_instr){for(var j=0;j<instruction.length;++j){var reg=instruction[j][0];var offset=instruction[j][1];registers[reg]+=offset}program_complete=false;break}}if(program_complete){document.getElementById("output").value+=regs_to_string(registers);stop();return}iterations++;if(timeout&&Date.now()-start_time>TIMEOUT_MILLISECS){error(ERROR_TIMEOUT);return}}setTimeout(function(){fractran_iter(ITERS_PER_SEC*(Date.now()-iter_start_time)/1E3)},0)};

<div style="font-size:12px;font-family:Verdana, Geneva, sans-serif;"><div style="float:left; width:50%;">Code:<br><textarea id="code" rows="4" style="overflow:scroll;overflow-x:hidden;width:90%;">47/10 3/5 106/47 3599/54272 53/61 2881/27136 2479/13568 2077/6784 1943/3392 1541/1696 1273/848 1139/424 871/212 737/106 469/53 142/3953 67/71 5/67 1/147 1/363 1/507 1/867 1/1083 1/1587 1/2523 1/2883 1/4107 1/5547 1/7 1/11 1/13 1/17 1/19 1/23 1/29 1/31 1/37 1/43</textarea><br>Input:<br><textarea id="input" rows="2" style="overflow:scroll;overflow-x:hidden;width:90%;">2^142857 * 5</textarea><p>Timeout:<input id="timeout" type="checkbox" checked="true"></input></p></div><div style="float:left; width:50%;">Output:<br><textarea id="output" rows="6" style="overflow:scroll;width:90%;"></textarea><p><input id="run" type="button" value="Run" onclick="run()"></input><input id="stop" type="button" value="Stop" onclick="interrupt()" disabled="true"></input><input id="clear" type="button" value="Clear" onclick="clear_output()"></input>&nbsp;<span id="stderr" style="color:red"></span></p></div></div>

Replace 142857 with another number. Output should be 3^1 if true, 1 = 3^0 if false. Takes a while for larger numbers (well, this is FRACTRAN...).

Answer 7

JavaScript - 23 Characters

As a function (ECMAScript 6):

f=x=>!/(.).*\1/.test(x)

Or taking input from a prompt (25 characters)

!/(.).*\1/.test(prompt())

Answer 8

C# 73 60 59

First golfing for me ...

Write a function that takes a nonnegative integer as input

bool f(int i){return(i+"").Distinct().SequenceEqual(i+"");}

Could strip another character by converting uint to int, but I rather take the task too literally than the other way around. Here we go ...

Answer 9

Mathematica, 35 25 characters

(27 if the function needs a name.)

Unequal@@IntegerDigits@#&

EDIT: Saved 8 characters thanks to belisarius!

Answer 10

Ruby (24 bytes)

Use a regular expression to match "some character, followed by zero or more characters, then the same character".

->(s){!!(s !~/(.).*\1/)}

If truthy or falsy values are accepted, rather than literal true or false, then we get 20 characters:

->(s){s !~/(.).*\1/}

Answer 11

C (87)

Since I can't win, I'll go for efficiency.

Function code:

int u(uint32_t d){short s=0,f;while(d){f=1<<d%10;if(s&f)return 0;s|=f;d/=10;}return 1;}

Answer 12

R, 53 51 48 34 Bytes

function(n)!grepl("(.).*\\1",n,,T)

Try it online!

Convert to a string and split. Convert to a table of counts minus 1, sum and negate

Inspired by Most common number answer by Alex and suggestion by Hugh.

A couple saved, thanks to @plannapus One more from @Gregor And a couple from making it an anonymous function

Now with wonderful regex goodness thanks to @J.Doe. This looks for any single char in the number that matches itself else where in the string. The grepl command returns a logical that is then returned. Perl style regexes is set to True.

Answer 13

J (9)

Assumes the value to be tested is in variable b (I know this can be made into a function, but don't have a clue on how. J is confusing. Any help on this is appreciated) Thanks Marinus!

(-:~.)@":

Checks if the lenght of the string rep of the number with all the duplicates removed is the same as the lenght of the regular string rep.

Answer 14

R (70, 60, 53, 52)

Thank you all for the useful comments! Your comments are incorporated in the answer.

### Version 70 chars
f=function(x)!any(duplicated(strsplit(as.character(x),split="")[[1]]))

### Version 60 chars
f=function(x)all(table(strsplit(as.character(x),"")[[1]])<2)

### Version 53 chars
f=function(x)all(table(strsplit(paste(x),"")[[1]])<2)

### Version 52 chars
f=function(x)all(table(strsplit(c(x,""),"")[[1]])<2)

f(48778584)
f(17308459)

Answer 15

Brachylog, 1 byte

≠

Try it online!

Answer 16

Mathematica (20 19)

(22 21 if function needs a name)

Max@DigitCount@#<2&

or

Max@DigitCount@#|1&

where | ist entered as [Esc]divides[Esc]

Answer 17

Haskell, 34 bytes

import Data.List
((==)=<<nub).show

Try it online!

Answer 18

C99, 59 chars

a(x){int r=1,f[10]={};for(;x;x/=10)r&=!f[x%10]++;return r;}

Answer 19

Groovy (36 chars)

f={s="$it" as List;s==s.unique(!1)}

Tested it using:

println f(args[0].toInteger())

Answer 20

C, 58 bytes

f;a(x){for(f=0;x;x/=10)f+=1<<x%10*3;return!(f&920350134);}

Can keep a tally of up to 7 identical digits before rolling over.

in test program (it's easier to see how it works with the constant in octal)

a(x){int f=0;for(;x;x/=10)f+=1<<x%10*3;return!(f&06666666666);}

main(){
scanf("%d",&r);
printf("%o\n",a(r));}

If you happen to have a large power of 2 handy the constant can be calculated like f&(1<<30)/7*6

Answer 21

R, 66 65 characters

f=function(x)!sum(duplicated((x%%10^(i<-1:nchar(x)))%/%10^(i-1)))

Separate the digits using integer division and modulo, then check if they are duplicates.

Usage:

> f(48778584)
[1] FALSE
> f(17308459)
[1] TRUE

Or, as @MickyT suggested, for 63 characters:

f=function(x)!anyDuplicated((x%%10^(i<-1:nchar(x)))%/%10^(i-1))

Answer 22

Haskell:

 import Data.List

 all ((== 1) . length) . group . sort . show

Answer 23

Java ( 131 59 57)

57 characters:

removed ^ and $ as @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ suggested

boolean u(int n){return !(n+"").matches(".*(.).*\\1.*");}

59 characters (works also with negative numbers!):

boolean u(int n){return !(n+"").matches("^.*(.).*\\1.*$");}

79 78 characters (thanks @n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ ):

Use for loop to save a few charachers and use int as a boolean array.

Use & instead of && to save 1 character (It turns out that java allows it).

boolean u(int n){for(int x=0;n>0&(x>>n%10&1)==0;n/=10)x|=1<<n%10;return n==0;}

131 characters (returns true for negative numbers):

boolean u(int n){int[] p=new int[]{2,3,5,7,11,13,17,19,32,29};double a=9001312320D;while(n>0){a/=p[n%10];n/=10;}return (long)a==a;}

with comments:

boolean unique(int n){
    int[] p=new int[]{2,3,5,7,11,13,17,19,32,29};//list of 10 first primes
    double a=9001312320D;//10 first primes multiplied
    while(n>0){
        a/=p[n%10];//divide by (n%10+1)th prime
        n/=10;//divide n by 10, next digit
    }
    return (long)a==a;//if a is integer then n has all digits unique
}

And answer that is technically correct (character count includes only the function, not global variables), but I think it's cheating, 29 characters:

boolean u(int i){return m[i];}

m[] is boolean array that contains correct answers for all 32-bit integers.

Answer 24

J (8)

Competely sepertae from my previous answer.

*/@~:@":

Answer 25

J, 8 bytes

-:~.&.":

Explanation:

         ": - converts the integer to a string
       &.   - "under" conjuction - uses the result from ": as operand to ~.
               then applies the inverse to the result, so converts the string back to integer
    ~.      - removes duplicates
  -:        - matches the operands

So, it first converts the number to a string, removes the duplicates and converts the resulting string to a number. Then the original number is matched against the transformed one. In fact it is a hook of two verbs: -: and ~.&.":

┌──┬──────────┐
│-:│┌──┬──┬──┐│
│  ││~.│&.│":││
│  │└──┴──┴──┘│
└──┴──────────┘

Try it online!

Answer 26

PARI/GP, 26 bytes

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

Try it online!

Answer 27

Perl 6, 12 bytes

{!/(.).*$0/}

Try it online!

Answer 28

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.

Answer 29

Javascript 73 chars

function f(n){return !~(n+'').split('').sort().join('').search(/(\d)\1/)}

Answer 30

Befunge 98, 17 bytes

This is a non-competing answer because Befunge does not have functions.

~:1g1`j@1\1p3j@.1

Prints a 1 if the number's digits are all unique; otherwise, it just ends.

This works by accessing a cell in the Funge space whose x coordinate is the ASCII value of the character inputted (takes input character by character) and whose y coordinate is 1. If the digit has not been seen before, the value of the cell is 32 (space character). If that is so, I set the value to 1.

As a bonus, this works for non-numbers as well.

Answer 31

PowerShell - 26

!($args|sls '(.)(?=.*\1)')

Answer 32

Perl 6 (19 bytes)

{.comb.uniq==.comb}

.comb splits a string into characters (for example, 42.comb gives "4", "2"). .uniq removes all non-unique characters. .comb characters in string (originally I used .chars, but .comb is shorter). == converts lists into number of elements in it, and compares the numbers. When . is used without object before, $_ which is default function parameter is assumed. {} are function literals.

Answer 33

POSIX sh and egrep (47, 43, 40)

f()([ ! `echo $1|egrep '([0-9]).*\1'` ])

• [-1 char]: Use ! instead of -z with test - Thanks DigitalTrauma
• [-1 char]: Use `CODE` instead of $(CODE) - Thanks DigitalTrauma
• [-2 chars]: Use fold -1 instead of grep -o .^₁ - Thanks DigitalTrauma.
• [-3 chars]: Check for repeated digits with a backreferenced regular expression.

If POSIX compliance is not important echo PARAM | can be replaced by <<<PARAM, reducing the functions length to 37:

f()([ ! `egrep '([0-9]).*\1'<<<$1` ])

Usage:

$ if f 48778584; then echo true; else echo false; fi
false
$ if f 17308459; then echo true; else echo false; fi
true

---

^{_{1 The fold -N notation is deprecated in some versions of fold.}}

Answer 34

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 :)

EDIT: With ProgramFOX's comments:

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

Answer 35

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.

Answer 36

Burlesque, 2 bytes

Using the Unique built-in:

blsq ) 17308459U_
1
blsq ) 48778584U_
0

Answer 37

Go, 138 bytes

Bad language choice.

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)}

Answer 38

05AB1E, 2 bytes

ÙQ

Try it online!

-1 thanks to Kaldo

---

{   # Sort.
 ¥  # Consecutive differences.
  P # Product.

---

Works on negatives, though OP never specified...

Answer 39

Add++, 7 bytes

L,BDdq=

Try it online!

Answer 40

Japt, 5 3 bytes

¶ìâ

Try it here

Answer 41

Python 3, 29 bytes

lambda x:10**len({*str(x)})>x

Inspired on this answer in Python 2; for Python 3 to only lag behind 2 by one byte is pretty rare :)

Answer 42

C# (Visual C# Interactive Compiler), 28 bytes

i=>(i+"").ToDictionary(c=>c)

Try it online!

This lambda uses presence or absence of an exception to determine whether a number has unique digits. If a digit occurs more than once, a duplicate key exception will be thrown when converting the string to a dictionary.

Answer 43

Braingolf, 12 bytes

dlMulMve1:0|

Try it online!

Explanation

dlMulMve1:0|  Implicit input from args

d             Split into digits
 lM           Push length of stack to next stack
   u          Keep only first occurrence of each unique digit
    lM        Push length of stack to next stack
      v       Switch to next stack
       e      If top 2 items are equal
        1      - Push 1
         :    Else
          0    - Push 0
           |  EndIf

              Implicit output of top of stack

Answer 44

8086 machine code, 9 bytes

51 AC 8B FE F2 AE 59 E0 F7

Unassembled listing:

    DU  MACRO
    SEARCH: 
51      PUSH CX             ; save loop counter 
AC      LODSB               ; load next digit from SI into AL, advance SI 
8B FE   MOV  DI, SI         ; DI is offset of next digit 
F2/ AE  REPNZ SCASB         ; search until match, ZF=1 if found 
59      POP  CX             ; restore loop counter 
E0 F7   LOOPNZ SEARCH       ; loop until match found or CX is 0
        ENDM

Implemented as a MACRO (most similar to function). Input is SI, length in CX. Output is ZF=0 if unique, ZF=1 if not unique. Also returned in CX is position from lowest order digit where first duplicate was found and SI the duplicated digit.

Below is a test program in IBM PC DOS that reads input from command line args, displays T or F:

Download and test DU.COM.

Answer 45

JavaScript, 22 bytes

s=>!s[new Set(s).size]

Try it online

Answer 46

C 151

Bytes = 181 or 151 without newline chars. Program keeps a tally of digits and exits if greater than 1.

#include <stdio.h>
int i,j[10],r;
int main()
{
scanf("%d",&i);
do{
r=i%10;
j[r]++;
if(j[r]>1){printf("false");return 0;}
i/=10;
}while(i);
printf("true");
return 0;
}

Answer 47

C# 72 characters

var a=i.ToString().ToCharArray();
return a.Distinct().Count()==a.Count();

Answer 48

Julia, 44

f(n)=(d=digits(n);length(Set(d))==length(d))

Answer 49

k4 (8)

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

inspired by a combination of the J and Golfscript answers

Answer 50

Cobra - 109

def f(n) as bool
    l=0
    m=n.toString
    for i in m,for j in m,if i==j,l+=1
    return if(l>m.length,false,true)

Makes me wish that LINQ would work properly in Cobra.

Answer 51

Perl, 63ish

$ echo 48778584| perl -F// -alpe '$"="";$_= 0<"@{[map {$a{$_}++} @F]}"?"false":"true"'
false
$ echo 17308459| perl -F// -alpe '$"="";$_= 0<"@{[map {$a{$_}++} @F]}"?"false":"true"'
true

Answer 52

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.

Answer 53

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!

Answer 54

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

Answer 55

Julia - 29 28

f(n)=unique("$n")=="$n".data

Old version:

f(n)=join(unique("$n"))=="$n"

Answer 56

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);}

Answer 57

COBOL 66 with Object Cobol extensions

Microfocus Cobol (721 659 592)

       method-id u.
       local-storage section.
       77  i pic s9(9) comp.
       77  c pic 9 occurs 10 value 0.
       77  g pic 9.
       01  s pic 9(9).
       01  t redefines s pic 9 occurs 9.
       linkage section.
       77  n pic s9(9) comp.
       77  d pic x.
       88  f value is 'y'.
       procedure division using by value n returning d.
           move n to s. move 'n' to d. perform p varying i from 1 by 1 until i greater than 9 or f. goback.
       p.  move t (i) to g. if c (g) is not zero then move 'y' to d. add 1 to c (g).
       end method.

Hmmm, I think I should go back to pitch & putt :(

EDIT: OMGZ I CAN USE LOWER CASE!!!

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

PHP 74 Chars

<?php $a=str_split($argv[1]);var_dump(count($a)==count(array_unique($a)));

Answer 62

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.

Answer 63

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))))

Answer 64

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)

Answer 65

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.

Answer 66

F# - 82

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

Answer 67

C++ 82

this code is at least legitimate

int a(uint32_t i,uint32_t r=0,uint32_t h=0)
{return i?r&h?0:a(i/10,r|h,1<<i%10):1;}        // thanks to DreamWarrior

C 47

this works if you compile it in debug mode or provide two more arguments set to 0

a(i,r,h){return i?r&h?0:a(i/10,r|h,1<<i%10):1;}

Answer 68

Regular Expression 17 bytes

^(?:(.)(?!.*\1))*$

Original expression taken from https://stackoverflow.com/a/12870549.

Answer 69

Clojure, 23

#(=(distinct x)(seq x))

Usage: (#(=(distinct x)(seq x)) "23563462464543")

Answer 70

Candy, 13 bytes

This is a late post, but...

(k&{1k1.|0})0

Invokation with -I , which is actually for string parameters.

The long form:

while    # while we are still consuming characters from the stack
  setSP  # pop the next stack-id from the stack.  It's the ord of the digit
  stackSz  # test for empty stack
  if
    digit1 setSP  # goto stack #1, and push the true indicator on
    digit1
    retSub        # and halt
  else
    digit0    # mark the stack as having been visited
  endif
endwhile
digit0    # push the false indicator if we got this far

Answer 71

Prolog, 49 bytes

Code:

p(N):-number_codes(N,L),sort(L,S),msort(L,T),S=T.

Explained:

p(N):-number_codes(N,L), % Convert int to list (of charcodes)
      sort(L,S),         % Sort list and remove duplicates
      msort(L,T),        % Sort list and keep duplicates
      S=T.               % Check if lists are equal

Example:

p(48778584).
false

p(17308459).
true