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

Trilangle, 34 bytes

UB is only disallowed in C, huh?

'0.<".@|>i\[email protected]'2#!>.(&,\,<.-

This works on x64, at least. Can't guarantee it'll work on other architectures.

Expects the number on STDIN, immediately followed by EOF (no trailing newline). Prints "0" (with trailing newline) iff the digits are unique.

Edit: there is now an online interpreter. This code seems to work in wasm, too (but I still haven’t tested native ARM or other processors).

High-level Overview

Roughly equivalent to this C code:

#include <stdio.h>

int main() {
    // A bitfield of the seen numbers:
    // e.g. if "12" is input, this is 0b0000'0000'0000'0110;
    // if "3" is input, this is 0b0000'0000'0000'1000; etc.
    int seen = 0;

    while (1) {
        // Get a number in. If EOF is read,
        // we haven't encountered a duplicate, so indicate success.
        int i = getchar();
        if (i == EOF) {
            puts("0");
            break;
        }

        // Check if the digit has already been seen.
        // If so, exit without printing anything.
        // If not, add this digit to the bitfield.
        i = 1 << (i - '0');
        if (i & seen) break;
        seen = seen | i;
    }
}

How does Trilangle work?

Trilangle is a 2-D programming language laid out in a triangular grid (hence the name). The IP can move in one of six directions; it starts heading southwest. If it walks off the edge of the grid, it continues one row or diagonal to its left.

The memory model is like a stack, but it's possible to index into it.

Explanation of the Code

The interpreter unfolds that code into this triangular grid:

       '
      0 .
     < " .
    @ | > i
   \ r e @ .
  . j . . 2 '
 2 # ! > . ( &
, \ , < . - . .

I hope this explanation makes any sense. Here goes nothing.

Below is an image of the code, with the IP's paths highlighted in various colors.

The IP starts on the red path, where it hits the following instructions:

• '0: Push the number 0 to the stack
• <: Change the direction of control flow
• i: Get a character from STDIN and push it to the stack; pushes -1 on EOF
• >: Branch on the sign of the top of the stack

If the value is negative, EOF was read, and the IP continues on the green path:

• e: Pop from the stack, and push 1 << value. This is the UB: 1 << -1 is undefined. At least on x64, it evaluates to 0.
• .: No-op
• !: Print the top of the stack as a decimal integer
• ,: Pop from the stack
• i: Attempt to read from STDIN again (always sees EOF on this path)
• @: End program

The ,i aren't strictly necessary on this code path, but they're harmless to leave in.

If a character was successfully read, control flow continues on the blue path:

• "0: Push the value of the character '0' (i.e. 48) to the stack
• -: Pop two values from the stack and push their difference
• ..: No-ops
• e: Pop from the stack and push 1 << value
• |: Redirect control flow
• ".,: Push the value of the character '.' and immediately pop it, effectively a no-op
• 2: Duplicate the top of the stack
• .: Another no-op
• \: Redirect control flow
• '2: Push the number 2 to the stack
• ..: Even more no-ops
• j: The indexing function. After this, the stack contains { seen, i, i, seen } (from the C code above), in that order.
• .: Yet another no-op
• &: Pop two values from the stack and push a & b (bitwise and)
• (: Decrement the top of the stack. If seen & i is 0, the top of the stack is now negative; otherwise, it's still positive
• .: Yep, still a no-op
• >: Branch

If i and seen have any bits in common, the IP follows the yellow path; otherwise, the IP follows the magenta path.

The yellow path contains a harmless r (bitwise or) instruction, before the IP hits @ and the program ends.

On the magenta path, the IP continues as follows:

• <: Redirect control flow
• ,: Pop from the stack
• \: Redirect control flow
• #: Ignore the next instruction (j)
• r: Pop two values from the stack and push a | b (bitwise or)
• |: Redirect control flow
• <: Redirect control flow, merging with the red path

From which it reads the next character from STDIN, and continues until it either reads the same character twice or reads EOF.

Disassembly

I recently added a --disassemble flag to the interpreter. The output is fairly straightforward if you understand the syntax:

0.0:    PSI #0
0.2:    GTC
0.3:    BNG 2.0
1.0:    PSC '0' ; 0x30
1.1:    SUB
1.4:    EXP
1.6:    PSC '.' ; 0x2e
1.7:    POP
1.8:    DUP
1.11:   PSI #2
1.14:   IDX
1.16:   AND
1.17:   DEC
1.19:   BNG 4.0
        JMP 3.0
2.0:    EXP
2.2:    PTI
2.3:    POP
2.4:    GTC
2.5:    EXT
3.1:    IOR
3.2:    EXT
4.1:    POP
4.4:    IOR
4.7:    JMP 0.2

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

Mathematica, 35 25 characters

(27 if the function needs a name.)

Unequal@@IntegerDigits@#&

EDIT: Saved 8 characters thanks to belisarius!

Answer 13

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 14

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 15

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 16

FRACTRAN, 15 fractions, 83 (or 32) bytes

This beats Sp3000's 38-fraction solution.

$$\frac25,\frac{875}{88},\frac78,\frac{25{\cdot}17^c}{92},\frac14,\frac{15}{38},\frac{69}{2},\frac{23}{17},\frac{19}{3^b},\frac43,\frac{7}{23},\frac{26}{19},\frac{11}{7^c},\frac{1}{91},\frac{8}{11}$$

The initial state is \$3^n\cdot 13\cdot 23\$. It halts at \$1\$ if every digit in the base-\$b\$ representation of \$n\$ occurs fewer than \$c\$ times, and at some other value otherwise. Plug in \$b=10, c=2\$ to get the answer to the challenge as posed.

The final state is actually \$13^{k - \mathrm{wt}_c\left(\sum c^{d_i}\right)}\$, where \$d_1\cdots d_k\$ are the base-\$b\$ digits of \$n\$, and \$\mathrm{wt}_c\$ is the base-\$c\$ digit sum. Adding a power of \$c\$ increases the digit sum by \$1\$ if there is no carry, and doesn't increase it if there is a carry, so the final state is \$1\$ iff each digit occurs fewer than \$c\$ times.

The length of the \$b=10, c=2\$ version is 83 bytes in the decimal coding (2/5,875/88,...) that seems to be standard here. With a more efficient coding (Elias δ), it's 32 bytes.

There's a fast online interpreter here. I see no way to encode a program in the URL, so you'll have to copy-paste this:

2%5 875%88 7%8 25*17^2%92 1%4 15%38 69%2 23%17 19%3^10 4%3 7%23 26%19 11%7^2 1%91 8%11

Then enter, e.g., [3,48778584],[13,1],[23,1] as the input.

Answer 17

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 18

Haskell, 34 bytes

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

Try it online!

Answer 19

Brachylog, 1 byte

≠

Try it online!

Answer 20

JavaScript, 22 bytes

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

Try it online

Answer 21

C99, 59 chars

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

Answer 22

Groovy (36 chars)

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

Tested it using:

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

Answer 23

Haskell:

 import Data.List

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

Answer 24

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 25

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 26

J (8)

Competely sepertae from my previous answer.

*/@~:@":

Answer 27

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 28

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 29

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 30

Add++, 7 bytes

L,BDdq=

Try it online!