Are all three integers distinct?

Question

You will be given 3 integers as input. The inputs may or may not be different from each other. You have to output 1 if all three inputs are different from each other, and 0 if any input is repeated more than once.

This is code-golf, so make your code as short as possible!

Answer 1

Python 3, 23 21 20 bytes

lambda*a:len({*a})>2

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Answer 2

Perl 6, 7 bytes

*.Set>2

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Answer 3

R, 13 bytes

A different solution to @Kirill by using mad() for an unintended purpose!

mad(scan())>0

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Answer 4

JavaScript, 22 bytes

If we can output boolean values then the last 2 bytes can be removed.

a=>new Set(a).size>2&1

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For the same byte count, this works on arrays of any size but assumes the input will never contain a 0 and output is a boolean.

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

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Answer 5

R, 24 22 20 bytes

all(table(scan())<2)

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Returns a boolean, but as folks have already discussed on the Python answer, this should be OK.

Thanks to digEmAll for saving 2 bytes.

Answer 6

Ruby, 16 bytes

->a{1-(a<=>a|a)}

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Answer 7

Cubix, 55 25 bytes

-29 thanks to Jo King

O@O1u|@O@II-!/;I-!/;u^?-p

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It should be possible to golf off quite a few bytes.

Answer 8

05AB1E, 2 bytes

ÙQ

Try it online or verify some more cases.

Explanation:

Ù     # Uniquify the (implicit) input
 Q    # Check if it's still equal to the (implicit) input

Answer 9

Mathematica, 13 bytes

Boole[E!=##]&

Pure function. Takes three integers as input and returns 0 or 1 as output. I know that this is rather similar to David G. Stork's answer, but it exploits SlotSequence to shave off a byte (as compared to Boole@*Unequal).

Answer 10

brainfuck, 91 bytes

,>,>,[-<-<->>]>>+++++++[>+++++++<-]+<<<<[>]>>[<<<[-<->]<[>]>>->[>.<<<->>-]<+]<+[>>>[>]<-.>]

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How it works

,>,>,                   'read input as A, B, and C
[-<-<->>]>>+            'compute A-C, B-C
++++++[>+++++++<-]+     'prepare output
<<<<[>]>>               'if A-C != 0 && B-C != 0
[
    <<<[-<->]           'compute A-B
    <[>]>>->            'if A-B != 0
    [>.<<<->>-]         'print 1
    <+
]
<+
[                       'else (this else is for both of the if statements, even though they are nested... wierd, I know)
    >>>[>]              
    <-.>                'print 0
]

Answer 11

Japt -N, 3 bytes

eUâ

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---

Explanation

Uâ deduplicates the input and e tests if it's equal to the original.

Answer 12

C (gcc), 25 26 bytes

f(a,b,c){a=a-b&&a-c&&b-c;}

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Answer 13

J, 4 bytes

-:~.

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Explanation:

Is the argument equal -: to itself after removing the duplicates ~.

Answer 14

Powershell, 27 25 bytes

-2 bytes thanks @AdmBorkBork

+!(($args|group).Count-3)

Test script:

$f = {
+!(($args|group).Count-3)
}

&$f 1 2 3
&$f 3 2 1
&$f 2 1 3
&$f 2 2 3
&$f 2 1 1
&$f 2 1 2

Explanation:

    $args|group           # Group arguments
   (           ).Count    # Count of groups 
  (                   -3) # is 0 if inputed integers are unique
 !                        # operator not converts int to boolean: true if integers are unique
+                         # converts boolean to int: 1 if integers are unique, otherwise 0

Answer 15

K (oK), 4 bytes

3=#?

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Does the count of the distinct elements equal 3?

Answer 16

Jelly, 2 bytes

QƑ

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Answer 17

Common Lisp, 25 2 bytes

/=

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-23 bytes thanks to @ceilingcat!

Answer 18

APL (Dyalog Unicode), 3 bytes^SBCS

Anonymous tacit prefix function. Takes list as argument.

∪≡⊢

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∪ does the set of unique elements from the argument

≡ match

⊢ the unmodified argument?

Answer 19

Clean, 32 bytes

import Data.List
$l|hasDup l=0=1

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Answer 20

Java 9, 43 27 bytes

thanks to @Olivier Grégoire

(a,b,c)->a!=b&b!=c&a!=c?1:0 

Previous attempt:

(a)->a[0]==a[1]||a[0]==a[2]||a[1]==a[2]?0:1

Answer 21

Attache, 10 bytes

`==#Unique

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This is a fork of the operator `== and Unique, equivalent to:

{ _ == Unique[_] }

Alternatives

{#_=#Unique[_]} (15 bytes)

Any##Same=>Pairs@Sort (21 bytes)

Any@{`=&>_[[0'1,1'2,2'0]]} (26 bytes)

&${not(x=y or y=z or x=z)} (26 bytes)

&${x/=y and y/=z and x/=z} (26 bytes)

{Any!Same=>Chop&2!_[0'1'1'2'2'0]} (33 bytes)

Answer 22

><>, 19 17 bytes

-2 bytes by Jo King.

:{:{:{=}=}=++0=n;

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Answer 23

Red, 21 bytes

func[b][b = unique b]

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Answer 24

T-SQL, 39 bytes

SELECT IIF(a=b OR b=c OR c=a,0,1)FROM s

Input is taken as separate columns a, b, c from a pre-existing table s, per our IO standards.

Tried a variation using COUNT DISTINCT from input taken as separate rows, but that was a couple bytes longer.

Answer 25

Jelly, 5 6 bytes

ɠḲQL=3

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From 5 to 6 bytes because this is my first time and I messed up (whoops) fixed it now

ɠḲQL=3
^^^^^
||||Is it equal to three?
|||How many unique numbers do we have? (length of unique numbers array)
||Sort By Unique
|Split by Spaces
Read Input

Answer 26

Pyth, 3 bytes

s{I

Takes input as a list.
Try it here

Explanation

s{I
 {IQ     Check if the (implicit) input is invariant under deduplication.
s        Cast to int.

If we're allowed to treat True and False as 1 and 0 (which they are under the hood in Pyth), we can drop the s to get down to 2 bytes.

Answer 27

SmileBASIC, 25 24 bytes

READ A,B,C?A-B&&B-C&&C-A

Answer 28

Brachylog, 6 bytes

d?∧1|0

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short explanation

d? deduplcates input an test if still equal to input(?)

∧1 if true return 1

|0 else return 0

Answer 29

JavaScript (Node.js), 67 bytes

f=a=>a.map((m,i)=>a.map((n,j)=>m==n&i!=j).every(z=>!z)).every(y=>y)

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Answer 30

q 14 bytes

{x~distinct x}

Technically this solution will return '1b' or '0b', which is the way a boolean value is distinguished from a numeric type, though it retains all arithmetic functionality, and so is in essence a 1 or 0:

q)1b +35
36

To return 1 or 0 non-boolean you have the below, which takes the byte count to 21

{$[x~distinct x;1;0]}

Answer 31

Triangularity, 17 bytes

..i..
.MiC.
}u)3=

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How it works

Excluding the mandatory triangular padding, we're left with:

iMiC}u)3=

Which can minimally be explained as: Push all STDIN as an array of strings (representing each line of input), count the occurrences of all of STDIN's elements in itself (which, in case all are distinct, should yield [1, 1, 1]), sum, then check whether the sum is equal to 3. There are 2 tricks here (which together save no less than 14 bytes):

• Using i and treating the inputs as strings rather than integers – which is quite uncommon, as )IE is frequently used instead, taking input as a list.
• Instead of checking whether they all are equal to 1 and then taking the product – or equivalently, taking the product and checking whether it is equal to 1 – this takes the sum and compares it with 3. Proving the validity of this method is obviously trivial: [1, 1, 1] is the only valid way to partition 3 into 3 positive integers summing to it and the contents of this list are always positive, as every element of a collection must occur at least once in it :P.

Answer 32

Charcoal, 14 bytes

≔Ｅ³ＮθＩ¬⊖⌈Ｅθ№θι

Try it online! Link is to verbose version of code. 5 bytes could be saved by using a slightly non-standard input format [[x, y, z]]. Explanation:

≔Ｅ³Ｎθ

Input 3 numbers and make a list.

Ｉ¬⊖⌈Ｅθ№θι

Count the number of occurrences of each number in the list. Take the maximum, subtract 1, take the logical not, then cast to string for implicit print.

Answer 33

PHP, 30 bytes

<?=array_unique($argv)==$argv;

takes input from command line arguments; prints 1 for truthy, empty string for falsy

Answer 34

Flobnar, 25 bytes

$-<*\\\!!@
::*<&&&
-$$>
$

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Given integers a,b,c, evaluates !!((c-b)*(b-a)*(c-a))

Answer 35

C (gcc), 50 41 bytes

-9 bytes thanks to GB !

m(int*a){a=*a==a[2]^*a==*++a^*a==*++a^1;}

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No built-in sets in C, and iterating is too long, but at least i can shave off a few bytes by taking advantage of the fact that *a == a[0] ...

Answer 36

SWIFT 37 Bytes

Set(Array(readLine()!)).count==3 ?1:0

Answer 37

Julia 1.0, 24 bytes

f(x...)=length(Set(x))>2

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Answer 38

Ruby, 11 bytes

->a{a==a|a}

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Assuming we can return a boolean. Which everyone is doing nowadays.

Answer 39

Pari/GP, 12 bytes

a->#Set(a)>2

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Answer 40

Stax, 3 bytes

ä▼Z

Run and debug it

Explanation

ux=         #Full program, unpacked, implicit input as array
u           #Keep only unique elements in array, maintaining first order of appearance
 x=         #Copy original input onto stack, is it equal to unique-ified array.

Short and sweet. Wish Stax would implicitly check against "X" register if there was nothing to check against, but it works out regardless.

Answer 41

Haskell, 26 bytes

g a b c=and$map(==)[a,b,c]

Answer 42

Math++, 40 29 bytes

?>a
?>b
?>c
(a-b)&(b-c)&(a-c)

Answer 43

Lua, 38 bytes

x,y,z=...print(x~=y and x~=z and y~=z)

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Answer 44

C (clang), 32 bytes

f(a,b,c){return a!=b&b!=a&a!=c;}

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Answer 45

JavaScript, 25 24 28 27 23 bytes

Using bitwise, comparison and logical operators Try it online!

(4 bytes saved courtesy of Jonathan Frech)

(a,b,c)=>a!=b&a!=c&b!=c

Answer 46

Pepe, 85 77 bytes

rEeEREeErrEEEEEeRrEeErrEEEEEerEEEEEEERerEEEEEerRrEEEEEEeErREEREEEEErEeReeReEE

Try it online! Input is a,b,c. Change the "separated by" box to , to make it work!

Answer 47

Rust, 29 bytes

|a,b,c|(a!=b&&b!=c&&a!=c)as _

as _ means casting to a type determined by type inference. The function result needs to be assigned to a variable declared to store integers.

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Answer 48

MathGolf, 2 bytes

▀=

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Explanation

▀    Get unique elements
 =   Check if equal to the input

Answer 49

Braingolf, 5 bytes

ul3-n

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Edit: fixed from identical -> distinct

Answer 50

Pyth, 5 bytes

s<2l{

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Looks like someone already did pyth but I didn't know about the I for testing invariants.

Answer 51

[Rust], 17 bytes

Anything XORd with itself yields zero. So. Any language with multiplication and XOR can do this:

(a^b)*(b^c)*(a^c)

Since input specifics were rather vague I chose a liberal interpretation. Or if you want to be more strict, we can define it as a closure (kind of like a lambda function),

24 bytes

|a,b,c|(a^b)*(b^c)*(a^c)

Try it on the Rust Playground

Answer 52

Python 2

def e(a,b,c):return 1 if len(set([a,b,c]))>2 else 0

...could be shortened to

def e(a,b,c):return len(set([a,b,c]))>2