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Challenges

Are All Elements Equal?

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Challenge

Given a list of integers >= 0 , check if all of them are equal.

Tests

[1,1,1,1,1] -> true
[0,1,1,6,7] -> false
[1]         -> true  
[]          -> undefined(you do not need to handle this)

You may take the integers in any form(strings, list, codepoints, etc.)

You can output two distinct values for true or false, or simply a truthy or falsy value in your language.

Trivial builtins are allowed, but it is encouraged to edit them into the trivial builtins answer, or try a different approach.

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2 comment threads

Trivial builtins (2 comments)
CW trivial builtins answer? (5 comments)

16 answers

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MATL, 4 bytes

tPX=

Try it online!

tPX=
t    - duplicate with implicit input
 P   - fliplr in MATLAB, reverses top of stack
  X= - isequal
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+6
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Zsh, 8 bytes

>$@
<^$1

Attempt This Online!

Outputs via exit code: 0 for not all the same; 1 for all the same

  • >$@: create a file named for each element in the input
    • effectively deduplicates because you can only have one file with a given name
  • <^$1: try to find a file with any name other than the first input. If there were at least two distinct inputs, this will succeed and return 0, otherwise it will fail with status 1
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+4
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APL (Dyalog Unicode), 3 bytes

⍋≡⍒

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computes the permutation vector that would sort the argument into ascending order

computes the permutation vector that would sort the argument into descending order

If they are identical (), all elements must be equal.

See also this talk by Conor Hoekstra.

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+4
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Japt, 3 bytes

e¡g

Try it

e¡g     :Implicit input of array U
e       :Is equal to
 ¡      :Map U
  g     :  First element of U
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+3
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Haskell, 16 bytes

f(a:b)=all(==a)b

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Python 3, 20 bytes

lambda l:len({*l})<2

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Rockstar, 76 bytes

Takes individual integers as input.

listen to f
let n be f
o's1
while n
let o be o and n is f
listen to n

say o

Try it (Code and input will need to be added manually, with each input integer on its own line)

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C (gcc), 46 bytes

f(int*a){return *a<0|a[1]<0||*a==a[1]&f(a+1);}

This takes an array that is terminated by a negative number (which is not part of the list content, just like the terminating \0 is not part of the string content in C strings).

Try it online!

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JavaScript, 20 bytes

Outputs false for truthy and true for falsey.

a=>a.some(x=>x-a[0])

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Outputs true and false as normal.

a=>new Set(a).size<2

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brainfuck, 40 bytes

Outputs ÿ if all elements are equal, and \x00 byte otherwise.

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

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brainfuck, 41 bytes

Different approach. Output is opposite.

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

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Ruby, 15 bytes

->a{a.all?a[0]}

->a{          }  # lambda taking array `a`
    a.all?       # do all items in the array match... 
          a[0]   # ...the first? 

Try it online!

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Ruby, 14 bytes

->{_1|[]in[_]}

Alternative solution:

->a{!(a|a)[1]}

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Couple of other solutions I developed:

->a{!a.uniq[1]}   # 15 bytes
->{!(_1|[])[1]}   # 15 bytes
->a{(a|a).one?}   # 15 bytes
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Jelly, 1 byte

E

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Boring builtin answer

Non builtin, the shortest I can get is 3 bytes:

ḷ\⁼

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Here, we replace each element with the first element, then check if that is equal to the original array

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+1
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Trivial answers

Husk, 1 byte

E

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You don't need signum (4 comments)
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J, 4 bytes

-:|.

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Checks if the input array matches itself reversed. This is a tacit form and is executed monadically as x f (g x).

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This requires parentheses to work, so shouldn't they be counted? (4 comments)
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J, 5 char

2>#~.

How it works:

'~.' produces list of unique elements of list on its right

'#' counts how many elements in the list on its right

'2>' tests if number on its right only has one unique element

Sample runs:

2>#~. 3 3 3 4

0

2>#~. 3 3 3 3

1

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