Isogram checker

Question

Challenge :

Given a word, check whether or not it is an isogram.

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What :

An isogram is a word consisting only of letters with no duplicates (case insensitive). The empty string is an isogram.

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Examples :

"Dermatoglyphics"     ---> true
"ab"                  ---> true
"aba"                 ---> false
"moOse"               ---> false
"abc1"                ---> false
""                    ---> true

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Input :

You may accept input in any reasonable format

The input will only contain letters and/or numbers, no spaces ([a-zA-Z0-9])

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Output :

• true or any truthy value if the input is an isogram
• false or any falsy value otherwise

This is code-golf so shortest code in bytes in each language wins.

Answer 1

Python 2/3, 36 52 48 bytes

lambda s:len(s)==len({*s.lower()}-{*str(56**7)})

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I take advantage of the fact that set contains only unique elements. By invoking the __len__ method of each, I can determine whether s also contains only unique elements (ignoring case).

EDIT: Updated to satisfy the previously-overlooked requirement to return False for numeric inputs. The set of all digits is encoded as set(str(56**7)).

EDIT 2: Following this user suggestion, I now take advantage of unpacking the arguments to set comprehension. This formally breaks compatibility with Python 2.

Answer 2

05AB1E, 5 bytes

lDáÙQ

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Explanation

l        # convert input to lower-case
 D       # duplicate
  á      # keep only letters
   Ù      # remove duplicates
    Q     # compare for equality

Answer 3

R, 41 bytes

!grepl("(.).*\\1|\\d",tolower(scan(,"")))

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Regex approach. !grepl(regex,scan(,""),F) didn't work so I guess capturing doesn't match case-insensitively in R? I'm bad at regex in general, though, so I won't be surprised if I'm just doing it wrong...

R, 58 bytes

!anyDuplicated(c(el(strsplit(tolower(scan(,"")),"")),0:9))

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Appends the digits 0:9 to the (lowercased) list of characters and tests if there are any duplicates.

Answer 4

Ruby, 25 23 21 bytes

-2 bytes on both thanks to Giuseppe

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

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-2 bytes thanks to Kirill L.

Ruby -n, 21 19 18 16 bytes

p !/(.).*\1|\d/i

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Answer 5

Brachylog, 4 bytes

ḷo⊆Ạ

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The predicate will succeed if the input is an isogram and fail if it is not, outputting the lowercase Latin alphabet if it does succeed. Since Brachylog's ⊆ built-in predicate doesn't exactly match the ordinary relationship between a subset and superset, I had to spend a byte on sorting the lowercased input, but saved a byte on not having to explicitly check for duplicates in it. (If it didn't need to fail with numbers, we could just use ḷ≠.)

Answer 6

Husk, 6 bytes

§=f√ü_

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Explanation

§=f√ü_  -- takes a string as argument, eg: "sAad2"
§       -- fork the argument..
  f√    -- | filter out non-letters: "sad"
    ü_  -- | deduplicate by lower-case: "sAd2"
 =      -- ..and compare: 0

Answer 7

Japt, 12 bytes

;v
oC ‰ eUq

Explanation:

;v
;         // Set alternative default vars, where C is the lowercase alphabet
 v        // Make the implicit input lowercase and reassign it
oC ‰ eUq
oC        // Remove all items from the input that are not in the alphabet
   ‰     // Split into chars and select unique array items
      eUq // Check if the result is equal to the input split into chars

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Answer 8

MATL, 9 bytes

kt2Y2X&X=

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k   % Lowercase implicit input
t   % Duplicate that
2Y2 % Push lowercase alphabet
X&  % Intersection of alphabet and duplicate lowercase input
X=  % Check for exact equality.

Answer 9

Python 3, 46 bytes

lambda s:s.isalpha()*len(s)==len({*s.lower()})

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Answer 10

Japt 2.0, 12 11 bytes

-1 byte thanks to Nit

v
f\l â eUq

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Answer 11

JavaScript (Node.js), 29 25 bytes

s=>!/(.).*\1|\d/i.test(s)

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Thanks for the update on answer to @BMO , @l4m2 , @KevinCruijssen

-4 bytes thanks to @KevinCruijssen

Answer 12

Retina, 16 bytes

Ci`(.).*\1|\d
^0

Returns 1 as Truthy and 0 as Falsey values.
Thanks @Neil for discovering and fixing a bug in my initial code.

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Explanation:

C             Check if the input matches part of the following regex:
 i`            Case insensitivity enabled
               Check if part of the input matches either:
  (.)           A character `C`
     .*         followed by zero or more characters
       \1       followed by the same character `C` again
         |     Or
          \d    A digit
^0             Invert Truthy/Falsey, basically replacing every 0 with a 1,
               and every other value with a 1

Answer 13

PowerShell, 91 bytes

param($b)($a=[char[]]$b.ToUpper()|group|sort c*)[0].Count-eq$a[-1].count-and$b-notmatch'\d'

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Naive solution, but I can't come up with a better algorithm. Takes input $b, converts it ToUppercase, casts it as a char-array. Pipes that array into Group-Object which constructs a object that has name/count pairs for each input letter. We then sort that based on the count and take the 0th one thereof. We check that its .Count is -equal to the .Count of the last [-1] pair. If so, then the counts are all equal, otherwise we have a different amount of letters.

We then -and that with checking whether the input -notmatches against \d to rule out any digits in the input. That Boolean result is left on the pipeline and output is implicit.

Answer 14

Jelly, 8 bytes

ŒufØA⁼QƲ

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Answer 15

Python 2, 57 56 bytes

x=input().lower()
print len(set(x)-set(`763**4`))/len(x)

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First it turn then input into a set, removing the duplicates, then remove the digits (encoded in `763**4`), then check if the length is the same as the original input

Answer 16

Java 8, 61 39 bytes

s->!s.matches("(?i).*((.).*\\2|\\d).*")

Explanation:

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s->  // Method with String parameter and boolean return-type
  !s.matches("(?i).*((.).*\\2|\\d).*")
     //  Return whether the input does not match the regex

Regex explanation:

String#matches implicitly adds ^...$.

^(?i).*((.).*\2|\d).*$
 (?i)                      Enable case insensitivity
^    .*                    Zero or more leading characters
       (       |  )        Followed by either:
        (.)                 Any character `C`
           .*               with zero or more characters in between
             \2             followed by that same character `C` again
               |           Or:
                \d          Any digit
                   .*$     Followed by zero or more trailing characters

Answer 17

APL (Dyalog Unicode), 12 bytes

Anonymous tacit function.

(∪≡~∘⎕D)819⌶

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819⌶ lowercase

(…) apply the following tacit function on that:

 ~∘⎕D remove Digits from the argument

 ∪≡ are the unique elements of the argument identical to that?

Answer 18

Perl 6, 22 bytes

{!(.uc~~/(.).*$0|\d/)}

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No matches for some character then later the same character. Implicit function as a code block, match implicitly on $_, invert book with !. Added |\d (ta Adam) but also needed .uc~~, which needed parentheses...

Alternative with Bags, 23 bytes

{.uc.ords.Bag⊆65..97}

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This one normalises case then makes a bag (set with incidence counts). Subset or equal only true if all members are members of the comparison Bag, and all incidence counts are less than or equal to those in the comparison Bag. So any repeats or digits would make the comparison false.

Answer 19

Visual Basic for Applications (32 bit), 102 bytes

s=LCase(InputBox(u)):j=1:For i=1To Len(s):k=Mid(s,i,1):j=j*0^Instr(i+1,s,k)*(k Like"[a-z]"):Next:?j<>0

Used the fact that in VBA 0^x yields 1 if x is zero and 0 otherwise. Run in immediate (debug) window.

Edit: as pointed out by Taylor in the comments this only works in 32 bit installs of MS Office.

Answer 20

Julia 0.6, 29 bytes

s->!ismatch(r"\d|(.).*\1"i,s)

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Answer 21

Perl 5 with -n -M5.010, 17 bytes

say!/\d|(.).*\1/i

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Perl 5 port of my Julia answer.

Answer 22

C (gcc), 87 85 83 bytes

• Saved two four bytes thanks to ceilingcat.

f(s,p,b,P)char*s,*p;{for(b=s;*s;++s)for(p=b*=*s>64;b&&p<s;b=(*s^*p++)&95?b:0);s=b;}

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Answer 23

Powershell, 32 bytes

param($s)$s-notmatch'\d|(.).*\1'

Answer 24

Japt v2.0a0, 8 bytes

v
¶r\L â

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v\n¶r\L â     :Implicit input of string U
v             :Lowercase
 \n           :Reassign to U
   ¶          :Is equal to?
    r         :Remove
     \L       :  Regex /[^a-z]/gi
        â     :Deduplicate

Answer 25

Zsh, 32 bytes

c=${#${(uL)${(s::)1}}};((c<$#1))

Try it online!. Isogram/truthy returns 1, otherwise 0

Answer 26

K (ngn/k), 18 bytes

{(a^,/$!10)~?a:_x}

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Answer 27

CJam, 11 bytes

qelA,s+_L|=

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Explanation

The basic idea is to append each digit then check for duplicates. Since the append ensures that each digit is already present once, any further presence of digits will be a duplicate, causing it to return false.

q      e# read the input:            | "MOoSE1"
el     e# convert to lowercase:      | "moose1"
A      e# push 10:                   | "moose1" 10
,      e# range [0,N):               | "moose1" [0 1 2 3 4 5 6 7 8 9]
s      e# string representation:     | "moose1" "0123456789"
+      e# concatenate:               | "moose10123456789"
_      e# duplicate:                 | "moose10123456789" "moose10123456789"
L|     e# union with the empty list: | "moose10123456789" "mose1023456789"
       e# (this gets rid of duplicates)
=      e# Equal to original:         | 0

Answer 28

Red, 76 bytes

func[s][a: charset[#"a"-#"z"#"A"-#"Z"]parse s[any[copy c a ahead not to c]]]

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Answer 29

Smalltalk, 57 bytes

Method to be defined in class String:

s^(self select:#isLetter)asUppercase asSet size=self size

This is most likely self-explanatory.

Answer 30

Pyth, 17 bytes

.Am&!t/rz0d}dGrz0

Test suite

Explanation:

.Am&!t/rz0d}dGrz0 # Code
  m           rz0 # Map the following over the lowercase input:
      /rz0d       #  Count occurrences of d in lowercase input
     t            #   minus 1
    !             #    inverted (0 -> True)
   &              #     and
           }dG    #      d is in the lowercase alphabet
.A                # Print whether all values are truthy

Python 3 translation:

z=input()
print(all(map(lambda d:not z.lower().count(d)-1and d in "abcdefghijklmnopqrstuvwxyz",z.lower())))