Isogram checker

Question

Challenge :

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

---

What :

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

---

Examples :

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

---

Input :

You may accept input in any reasonable format

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

---

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)})

Try it online!

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

Try it online!

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(,"")))

Try it online!

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))

Try it online!

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}

Try it online!

---

-2 bytes thanks to Kirill L.

Ruby -n, 21 19 18 16 bytes

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

Try it online!

Answer 5

Brachylog, 4 bytes

ḷo⊆Ạ

Try it online!

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

JavaScript (Node.js), 29 25 bytes

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

Try it online!

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

-4 bytes thanks to @KevinCruijssen

Answer 7

Husk, 6 bytes

§=f√ü_

Try it online!

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 8

Japt 2.0, 12 11 bytes

-1 byte thanks to Nit

v
f\l â eUq

Test it online!

Answer 9

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.

Try it online.

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 10

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

Try it here.

Answer 11

MATL, 9 bytes

kt2Y2X&X=

Try it online!

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 12

Python 3, 46 bytes

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

Try it online!

Answer 13

PowerShell, 91 bytes

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

Try it online!

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

Try it online!

Answer 15

Python 2, 57 56 bytes

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

Try it online!

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:

Try it online.

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

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 18

APL (Dyalog Unicode), 12 bytes

Anonymous tacit function.

(∪≡~∘⎕D)819⌶

Try it online!

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 19

Perl 6, 22 bytes

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

Try it online!

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}

Try it online!

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 20

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;}

Try it online!

Answer 21

Swift, 81 bytes

let s=readLine()!.lowercased().characters;print(s.count<1||Set(s).count==s.count)

Try it online!

Answer 22

05AB1E, 4 bytes

lDÔQ

Try it online!

Explanation

l      # convert input to lowercase
 D     # duplicate and push to stack
  Ô    # uniquify the list of characters
   Q   # check for equality

Answer 23

K (ngn/k), 18 bytes

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

Try it online!

Answer 24

APL (Dyalog Unicode), 25 20 22 bytes

''≡'(.).*\1|\d'⎕S'&'⍠1

Try it online!

Returns 1 for true, else 0.

Saved 5 bytes thanks to @H.PWiz

Fixed, and saved another byte thanks to @Adám

How?

''≡'(.).*\1|\d'⎕S'&'⍠1 ⍝ Tacit fn
                    ⍠1 ⍝ Ignore case
               ⎕S'&'   ⍝ Search and return the match(es)
   '(.).*\1|\d'        ⍝ For this regex
''≡                    ⍝ And compare to the empty string

Answer 25

CJam, 11 bytes

qelA,s+_L|=

Try it online!

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 26

Red, 76 bytes

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

Try it online!

Answer 27

C#, 82 bytes

bool f(string s)=>!!(s.GroupBy(c=>c).Any(c=>c.Count()>1|(!Char.IsLetter(c.Key))));

edit: added test for char

edit: using GroupBy to shorten it by 5 byte

Answer 28

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 29

Tcl, 114 bytes

proc I w {lmap c [set L [split $w ""]] {if {[regexp -all -nocase $c $w]>1|![string is alp $c]} {return 0}}
expr 1}

Try it online!

Tcl, 121 bytes

proc I w {lmap c [set L [split $w ""]] {if {[llength [lsearch -al -noc $L $c]]>1|![string is alp $c]} {return 0}}
expr 1}

Try it online!

Still too long for my taste!

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())))

Answer 31

Excel VBA, 82 bytes

An Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window.

[B1:B26]="=Len(A$1)-Len(Substitute(Lower(A$1),Char(Row()+95),""""))":?[Max(B:B)]<2

Answer 32

PHP, 137 Bytes

Try it online!

Code, using ctype_alpha

function f($s){if(ctype_alpha($s)){echo(count(array_unique(count_chars
(strtolower($s))))>2)?0:1;}
else{echo(strlen($s))?0:1;}}

Explanation

(I am absolutely sure this can be shortened)

function f($s){
 if(ctype_alpha($s)){     #ctype_alpha check if the string contains only letters
  echo
   (array_values(array_unique(count_chars(strtolower($s))))!= 
      [0,1])?0:1;
   #count_chars, retuns an array with the times each letter is appear in the string
   #array_unique, returns an array with only the unique values
   #the array values will be [0,1] if no letter repeated
 }else{
    #here the empty string gets controlled, and if the string contained numbers
  echo(strlen($s))?0:1;
 }
}

Yes, i will try to reduce the bytes count :D

Answer 33

Perl 5 -p, 20 bytes

$_=!/\d/*!/(.).*\1/i

Try it online!

Answer 34

Julia 0.6, 29 bytes

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

Try it online!

Answer 35

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

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

Try it online!

Perl 5 port of my Julia answer.

Answer 36

C# (Visual C# Interactive Compiler), 42 bytes

s=>s.ToDictionary(c=>c<65?c/0:c<97?c:c-32)

Try it online!

Tests for success by presence or absence of an exception.

When a character is a number, a divide-by-zero exception is thrown. Uppercase and lowercase letters are converted to uppercase and used to construct a dictionary. A duplicate key exception will be thrown if the same character is used more than once.

Answer 37

C# (.NET Core), 112 bytes

Here's my (super long) take on this:

c=>{var k=c.ToLower().ToCharArray();return!Enumerable.SequenceEqual(k.Distinct(),k)|!c.Any(m=>char.IsDigit(m));}

It checks if the distinct version of the string is the same as the regular version, and if it is so, return false. Also return false if there is a digit in the string

Try it online!

Answer 38

Python 2, 43 bytes

t=input().lower()
print len(set(t))==len(t)

Try it online!

Expects a quote-delimited string from STDIN as input.

Converts the input to lowercase to make comparisons case-insensitive, then compares the number of unique values to the total number of

Answer 39

Powershell, 32 bytes

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

Answer 40

C (gcc) -Do=1<<*s%32, 53+12 47+12 = 59 bytes

i;f(char*s){for(i=o;i&o&&*s/58;i^=o)s++;s=!*s;}

Try it online!

---

Uses the bits of i to store whether a letter has been encountered.

#define o 1<<*s%32  //byte corresponding to letter *s
i;f(char*s){
    for(i=o;        //mark first character
        i&o&&*s/58; //until character duplicate (i&o) or end of string or number (*s/58)
        i^=o)       //toggle the character
        s++;        //step through string
    s=!*s;          //return
}

Answer 41

MBASIC, 202 bytes

1 DIM A(26):INPUT S$:FOR I=1 TO LEN(S$):P=ASC(MID$(S$,I,1))
2 IF P>96 THEN P=P-96
3 IF P>64 THEN P=P-64
4 IF P>26 THEN 7
5 A(P)=A(P)+1:NEXT:FOR I=1 TO 26:IF A(I)>1 THEN 7
6 NEXT:PRINT"true":END
7 PRINT"false"

It's ugly but it works. Merry Christmas!

Answer 42

Swift, 106 bytes

func j(s:String){var a:Set<Character>=[];for c in s{if c<"0"||c>"9"{a.insert(c)}};print(s.count==a.count)}

Try it online!