Is it a pangram?

Question

Write a function or program that takes as its input a string and prints a truthy value if the string is a pangram (a sequence of letters containing at least one of each letter in the English alphabet) and a falsey value otherwise.

Case of letters should be ignored; If the string is abcdefghijklmnopqrstuvwXYZ, then the function should still return a truthy value. Note that the string can contain any other characters in it, so 123abcdefghijklm NOPQRSTUVWXYZ321 would return a truthy value. An empty input should return a falsey value.

Test cases

AbCdEfGhIjKlMnOpQrStUvWxYz

==> True


ACEGIKMOQSUWY
BDFHJLNPRTVXZ

==> True


public static void main(String[] args)

==> False


The quick brown fox jumped over the lazy dogs. BOING BOING BOING

==> True

This is code golf. Standard rules apply. Shortest code in bytes wins.

Question name request: Did the quick brown fox jump over the lazy dog? — user54200, Aug 4 '16 at 11:59

Community · Accepted Answer · 2020-06-17 09:04:33Z

26

Pyth, 7 bytes

L!-Grb0

Explanation:

L             lambda (implicit b:)
    rb0       Convert b to lowercase
   G          Lowercase alphabet, "abcd...z"
  -           Set difference, all elts of first that aren't in second
 !            Logical NOT (The empty string is falsey)

Try the full-program, single-line version here.

edited Jun 17 '20 at 9:04

Community♦

1

answered Dec 10 '15 at 2:30

lirtosiast

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\$\begingroup\$ I think the shortest way to fix this for newlines in the input is to make a function: L!-Grb0. !-Grs.z0 would also work but is longer. \$\endgroup\$ – FryAmTheEggman Dec 10 '15 at 16:13
\$\begingroup\$ Oh, I didn't see the question updated to include \n in the string. Thanks. \$\endgroup\$ – lirtosiast Dec 10 '15 at 22:30
\$\begingroup\$ 6 bytes: pyth.herokuapp.com/… \$\endgroup\$ – Maltysen Dec 11 '15 at 3:56
\$\begingroup\$ @Maltysen While there is a (weak) consensus on allowing strings from input to be delimited by quotes, I'm not sure about this as it goes further in requiring Python string syntax. \$\endgroup\$ – lirtosiast Dec 11 '15 at 16:04
\$\begingroup\$ I never would have thought an alphabet built-in would be useful... \$\endgroup\$ – Cyoce Dec 12 '15 at 8:43

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Brad Gilbert b2gills · Accepted Answer · 2015-12-10 05:48:34Z

Perl 6, 20 bytes

'a'..'z'⊆*.lc.comb

usage:

my &code = 'a'..'z'⊆*.lc.comb;
#  the parameter is ^ there

say code '123abcdefghijklm NOPQRSTUVWXYZ321' # True
say code '123abcdefghijklm NOPQRSTUVWXY'     # False

I used the 3 byte "french" version (⊆) of U+2286 SUBSET OF OR EQUAL TO operator instead of the 4 byte "texas" version ((<=)) which would have also required an extra space in front of it.

Dennis · Accepted Answer · 2016-07-18 05:14:59Z

12

GS2, 11 9 bytes

☺ 6ΘàB1."

Thanks to @MitchSchwartz for golfing off 2 bytes!

The source code uses the CP437 encoding. Try it online!

How it works

☺              Push 32 (code point of space).
  6            Bitwise OR.
   Θ           Make a block of these two instructions and map it over the input.
               This turns uppercase letters into their lowercase counterparts.
      à        Push the lowercase alphabet.
       B1      Swap and apply set difference.
         ."    Push the logical NOT of the length of the result.

edited Jul 18 '16 at 5:14

answered Dec 10 '15 at 4:41

Dennis

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\$\begingroup\$ quick block m2 (\xe9) saves 2 bytes. \$\endgroup\$ – Mitch Schwartz Dec 10 '15 at 5:24
\$\begingroup\$ @MitchSchwartz Oh, so that's how you use those. Thanks! \$\endgroup\$ – Dennis Dec 10 '15 at 5:29

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edc65 · Accepted Answer · 2015-12-12 11:11:28Z

11

JavaScript ES6, 51 57

Edit 6 bytes save thx @user81655

a=>new Set(a.toUpperCase().match(/[A-Z]/g)).size>25

Test snippet

F=a=>new Set(a.toUpperCase().match(/[A-Z]/g)).size>25

function update() {  O.innerHTML=F(I.value) }
I.value='qwertyuiopasdfghjklzxcvbnm';update()

input { width: 70% }

<input id=I oninput='update()'>
<pre id=O></pre>

edited Dec 12 '15 at 11:11

answered Dec 10 '15 at 9:20

edc65

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\$\begingroup\$ Would a.replace(/[^A-Z]|[^a-z]/g,'') or a.replace(/[^A-Z]/gi,'') work? \$\endgroup\$ – user46167 Dec 11 '15 at 1:12
2

\$\begingroup\$ @ev3commander no. A and a must become the same character, else the set will keep them as distinct and the size will be > 26 \$\endgroup\$ – edc65 Dec 11 '15 at 6:21
\$\begingroup\$ What if you use the spread operator with [...a.toUpperCase().replace(/[^A-Z]/g,'')].length>25 ? \$\endgroup\$ – Scott Dec 11 '15 at 16:55
\$\begingroup\$ @ScottKaye obviously no. Try it with 'AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA' \$\endgroup\$ – edc65 Dec 11 '15 at 17:35
1

\$\begingroup\$ @user81655 right, it works, great. Thanks. I should not answer comments while asleep \$\endgroup\$ – edc65 Dec 12 '15 at 11:07

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Community · Accepted Answer · 2020-06-17 09:04:33Z

9

R 50 ,46 39 bytes

all(sapply(letters,grepl,readline(),T))

Edit drops the need for tolower by adding ignore.case=TRUE (T)

edited Jun 17 '20 at 9:04

Community♦

1

answered Dec 11 '15 at 0:05

mnel

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\$\begingroup\$ Not too familiar with R, but shouldn't ignore.case=TRUE (T) be included in the count as well then? \$\endgroup\$ – Ruslan Dec 11 '15 at 13:30
2

\$\begingroup\$ @Ruslan It is! It is the Tas the end, thanks to argument placement matching there is no need to actually specify the name of the argument (and T is the default alias for TRUE). The code written here performs the needed action as is, without any need to add anything. \$\endgroup\$ – plannapus Dec 11 '15 at 14:29

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Community · Accepted Answer · 2020-06-17 09:04:33Z

8

O, 11 bytes

GQ_s{n-}dS=

Try it online.

Sadly, O does not have set difference :/

Explanation

G            Pushes the alphabet to the stack
 Q           Pushes input to the stack
  _          Converts the string to lowercase
   s         Split string into char array
    {  }d    Iterate through array
     n       Pushes current element to the stack
      -      String subtraction
         S   Pushes a blank string to the stack
          =  Equals

edited Jun 17 '20 at 9:04

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1

answered Dec 12 '15 at 1:42

a spaghetto

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Community · Accepted Answer · 2017-04-13 12:39:03Z

7

Python 2, 53 51 bytes

f=lambda s,c=65:c>90or(chr(c)in s.upper())*f(s,c+1)

Alternate solutions:

lambda s:all(chr(c)in s.upper()for c in range(65,91))

lambda s:not set(range(65,91))-set(map(ord,s.upper()))

Thanks to xnor for pointing out that sets have an <= operator, for an alternate 51:

lambda s:set(range(65,91))<=set(map(ord,s.upper()))

edited Apr 13 '17 at 12:39

Community♦

1

answered Dec 10 '15 at 4:16

Mitch Schwartz

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1

\$\begingroup\$ If I'm not mistaken, the last expression is the same as lambda s:set(range(65,91))<=set(map(ord,s.upper())), also for 51. \$\endgroup\$ – xnor Dec 10 '15 at 5:06
\$\begingroup\$ Python 3.5 can save bytes here: p=lambda s:{*range(65,91)}<={*map(ord,s.upper())}. By the way, I can't seem to find any rules on whether a lambda needs to be assigned (as in your first case) or not (as in your later ones). Help? \$\endgroup\$ – Tim Pederick Dec 11 '15 at 16:30
\$\begingroup\$ @TimPederick Naming the lambda is unnecessary unless you need to use the function elsewhere, like in the first recursive solution. \$\endgroup\$ – FryAmTheEggman Dec 11 '15 at 16:57
\$\begingroup\$ @TimPederick Thanks for pointing that out. I renamed my answer as Python 2 instead of just Python. You have my blessing to post that as a new answer if you want, which I think would be ok by community norms although I'm not sure. \$\endgroup\$ – Mitch Schwartz Dec 11 '15 at 17:33
\$\begingroup\$ @FryAmTheEggman: Thanks for clarifying. That distinction hadn't occurred to me! I've also found a meta post explaining the rule. There goes two bytes from a few things I've written... \$\endgroup\$ – Tim Pederick Dec 12 '15 at 13:51

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Glen O · Accepted Answer · 2015-12-10 03:47:47Z

Julia, 38 bytes

s->endof(∩('a':'z',lowercase(s)))>25

This is simple - lowercase deals with the uppercase/lowercase issue, 'a':'z' holds all of the lowercase letters, ∩ is intersection, removes any character that isn't a letter and, because 'a':'z' comes first, will only have one of each letter that appears in s. endof is the shortest way to get the length of the resulting array, and if it's 26, then it's a pangram (it can't be more than 26, and >25 saves a byte relative to ==26).

Martin Ender · Accepted Answer · 2015-12-10 16:07:44Z

Retina, 22 bytes

Msi`([a-z])(?!.*\1)
26

Try it online.

The first line matches any letter which does not appear again later in the string. That ensures that we don't match each letter at most once, no matter how often it occurs. Match mode will by default replace the string with the number of matches found. So in the second stage, we match 26 against the result of the first input, which will give either 0 or 1, depending on whether we found the maximum of 26 matches or not.

Alexis Andersen · Accepted Answer · 2015-12-13 21:14:55Z

5

Ruby, 41 33

->s{(?a..?z).all?{|c|s[/#{c}/i]}}

Usage

p=->s{(?a..?z).all?{|c|s[/#{c}/i]}}
p["AbCdEfGhIjKlMnOpQrStUvWxYz"] 
  #=> true
p["ACEGIKMOQSUWY
BDFHJLNPRTVXZ"]
  #=> true
p["public static void main(String[] args)"]
  #=> false
p["The quick brown fox jumped over the lazy dogs. BOING BOING BOING"]
  #=> true

Thanks to Vasu Adari for saving me 8 bytes

edited Dec 13 '15 at 21:14

answered Dec 10 '15 at 17:31

Alexis Andersen

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2

\$\begingroup\$ You can save 8 bytes by making your regex to ignorecase. \$\endgroup\$ – Vasu Adari Dec 11 '15 at 7:49

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xnor · Accepted Answer · 2019-03-16 15:59:11Z

5

Haskell, 43 bytes

f s=until(all(`notElem`s))(succ<$>)"Aa">"["

Try it online!

Iterates through the lowercase/uppercase pairs until it finds one where both are not elements of the input, and checks that this failure is past the alphabet.

answered Mar 16 '19 at 15:59

xnor

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Minkolang 0.14, 18 bytes

$o7$ZsrlZ'26'$ZN.

Try it here.

Explanation

$o                    Read in whole input as characters
  7$Z                 Uppercase every letter
     s                Sort
      r               Reverse
       lZ             Alphabet - uppercase and lowercase
         '26'         Pushes 26 on the stack
             0$Z      Count how often the top 26 numbers of the stack appear in the stack
                N.    Output as number and stop.

Community · Accepted Answer · 2017-04-13 12:39:17Z

Python 3.5, 47 bytes

lambda s:{*map(chr,range(65,91))}<={*s.upper()}

Same principle as Mitch Schwartz's answer, but using the PEP 0448 enhancements to * unpacking, first introduced in Python 3.5.

This version differs slightly from what I wrote in my comment to Mitch's post, in that I turn the numbers into letters rather than vice versa. That's because that's how I wrote my original attempts at a solution, before discovering that I couldn't out-golf Mitch without outright copying his approach. So consider that tweak my one remaining shred of originality!

Community · Accepted Answer · 2020-06-17 09:04:33Z

R, 53 45 bytes

all(97:122%in%utf8ToInt(tolower(readline())))

Old version at 53 bytes:

all(letters%in%strsplit(tolower(readline()),"")[[1]])

Usage:

> all(97:122%in%utf8ToInt(tolower(readline())))
The quick brown fox jumps over the lazy dog
[1] TRUE
> all(97:122%in%utf8ToInt(tolower(readline())))
Write a function or program that takes as its input a string and prints a truthy value if the string is a pangram and a falsey value otherwise.
[1] FALSE
> all(97:122%in%utf8ToInt(tolower(readline())))
123abcdefghijklm NOPQRSTUVWXYZ321
[1] TRUE
> all(97:122%in%utf8ToInt(tolower(readline())))
Portez ce vieux whisky au juge blond qui fume
[1] TRUE

Magic Octopus Urn · Accepted Answer · 2016-11-01 11:52:46Z

4

05AB1E, 4 bytes (Probably Non-competing)

lêAå

Try it online!

l    # Push lowercase input.
 ê   # Push sorted, uniquified lowercase input.
  A  # Push lowercase alphabet.
   å # Is lowercase alphabet in sorted, uniquified, lowercase input?
     # True if panagram, false if not.

answered Nov 1 '16 at 11:52

Magic Octopus Urn

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Community · Accepted Answer · 2020-06-17 09:04:33Z

MATLAB / Octave, 35 33 bytes

@(x)~nnz(setdiff(65:90,upper(x)))

Try it online!

The anonymous function returns a logical 1 if the input x is a pangram, or a logical 0 if it isn't.

Essentially it uses the same approach as @ThomasKwa's Pyth solution. The set difference between all characters in the upper case alphabet range (65:91) and the input string (converted to upper case). Any characters that are in the alphabet but not in the input string are returned by setdiff. Only if the array returned by the set difference is empty is the string a pangram.

Using upper case instead of lower case saves a couple of bytes compared with 'a':'z' because the ASCII value can be used instead to make the range.

\$\begingroup\$ Great answer! Mine was 10 bytes longer \$\endgroup\$ – Luis Mendo Dec 10 '15 at 23:37 — Luis Mendo, Dec 10 '15 at 23:37

Unrelated String · Accepted Answer · 2019-03-10 11:05:42Z

4

Brachylog, 4 bytes

ḷo⊇Ạ

Try it online!

        The input
ḷ       lowercased
 o      sorted
  ⊇     is a superlist of
   Ạ    the lowercase alphabet.

answered Mar 10 '19 at 11:05

Unrelated String

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Laikoni · Accepted Answer · 2019-03-16 07:36:44Z

Haskell, 59 56 53 51 bytes

p s=and[any(`elem`map toEnum[a,a+32])s|a<-[65..90]]

Try it online!

Explanation:

Give an input string s, for each a in range 65 to 90 (the ASCII codes for A to Z) it is checked whether any character in s is equal to either a (the upper case character) or a+32 (the lower case character), converted to a character by toEnum. This generates a list of booleans. and checks if they're all True.

Old version:

import Data.Char
p s=and[any((==)a.toUpper)s|a<-['A'..'Z']]

For every upper case alphabet letter, check whether some letter from s in upper case is equal to it. any(==a)s is the same as elem a s but allows to modify the elements of s before the comparison - in this case, covert them to upper case.

Community · Accepted Answer · 2020-06-17 09:04:33Z

3

Japt, 14 bytes

#ao#{ e@Uv fXd

Try it online!

How it works

        // Implicit: U = input string
#ao#{   // Generate a range of integers from charCode("a") to charCode("{").
e@      // Check if every item X in this range returns truthily to:
Uv fXd  //  convert U to lowercase, and put all instances of X.toCharCode() in an array.
        // This returns false if U does not contain one of the characters.
        // Implicit: output last expression

edited Jun 17 '20 at 9:04

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1

answered Dec 10 '15 at 4:47

ETHproductions

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Reto Koradi · Accepted Answer · 2015-12-10 04:53:53Z

3

CJam, 11 bytes

'[,65>qeu-!

This is a complete program. Try it online.

Explanation:

'[,65>  Build upper case alphabet (see CJam tips thread).
q       Get input.
eu      Convert to all upper case.
-       Set difference between alphabet and upper cased input.
!       Negate.

answered Dec 10 '15 at 4:53

Reto Koradi

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SuperJedi224 · Accepted Answer · 2015-12-28 17:30:18Z

3

Javascript, 110 109 99 95 93 bytes

a=prompt(b=0).toUpperCase();for(i=65;i++<91;)b+=!~a.indexOf(String.fromCharCode(i));alert(!b)

Saved 6 bytes thanks to Thomas Kwa, and 10 thanks in part to ev3.

edited Dec 28 '15 at 17:30

answered Dec 10 '15 at 2:40

SuperJedi224

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\$\begingroup\$ Would b=0 work for b=[]? \$\endgroup\$ – user46167 Dec 11 '15 at 1:13
\$\begingroup\$ Not with this approach. But I may be able to make that work. \$\endgroup\$ – SuperJedi224 Dec 11 '15 at 1:32
\$\begingroup\$ I don't know Javascript, but can you do for(i=65;i++<91;)b+=!~a.indexOf(String.fromCharCode(i));alert(!b)? \$\endgroup\$ – lirtosiast Dec 11 '15 at 1:40
\$\begingroup\$ Wow. That's even shorter than what I just did. \$\endgroup\$ – SuperJedi224 Dec 11 '15 at 1:50

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Community · Accepted Answer · 2017-04-13 12:38:59Z

3

PowerShell v3+, 65 56 52 Bytes

($args.ToLower()-split''|sls [a-z]|group).Count-eq26

Thanks to TessellatingHeckler for the 9-byte golf.

Takes the input string, converts it .ToLower()case, then -splits on every character
Those are fed into an alias sls for Select-String which matches based on a regex [a-z] to pull out only the letters
Those are then fed into Group-Object, so we're only selecting one individual instance of each letter
That is then .Counted to see if it's -equal to 26, and prints True or False accordingly
Requires PowerShell v3 or newer for the sls alias

edited Apr 13 '17 at 12:38

Community♦

1

answered Dec 10 '15 at 19:41

AdmBorkBork

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\$\begingroup\$ It's a lot easier to pinch someone else's effort and try and shorten it a bit, than to read the questions from scratch - and more fun to compete within PowerShell than to go directly up against CJam and friends. But I can stop chasing your answers if it's annoying, sorry. \$\endgroup\$ – TessellatingHeckler Dec 15 '15 at 16:58
\$\begingroup\$ @TessellatingHeckler Not annoying! :D I'm just trying to give you more encouragement! \$\endgroup\$ – AdmBorkBork Dec 15 '15 at 18:00

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Community · Accepted Answer · 2020-06-17 09:04:33Z

2sable, 6 5 bytes

6 byte version:

AIl-g_

Try it online!

Explanation:

A        Push alphabet
 Il      Push lowercase input
   -     Remove all chars of input from alphabet
    g    Get length of the remainder
     _   Print negative bool, where length < 1 = 1 (true), length > 0 = 0 (false)

5 byte version, inspired by carusocomputing's 05AB1E answer:

lÙ{Aå

Try it online!

Explanation:

l        Push lowercase input
 Ù{      Push sorted uniquified input
   A     Push alphabet
    å    Is alphabet in sorted, uniquified input?

voidhawk · Accepted Answer · 2019-03-10 19:09:49Z

3

+100

APL (Dyalog Extended), 6 bytes

∧/⎕A∊⌈

Try it online!

Explanation:

∧/⎕A∊⌈  ⍝ Monadic function train
      ⌈  ⍝ Convert the input to uppercase
  ⎕A∊   ⍝ For each item in the set of uppercase
         ⍝ letters, determine if it exists in the
         ⍝ uppercased input
∧/       ⍝ And-reduce: test if all letters exist

answered Mar 10 '19 at 19:09

voidhawk

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Retina, 20 bytes

T`L`l
D`.
C`[a-z]
26

Try it online!

Explained

T`L`l        Transliterate stage. Replaces uppercase letters with lowercase letters.
D`.          Deduplicate stage - keep one copy of every match (meaning every character in this case), discarding suplicates.
C`[a-z]      Count stage - Count lowercase letters
26           (implicit) Count stage - Match the string "26"
             (Since there can't be duplicates, a match means each character occurred once.)
             (output the result of the last stage - 1 for pangrams, 0 for non-pangrams)

Deadcode · Accepted Answer · 2019-12-30 18:33:42Z

These have the same regex as in my regex answer. I felt it was worth posting a standalone answer showing its use in languages that don't require an import to access regex functions (resulting in very effective golf). In order of how favorably it compares against the current best non-regex answer:

JavaScript ES6, 37 bytes

a=>/(.*([A-Z])(?!.*\2)){26}/i.test(a) (37 bytes, always slow) - Try it online!
a=>/(.*?([A-Z])(?!.*\2)){26}/i.test(a) (38 bytes, slow for non-matches) - Try it online!
a=>/((?=(.*?([A-Z])(?!.*\3)))\2){26}/i.test(a) (46 bytes, fairly reasonable speed) - Try it online!
a=>/^((?=(.*?([A-Z])(?!.*\3)))\2){26}/i.test(a) (47 bytes, very reasonable speed) - Try it online!

When looping this set of test cases, it can be seen that the 47 byte version is about 12 times as fast as the 46 byte version (in SpiderMonkey's regex engine).

PHP, 49 or 51 bytes

PHP imposes a strict backtracking limit (or time limit?) on its regexes, so the version that would be 48 bytes doesn't work at all; it always returns an empty result (except for very short non-pangram strings, for which it prints 0).

<?=preg_match('/(.*?(\pL)(?!.*\2)){26}/i',$argn); (49 bytes) - Try it online!

The 49 byte version, however, actually runs fast; it prints an empty result or 0 for false, and prints 1 for true. However, with sufficiently long pangram strings having a long separation between occurrences of new letters, it returns a false negative, due to taking too long to process the string.

<?=preg_match('/(?>.*?(\pL)(?!.*\1)){26}/i',$argn); (51 bytes) - Try it online!
<?=preg_match('/^(?>.*?(\pL)(?!.*\1)){26}/i',$argn); (52 bytes) - Try it online!

These versions print 0 for false and 1 for true.

Perl 5 `-p`, 27 bytes

$_=/(.*(\pL)(?!.*\2)){26}/i (27 bytes, always slow) - Try it online!
$_=/(.*?(\pL)(?!.*\2)){26}/i (28 bytes, slow for non-matches) - Try it online!
$_=/(?>.*?(\pL)(?!.*\1)){26}/i (30 bytes, fairly reasonable speed) - Try it online!
$_=/^(?>.*?(\pL)(?!.*\1)){26}/i (31 bytes, very reasonable speed) - Try it online!

Perl 5, 33 bytes

Ties with the non-regex answer.

say<>=~/(.*(\pL)(?!.*\2)){26}/i+0 (33 bytes, always slow) - Try it online!
say<>=~/(.*?(\pL)(?!.*\2)){26}/i+0 (34 bytes, slow for non-matches) - Try it online!
say<>=~/(?>.*?(\pL)(?!.*\1)){26}/i+0 (36 bytes, fairly reasonable speed) - Try it online!
say<>=~/^(?>.*?(\pL)(?!.*\1)){26}/i+0 (37 bytes, very reasonable speed) - Try it online!

Ruby `-n`, 29 bytes

Prints nil for false and 0 for true. If printing 0 or 1 is desired, replace ~ with !! (+1 byte).

p ~/(.*([A-Z])(?!.*\2)){26}/i (29 bytes, always slow) - Try it online!
p ~/(.*?([A-Z])(?!.*\2)){26}/i (30 bytes, slow for non-matches) - Try it online!
p ~/(?>.*?([A-Z])(?!.*\1)){26}/i (32 bytes, fairly reasonable speed) - Try it online!
p ~/^(?>.*?([A-Z])(?!.*\1)){26}/i (33 bytes, very reasonable speed) - Try it online!

For comparison, the 33 byte Ruby answer turns into a 28 byte answer when using -n instead of a lambda (and prints false or true):

p (?a..?z).all?{|c|~/#{c}/i} - Try it online!

Ruby, 34 bytes

Curiously, this is only 1 byte longer than the mixed code/regex answer:

->s{s[/(.*([A-Z])(?!.*\2)){26}/i]} (34 bytes) - Try it online!
->s{s[/(.*?([A-Z])(?!.*\2)){26}/i]} (35 bytes) - Try it online!
->s{s[/(?>.*?([A-Z])(?!.*\1)){26}/i]} (37 bytes) - Try it online!
->s{s[/^(?>.*?([A-Z])(?!.*\1)){26}/i]} (38 bytes) - Try it online!

When looping this set of test cases, the same result is seen: the 38 byte version is 12 times as fast as the 37 byte version.

Downgoat · Accepted Answer · 2015-12-10 05:58:40Z

2

TeaScript, 12 bytes

Sz.e»xL.I(l©

First TeaScript post since I killed TeaScript :p

Try it online

Ungolfed

Sz.e(#xL.I(l))

Sz   // Lower case alphabet
.e(#   // Loop through alphabet, ensure
       // for every character, below returns true
    xL    // Input lowercased
    .I(l) // Checks if above contains current char
)

answered Dec 10 '15 at 5:58

Downgoat

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1

\$\begingroup\$ ;-; I feel bad now. TBH I like TeaScript the most. \$\endgroup\$ – Conor O'Brien Dec 10 '15 at 19:24

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Conor O'Brien · Accepted Answer · 2015-12-10 19:37:17Z

2

JavaScript ES6, 124 114 113 bytes

I'm sure this can be golfed more.

v=(Q,s=[...Array(26)].map((x,i)=>String.fromCharCode(i+97)))=>s.length-1?Q.search(RegExp(s.pop(),"i"))+1&&v(Q,s):1

Generates an anonymous function.

v=(Q,s=[...Array(26)].map((x,i)=>String.fromCharCode(i+97)))=>s.length-1?Q.search(RegExp(s.pop(),"i"))+1&&v(Q,s):1

alert(v(prompt("Enter pangram:")));

edited Dec 10 '15 at 19:37

answered Dec 10 '15 at 2:40

Conor O'Brien

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\$\begingroup\$ @apsillers I think I found the problem. Please test it again (my browser does not support ES6 atm) \$\endgroup\$ – Conor O'Brien Dec 10 '15 at 19:21
\$\begingroup\$ Yep, looks good now! \$\endgroup\$ – apsillers Dec 10 '15 at 19:30

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RCB · Accepted Answer · 2015-12-10 21:55:39Z

2

C, 107 bytes

#include<string.h>
int p(char*i){int a=64;while(++a<91)if(!strchr(i,a)&!strchr(i,a+32))return 0;return 1;}

answered Dec 10 '15 at 21:55

RCB

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Neil · Accepted Answer · 2015-12-11 00:42:38Z

2

ES6, 68 bytes

s=>[..."abcdefghijklmnopqrstuvwxyz"].every(x=>RegExp(x,"i").test(s))

That string looks awfully wasteful, but I don't know any better way.

answered Dec 11 '15 at 0:42

Neil

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\$\begingroup\$ Maybe using a range of charcodes? \$\endgroup\$ – Cyoce Dec 12 '15 at 8:51
\$\begingroup\$ @Cyoce That got me thinking and I tried a range of base 36 digits but so far it still takes 70 bytes: s=>[...Array(x=9,26)].every(z=>RegExp((++x).toString(36),"i").test(s)) \$\endgroup\$ – Neil Dec 13 '15 at 0:29

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Is it a pangram?

75 Answers 75

Pyth, 7 bytes

Perl 6, 20 bytes

GS2, 11 9 bytes

How it works

JavaScript ES6, 51 57

R 50 ,46 39 bytes

O, 11 bytes

Explanation

Python 2, 53 51 bytes

Julia, 38 bytes

Retina, 22 bytes

Ruby, 41 33

Usage

Haskell, 43 bytes

Minkolang 0.14, 18 bytes

Explanation

Python 3.5, 47 bytes

R, 53 45 bytes

05AB1E, 4 bytes (Probably Non-competing)

MATLAB / Octave, 35 33 bytes

Brachylog, 4 bytes

Haskell, 59 56 53 51 bytes

Japt, 14 bytes

How it works

CJam, 11 bytes

Javascript, 110 109 99 95 93 bytes

PowerShell v3+, 65 56 52 Bytes

2sable, 6 5 bytes

APL (Dyalog Extended), 6 bytes

Retina, 20 bytes

Explained

JavaScript ES6, 37 bytes

PHP, 49 or 51 bytes

Perl 5 -p, 27 bytes

Perl 5, 33 bytes

Ruby -n, 29 bytes

Ruby, 34 bytes

TeaScript, 12 bytes

Ungolfed

JavaScript ES6, 124 114 113 bytes

C, 107 bytes

ES6, 68 bytes

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