Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.
Write a function or method to check a sentence to see if it is a pangram or not and show its use.
A pangram is a sentence that contains all the letters of the English alphabet at least once, for example: The quick brown fox jumps over the lazy dog.
C
<lang C>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
int isPangram( char *string ) {
static char *alphabet="abcdefghijklmnopqrstuvwxyz"; char wasused[26], *sp, *ap; int j;
for (j=0; j<26; j++) wasused[j]=0;
for (sp=string;*sp; sp++) {
ap = strchr(alphabet, tolower(*sp));
if (ap != NULL) {
wasused[ap-alphabet] = 1;
}
}
for (j=0; (j<26) && wasused[j]; j++);
// if (j<26) printf("Missing character %c\n", alphabet[j]);
return (j==26);
}
main( int argc, char *argv[]) {
char *pangramTxt;
if (argc < 2) {
printf("usage %s (text to check for pangram)\n", argv[0]);
exit(1);
}
pangramTxt = argv[1];
printf("%s\n",pangramTxt);
printf("Is pangram? %s\n", (isPangram(pangramTxt)?"Yes":"No"));
return 0;
}</lang> Usage:
>pangram "The quick brown fox jumps over lazy dogs." Is Pangram? Yes
C++
<lang cpp>#include <algorithm>
- include <cctype>
- include <string>
using namespace std;
const string kAlphabet("abcdefghijklmnopqrstuvwxyz");
bool is_pangram(string s) {
// Convert to lower case.
transform(s.begin(), s.end(), s.begin(), (int(*)(int))tolower);
// Convert to a sorted sequence of unique characters.
sort(s.begin(), s.end());
s.erase(
unique(s.begin(), s.end()),
s.end());
// Remove non-alphabetic characters.
s.erase(
set_intersection(
s.begin(), s.end(),
kAlphabet.begin(), kAlphabet.end(),
s.begin()),
s.end());
// For a pangram, we're left with a..z.
return s == kAlphabet;
}</lang>
D
There are several ways to write this function, this is a low-level one. <lang D>/*pure*/ nothrow bool isPangram(string txt) {
pure nothrow static uint fullBitSet() { // compile-time function
uint result;
foreach (c; "abcdefghijklmnopqrstuvwxyz")
result |= (1u << (c - 'a'));
return result;
}
enum uint FULL_BIT_SET = fullBitSet(); // 67_108_863;
uint charset;
foreach (c; txt) {
if (c >= 'a' && c <= 'z')
charset |= (1u << (c - 'a'));
else if (c >= 'A' && c <= 'Z')
charset |= (1u << (c - 'A'));
}
return charset == FULL_BIT_SET;
}
void main() {
assert(isPangram("the quick brown fox jumps over the lazy dog")); // true
assert(!isPangram("the quick brown fox jumped over the lazy dog")); // false, no s
assert(isPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ")); // true
assert(!isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ")); // false, no r
assert(!isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ")); // false, no m
assert(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")); // true
assert(!isPangram("")); // false
}</lang>
Clojure
<lang lisp>(defn pangram? [s]
(let [letters (into #{} "abcdefghijklmnopqrstuvwxyz")]
(= (->> s .toLowerCase (filter letters) (into #{})) letters)))</lang>
E
<lang e>def isPangram(sentence :String) {
return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0
}</lang>
&! is the “but-not” or set difference operator.
F#
If the difference between the set of letters in the alphabet and the set of letters in the given string (after conversion to lower case) is the empty set then every letter appears somewhere in the given string: <lang fsharp>let isPangram (str: string) = (set['a'..'z'] - set(str.ToLower())).IsEmpty</lang>
Factor
<lang factor>: pangram? ( str -- ? )
[ "abcdefghijklmnopqrstuvwxyz" ] dip >lower diff length 0 = ;
"How razorback-jumping frogs can level six piqued gymnasts!" pangram? .</lang>
Forth
<lang forth>: pangram? ( addr len -- ? )
0 -rot bounds do
i c@ 32 or [char] a -
dup 0 26 within if
1 swap lshift or
else drop then
loop
1 26 lshift 1- = ;
s" The five boxing wizards jump quickly." pangram? . \ -1</lang>
Fortran
<lang fortran>module pangram
implicit none private public :: is_pangram character (*), parameter :: lower_case = 'abcdefghijklmnopqrstuvwxyz' character (*), parameter :: upper_case = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
contains
function to_lower_case (input) result (output)
implicit none character (*), intent (in) :: input character (len (input)) :: output integer :: i integer :: j
output = input
do i = 1, len (output)
j = index (upper_case, output (i : i))
if (j /= 0) then
output (i : i) = lower_case (j : j)
end if
end do
end function to_lower_case
function is_pangram (input) result (output)
implicit none character (*), intent (in) :: input character (len (input)) :: lower_case_input logical :: output integer :: i
lower_case_input = to_lower_case (input)
output = .true.
do i = 1, len (lower_case)
if (index (lower_case_input, lower_case (i : i)) == 0) then
output = .false.
exit
end if
end do
end function is_pangram
end module pangram</lang> Example: <lang fortran>program test
use pangram, only: is_pangram
implicit none character (256) :: string
string = 'This is a sentence.' write (*, '(a)') trim (string) write (*, '(l1)') is_pangram (string) string = 'The five boxing wizards jumped quickly.' write (*, '(a)') trim (string) write (*, '(l1)') is_pangram (string)
end program test</lang> Output: <lang>This is a sentence. F The five boxing wizards jumped quickly. T</lang>
Haskell
<lang haskell>import Data.Char (toLower) import Data.List ((\\))
pangram :: String -> Bool pangram = null . (['a' .. 'z'] \\) . map toLower
main = print $ pangram "How razorback-jumping frogs can level six piqued gymnasts!"</lang>
Icon and Unicon
Icon
A panagram procedure: <lang Icon>procedure panagram(s) #: return s if s is a panagram and fail otherwise if (map(s) ** &lcase) === &lcase then return s end</lang>
And a main to drive it: <lang Icon>procedure main(arglist)
if *arglist > 0 then
every ( s := "" ) ||:= !arglist || " "
else
s := "The quick brown fox jumps over the lazy dog."
writes(image(s), " -- is") writes(if not panagram(s) then "n't") write(" a panagram.") end</lang>
Unicon
This Icon solution works in Unicon.
Ioke
<lang ioke>Text isPangram? = method(
letters = "abcdefghijklmnopqrstuvwxyz" chars text = self lower chars letters map(x, text include?(x)) reduce(&&)
)</lang>
Here is an example of it's use in the Ioke REPL:
<lang ioke> iik> "The quick brown fox jumps over the lazy dog" isPangram? "The quick brown fox jumps over the lazy dog" isPangram? +> true
iik> "The quick brown fox jumps over the" isPangram? "The quick brown fox jumps over the" isPangram? +> false</lang>
J
Solution: <lang j>require 'strings' isPangram=: (a. {~ 97+i.26) */@e. tolower</lang>
Example use: <lang j> isPangram 'The quick brown fox jumps over the lazy dog.' 1
isPangram 'The quick brown fox falls over the lazy dog.'
0</lang>
Java
<lang java5>public class Pangram {
public static boolean isPangram(String test){
boolean pangram = true;
String lcTest = test.toLowerCase();
for(char a = 'a'; a <= 'z'; a++){
pangram &= lcTest.contains("" + a);
}
return pangram;
}
public static void main(String[] args){
System.out.println(isPangram("the quick brown fox jumps over the lazy dog"));//true
System.out.println(isPangram("the quick brown fox jumped over the lazy dog"));//false, no s
System.out.println(isPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ"));//true
System.out.println(isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ"));//false, no r
System.out.println(isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ"));//false, no m
System.out.println(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"));//true
System.out.println(isPangram(""));//false
}
}</lang> Output:
true false true false false true false
JavaScript
<lang javascript>function is_pangram(str) {
var s = str.toLowerCase(); // sorted by frequency ascending (http://en.wikipedia.org/wiki/Letter_frequency) var letters = "zqxjkvbpygfwmucldrhsnioate"; for (var i = 0; i < 26; i++) if (s.indexOf(letters.charAt(i)) == -1) return false; return true;
}
print(is_pangram("is this a pangram")); // false print(is_pangram("The quick brown fox jumps over the lazy dog")); // true</lang>
Logo
<lang logo>to remove.all :s :set
if empty? :s [output :set] if word? :s [output remove.all butfirst :s remove first :s :set] output remove.all butfirst :s remove.all first :s :set
end to pangram? :s
output empty? remove.all :s "abcdefghijklmnopqrstuvwxyz
end
show pangram? [The five boxing wizards jump quickly.] ; true</lang>
Lua
<lang lua>require"lpeg" S, C = lpeg.S, lpeg.C function ispangram(s)
return #(C(S(s)^0):match"abcdefghijklmnopqrstuvwxyz") == 26
end
print(ispangram"waltz, bad nymph, for quick jigs vex") print(ispangram"bobby") print(ispangram"long sentence")</lang>
OCaml
<lang ocaml>let pangram str =
let ar = Array.make 26 false in String.iter (function | 'a'..'z' as c -> ar.(Char.code c - Char.code 'a') <- true | _ -> () ) (String.lowercase str); Array.fold_left ( && ) true ar</lang>
<lang ocaml>let check str =
Printf.printf " %b -- %s\n" (pangram str) str
let () =
check "this is a sentence"; check "The quick brown fox jumps over the lazy dog.";
- </lang>
outputs:
false -- this is a sentence true -- The quick brown fox jumps over the lazy dog.
Oz
<lang oz>declare
fun {IsPangram Xs}
{List.sub
{List.number &a &z 1}
{Sort {Map Xs Char.toLower} Value.'<'}}
end
in
{Show {IsPangram "The quick brown fox jumps over the lazy dog."}}</lang>
Perl
<lang perl>use List::MoreUtils 'all';
sub pangram {all {$_[0] =~ /$_/i} 'a' .. 'z';}
print "Yes.\n" if pangram 'Cozy lummox gives smart squid who asks for job pen.';</lang>
PicoLisp
<lang PicoLisp>(de isPangram (Str)
(not
(diff
'`(chop "abcdefghijklmnopqrstuvwxyz")
(chop (lowc Str)) ) ) )</lang>
PHP
<lang php>function pangram($str){ $alphabet = Array("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"); foreach($alphabet as $val){ if(stripos($str,$val) === false){ return false; } } return true; }
echo pangram("The quick brown fox jumps over the lazy dog"); // 1 echo pangram("Am I a pangram?"); // 0</lang>
PL/I
<lang PL/I> test_pangram: procedure options (main);
is_pangram: procedure() returns (bit(1) aligned);
declare text character (200) varying; declare c character (1);
get edit (text) (L); put skip list (text);
text = lowercase(text);
do c = 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',
'v', 'w', 'x', 'y', 'z';
if index(text, c) = 0 then return ('0'b);
end;
return ('1'b);
end is_pangram;
put skip list ('Please type a sentence');
if is_pangram() then
put skip list ('The sentence is a pangram.');
else
put skip list ('The sentence is not a pangram.');
end test_pangram; </lang>
Output:
<lang> Please type a sentence
the quick brown fox jumps over the lazy dog The sentence is a pangram. </lang>
PureBasic
<lang PureBasic>Procedure IsPangram_fast(String$)
String$ = LCase(string$)
char_a=Asc("a")
; sets bits in a variable if a letter is found, reads string only once
For a = 1 To Len(string$)
char$ = Mid(String$, a, 1)
pos = Asc(char$) - char_a
check.l | 1 << pos
Next
If check & $3FFFFFF = $3FFFFFF
ProcedureReturn 1
EndIf
ProcedureReturn 0
EndProcedure
Procedure IsPangram_simple(String$)
String$ = LCase(string$)
found = 1
For a = Asc("a") To Asc("z")
; searches for every letter in whole string
If FindString(String$, Chr(a), 0) = 0
found = 0
EndIf
Next
ProcedureReturn found
EndProcedure
Debug IsPangram_fast("The quick brown fox jumps over lazy dogs.") Debug IsPangram_simple("The quick brown fox jumps over lazy dogs.") Debug IsPangram_fast("No pangram") Debug IsPangram_simple("No pangram")</lang>
Python
Using set arithmetic: <lang python>import string, sys if sys.version_info[0] < 3:
input = raw_input
def ispangram(sentence, alphabet=string.ascii_lowercase):
alphaset = set(alphabet) return not alphaset - set(sentence.lower())
print ( ispangram(input('Sentence: ')) )</lang>
Sample output:
Sentence: The quick brown fox jumps over the lazy dog True
Ruby
<lang ruby>def pangram?(sentence)
unused_letters = ('a'..'z').to_a - str.downcase.chars.to_a
unused_letters.empty?
end
p pangram?('this is a sentence') # ==> false p pangram?('The quick brown fox jumps over the lazy dog.') # ==> true</lang>
Scala
<lang scala>def is_pangram(sentence: String) = sentence.toLowerCase.filter(c => c >= 'a' && c <= 'z').toSet.size == 26 </lang>
<lang scala> scala> is_pangram("This is a sentence") res0: Boolean = false
scala> is_pangram("The quick brown fox jumps over the lazy dog") res1: Boolean = true </lang>
Tcl
<lang tcl>proc pangram? {sentence} {
set letters [regexp -all -inline {[a-z]} [string tolower $sentence]]
expr {
[llength [lsort -unique $letters]] == 26
}
}
puts [pangram? "This is a sentence"]; # ==> false puts [pangram? "The quick brown fox jumps over the lazy dog."]; # ==> true</lang>
Ursala
<lang Ursala>
- import std
is_pangram = ^jZ^(!@l,*+ @rlp -:~&) ~=`A-~ letters </lang> example usage: <lang Ursala>
- cast %bL
test =
is_pangram* <
'The quick brown fox jumps over the lazy dog', 'this is not a pangram'>
</lang> output:
<true,false>
VBScript
Implementation
<lang vb>function pangram( s ) dim i dim sKey dim sChar dim nOffset sKey = "abcdefghijklmnopqrstuvwxyz" for i = 1 to len( s ) sChar = lcase(mid(s,i,1)) if sChar <> " " then if instr(sKey, sChar) then nOffset = asc( sChar ) - asc("a") + 1 if nOffset > 1 then sKey = left(sKey, nOffset - 1) & " " & mid( sKey, nOffset + 1) else sKey = " " & mid( sKey, nOffset + 1) end if end if end if next pangram = ( ltrim(sKey) = vbnullstring ) end function
function eef( bCond, exp1, exp2 ) if bCond then eef = exp1 else eef = exp2 end if end function</lang>
Invocation
<lang vb>wscript.echo eef(pangram("a quick brown fox jumps over the lazy dog"), "is a pangram", "is not a pangram") wscript.echo eef(pangram(""), "is a pangram", "is not a pangram")"</lang>