37
\$\begingroup\$

Write a function/program that accepts a string of lower/uppercase letters [A-Za-z] as input, that checks whether the occuring letters are unique and in alphabetical order (ignoring lower and uppercase) or not. The output must be truthy if they are unique and in alphabetical order and falsy if not.

Here some testcases

a                           true
abcdefGHIjklmnopqrSTUVWXyz  true     
aa                          false
puz                         true
puzz                        false
puzZ                        false
puZ                         true
PuZ                         true
pzu                         false
pzU                         false
abcdABCD                    false
dcba                        false

If you want, run your program on all words of a wordlist like this one and and post some interesting ones =).

Score

Lowest number of bytes wins.

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  • 3
    \$\begingroup\$ Weak test cases. (See my comment on Richard A's PHP answer.) \$\endgroup\$ – manatwork Mar 11 '15 at 17:20
  • \$\begingroup\$ Does the alphabet loop? Should za be a truthy value? \$\endgroup\$ – MayorMonty Sep 24 '15 at 1:06
  • \$\begingroup\$ No, the alphabet begins with a and ends with z. \$\endgroup\$ – flawr Sep 24 '15 at 12:17
  • \$\begingroup\$ You should have some test cases that aren't in alphabetical order \$\endgroup\$ – Jo King Mar 3 at 21:43
  • 1
    \$\begingroup\$ @JoKing I added some. \$\endgroup\$ – flawr Mar 5 at 9:02

52 Answers 52

28
\$\begingroup\$

CJam, 8 bytes

lel_$_&=

Here is a test harness for all examples in the challenge. This returns 0 or 1 (which are falsy and truthy, respectively, in CJam).

And here is a script to filter the word list in the question (takes a few seconds to run). You'll have to copy the word list into the input field manually, because it's too long for a permalink.

Explanation

l        "Read input.";
 el      "Convert to lower case.";
   _$    "Get a copy and sort it.";
     _&  "Remove duplicates (by computing the set intersection with itself).";
       = "Check for equality with original (lower case) word.";
\$\endgroup\$
21
\$\begingroup\$

Regex (any flavor), 55 bytes

Some people don't consider regex to be a programming language, but it's been used before, and it's not close to being the shortest.

^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?$

I've added one byte for the i (case-insensitive) flag. This is very straightforward and might be shorter to generate on the fly.

If regex alone are not allowed, you can use this 56-byte Retina program suggested by Martin Büttner:

i`^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?$

Running this on the wordlist linked above yielded 10 6-letter words in alphabetical order.

["abhors", "almost", "begins", "begirt", "bijoux", "biopsy", "chimps", "chinos", "chintz", "ghosty"]

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  • 2
    \$\begingroup\$ You can use Retina instead of ES6 if someone complains that regex is not a language: i`^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?$ \$\endgroup\$ – Martin Ender Mar 1 '15 at 17:12
  • \$\begingroup\$ @MartinBüttner I'd forgotten about Retina. Thanks! \$\endgroup\$ – NinjaBearMonkey Mar 1 '15 at 17:17
  • \$\begingroup\$ @MartinBüttner According to the META (meta.codegolf.stackexchange.com/questions/2028/…) Regexes can be 'seen' somewhat as a programming language. \$\endgroup\$ – Ismael Miguel Mar 2 '15 at 9:24
  • \$\begingroup\$ @IsmaelMiguel I know. And in fact that definition was specifically chosen to make sure it doesn't rule out regex. But some people still regularly complain, because you can't use regex like any other language. \$\endgroup\$ – Martin Ender Mar 2 '15 at 12:45
  • \$\begingroup\$ @MartinBüttner Those who complain can go to a place called META and look for it. Why no one visits such a beautiful place full of questions that solve most issues? \$\endgroup\$ – Ismael Miguel Mar 2 '15 at 12:54
19
\$\begingroup\$

Python 3, 44 bytes

*s,=input().lower()
print(sorted(set(s))==s)

A simple approach - check uniqueness, check sortedness.

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  • \$\begingroup\$ Can you explain what *s,=... does? \$\endgroup\$ – flawr Mar 1 '15 at 13:22
  • \$\begingroup\$ @flawr This is called 'starred assignment'. In this code, it simply converts the right side into a list. It's the same as s=list(input().lower()). \$\endgroup\$ – Jakube Mar 1 '15 at 13:40
  • 1
    \$\begingroup\$ @flawr As Jakube says, here it's just converting input into a list of chars. In general it's a special assignment syntax which lets you do things like x,*y = [1, 2, 3, 4], which assigns 1 to x and [2, 3, 4] to y. \$\endgroup\$ – Sp3000 Mar 1 '15 at 13:42
  • \$\begingroup\$ @mbomb007 *s,= is list(s)... link \$\endgroup\$ – Sp3000 Mar 2 '15 at 22:43
  • \$\begingroup\$ You can do {*s} instead of set(s) to save 2 bytes. \$\endgroup\$ – mbomb007 Jan 17 '17 at 19:53
12
\$\begingroup\$

><>, 52 42 39 bytes

0>i:1+?v1n;
? )'`':/'@'v
0v?){:-<'`'/;n

This type of question is one of the few types that ><> is pretty comfortable with, since we only need to deal with one char at a time.

Explanation

Don't get lost! There's a lot of wrapping here.

0            Push 0. We'll be mapping a-z to 1-26, so 0 will be smaller than everything

(loop)
i            Read a char of input
:1+? 1n;     If there's no more input, print 1
:'`')?       If the char is bigger than backtick...
  '`'          Push backtick  (which is one before 'a'), else...
  '@'          Push an @ sign (which is one before 'A')
-            Subtract, mapping a-z to 1-26
:{)?         If the new char is bigger than the previous char...
               Repeat from the beginning of the loop, else...
  0n;          Print 0

Previous solution, 42 bytes

0i:1+?v1n;n0/\!
?)'`':/'@'v
? ){:-<'`'/ vv

The interesting thing is that, despite appearing to have the same functionality, the alternative

0i:1+?v1n;n0\/!
?)'`':/'@'v
? ){:-<'`'/ ^^

(The change is in the arrows and mirrors on the far right)

actually gives incorrect results, due to ><>'s interpreter using a Python defaultdict. What happens is that, by traversing through the empty space at the end of the second row, 0s are implicitly placed into the blank spaces when ><> tries to access the cell. This then messes with the ? conditional trampoline at the beginning of the same row, as the newly placed 0s are skipped rather than the v at the end.

\$\endgroup\$
  • \$\begingroup\$ I feel like you could save some bytes by only substracting 32 from lowercase letters rather than getting alphabetic index for all letters \$\endgroup\$ – Aaron Sep 23 '15 at 15:30
9
\$\begingroup\$

Haskell, 52 Bytes

import Data.Char
and.(zipWith(>)=<<tail).map toLower

Usage: (and.(zipWith(>)=<<tail).map toLower) "abcd" which outputs True.

\$\endgroup\$
9
\$\begingroup\$

C, 67 65 57 54 (52) characters

f(char*s){int c,d=0;for(;(c=*s++)&&(c&~32)>(d&~32);d=c);return!c;}

a little shorter:

f(char*s){int c,d=0;for(;(c=*s++)&&(c&~32)>d;d=c&~32);return!c;}

and even shorter:

f(char*s){int d=32;for(;(*s|32)>d;d=*s++|32);return!*s;}

Here's a little test: http://ideone.com/ZHd0xl

After the latest suggestions here are still two shorter versions:

// 54 bytes
f(char*s){int d=1;for(;(*s&=95)>d;d=*s++);return!*s;}

// 52, though not sure if valid because of global variable
d;f(char*s){d=1;for(;(*s&=95)>d;d=*s++);return!*s;}

Also this code relies on the fact, that in ASCII lowercase and uppercase only differ by the 5th bit (32) which I filter out. So this might not work with other encodings obviously.

EDIT: The latest version always sets the 5th bit as |32 is shorter than &~32.

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  • \$\begingroup\$ Good use of domain knowledge to handle the case sensitivity issue. \$\endgroup\$ – RomSteady Mar 2 '15 at 22:55
  • \$\begingroup\$ Save 2 by replacing the for loop with for(;(*s&=95)>d;d=*s++);. And you can initialize d to 1 without changing the result, saving 1 more. See. \$\endgroup\$ – AShelly Mar 3 '15 at 2:53
  • 1
    \$\begingroup\$ I'm not sure if this is considered legal in code golf, but d;f(char*s){d=32;for...} works, declaring d implicitly as a global int (which, in GCC, is a warning—"data definition has no type or storage class"—but not an error). This saves two bytes. \$\endgroup\$ – wchargin Mar 3 '15 at 3:13
  • \$\begingroup\$ AShelly hm, didn't consider that. Your suggestion changes the original string though. But whatever, it's code golf :D Also I'm not sure about WChargin's hint as d as a global variable would not really be part of the function. \$\endgroup\$ – Felix Bytow Mar 3 '15 at 6:22
  • 1
    \$\begingroup\$ Why not initialize d in the for loop rather than its own statement? That way you save a ;. \$\endgroup\$ – Josh Mar 3 '15 at 16:38
6
\$\begingroup\$

Ruby, 33

->s{c=s.upcase.chars
c==c.sort|c}

Checks to see if the sorted unique characters are the same as all the characters.

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  • 1
    \$\begingroup\$ Think you can get it a little shorter with c==c.sort|c \$\endgroup\$ – histocrat Mar 2 '15 at 20:59
  • \$\begingroup\$ Ooh, I like that, that's clever. Thanks. \$\endgroup\$ – britishtea Mar 2 '15 at 21:10
5
\$\begingroup\$

Javascript (ES5), 101

function i(s){b=0;l=''.a
s.toUpperCase().split('').forEach(function(c){if(c<=l)b=1
l=c})
return!b}

Improved to 87 by edc95:

upvote his comment :)

function i(s){return!s.toUpperCase().split(l='').some(function(c){return(u=l,l=c)<=u})}

Btw, the test cases currently in OP are fulfilled if a program is only checking uniqueness, disregarding order.


I cant write comments yet, so I'll answer some remarks here:

@edc65: Thanks! I tried rewriting it using some(), but I couldn't get a shorter solution, because even though it looks like it would enable me to get rid of the superflous b variable, you need to type "return" twice (same with reduce()), and you can't just return the comparison's result directly, because the last character needs to be saved after the comparison with it.

@edc65: That's a nice use of the comma operator for 87! I edited it into my answer for more visibility.

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  • \$\begingroup\$ That's a better idea than mine. Using .some could be even better (52 with ES6) \$\endgroup\$ – edc65 Mar 1 '15 at 17:25
  • \$\begingroup\$ You can remove the space between return and !b to save a char. \$\endgroup\$ – ProgramFOX Mar 1 '15 at 19:20
  • \$\begingroup\$ As is, just caring white space,96:function i(s){b=0;l='';s.toUpperCase().split('').forEach(function(c){if(c<=l)b=1;l=c});return!b} \$\endgroup\$ – edc65 Mar 1 '15 at 20:02
  • \$\begingroup\$ The same, golfed more,92:function i(s){s.toUpperCase(b=0).split(l='').forEach(function(c){if(c<=l)b=1;l=c});return!b} \$\endgroup\$ – edc65 Mar 1 '15 at 20:03
  • 1
    \$\begingroup\$ Using some(or every, same score),87:function i(s){return!s.toUpperCase().split(l='').some(function(c){return(u=l,l=c)<=u})} \$\endgroup\$ – edc65 Mar 1 '15 at 20:04
4
\$\begingroup\$

Haskell, 90 bytes

Supplies the function f :: String -> Bool

import Data.List
import Distribution.Simple.Utils
f l=g$lowercase l
g l=sort l==l&&l==nub l

Usage (assuming it is saved as golf.hs). ... is used to replace ghci's verbose loading messages.

$ ghci golf.hs
...
*Main> f "as"
...
True
*Main> f "aa"
False

If someone has a lowercase method shorter than import Distribution.Simple.Utils then please comment.

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  • 1
    \$\begingroup\$ Use map toLower from Data.Char instead of lowercase \$\endgroup\$ – nimi Mar 1 '15 at 17:11
  • 1
    \$\begingroup\$ Also: you can remove the parameter l at f, i.e. f=g.lowercase (or f=g.map toLower if you switch to toLower). Within g one comparison is enough: g l=nub(sort l)==l. \$\endgroup\$ – nimi Mar 1 '15 at 18:45
4
\$\begingroup\$

Wolfram Mathematica, 49 37 bytes

f[x_]:=(l=Characters[ToLowerCase[x]];Union[l]==l)

P.S. Shorter solution by Martin Büttner:

Union[l=Characters@ToLowerCase@#]==l&
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  • 2
    \$\begingroup\$ #⋃#==#&@*Characters@*ToLowerCase \$\endgroup\$ – alephalpha Mar 2 '15 at 3:16
  • 1
    \$\begingroup\$ @alephalpha That is beautiful! \$\endgroup\$ – Martin Ender Mar 2 '15 at 17:56
4
\$\begingroup\$

J, 17 bytes

Checks if the lowercase sorted /:~ string equals -: the lowercase nub ~. string.

   (/:~-:~.)@tolower

   NB. testing with the example inputs
   ((/:~-:~.)@tolower) every (1$'a');'abcdefGHIjklmnopqrSTUVWXyz';'aa';'puz';'puzz';'puzZ';'puZ';'PuZ'
1 1 0 1 0 0 1 1

As in J a 1-charater long "string" represented as a regular string (with quotes) is just a character atom not a real string I formatted the input appropriately so all input would be real strings. (In the example above I used 1$'a'.)

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4
\$\begingroup\$

MATLAB, 29 27 bytes

Now for a one-liner which even makes sense outside of code-golf.

As an anonymous function (use as o('yourstring'))

o=@(s)all(diff(lower(s))>0)

I guess this function is pretty self-explanatory since it reads like a newspaper ad.

Previous version (29 bytes):

all(diff(lower(input('')))>0)

Input must be presented between ' marks, e.g. 'Potato'.

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4
\$\begingroup\$

Brachylog, 3 bytes

ḷ⊆Ạ

Try it online!

The predicate succeeds if the input meets the requirements outlined and fails if it does not, printing true. or false. if run as a program.

       The input,
ḷ      lowercased,
 ⊆     is a not-necessarily-contiguous sub-list of
  Ạ    "abcdefghijklmnopqrstuvwxyz".

The first version I came up with, not explicitly referencing the alphabet:

Brachylog, 4 bytes

ḷ≠.o

Try it online!

        The input,
ḷ       lowercased,
 ≠      in which every character is distinct,
  .     is the output variable,
   o    which sorted,
        is still the output variable.
\$\endgroup\$
3
\$\begingroup\$

J, 21 characters

This is too long. The argument must have rank 1, i.e. it must be a string or vector.

*/@(<=~.;/:~)@tolower
  • tolower yy in lower case.
  • /:~ yy in lexical order.
  • ~. y – the nub of y, that is, y with duplicates removed.
  • x ; yx and y put into boxes and then concatenated.
  • < yy put into a box.
  • x = yx compared element-wise with y.
  • (< y) = (~. y) ; (/:~ y) – a vector indicating if y is equal to its nub and itself sorted.
  • */ y – the product of the items of y, or its logical and if the items are booleans.
  • */ (< y) = (~. y) ; (/:~ y) – a boolean indicating the desired property for lowercase y.
\$\endgroup\$
3
\$\begingroup\$

Julia, 44 bytes

s->(l=lowercase(s);l==join(sort(unique(l))))

This creates an anonymous function that takes a single argument s, converts it to lower case, and compares it to the unique sorted version of the string. It returns a boolean, i.e. true or false. If you want to test it out, assign it like f=s->... and then call f("PuZ"), etc.

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  • \$\begingroup\$ Amen to that, @flawr. Thanks for the support. \$\endgroup\$ – Alex A. Mar 2 '15 at 16:28
3
\$\begingroup\$

Pure Bash 4.x, 37

[[ ${1,,} =~ ^`printf %s? {a..z}`$ ]]

Input taken as a command-line parameter. As per standard shell semantics, exit code 0 means true (alphabetic) and exit code != 0 means false (not alphabetic).

The printf creates the regex as in @hsl's solution. The input string is expanded to lowercase and compared against the regex.


Previous answer:

Bash + coreutils, 52

Straightforward solution:

a=`fold -1<<<${1,,}`
cmp -s <(sort -u<<<"$a")<<<"$a"
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  • \$\begingroup\$ Note that this requires bash 4.x. \$\endgroup\$ – Mark Reed Mar 2 '15 at 20:09
  • \$\begingroup\$ @MarkReed Yes. Noted. \$\endgroup\$ – Digital Trauma Mar 2 '15 at 20:38
3
\$\begingroup\$

C# 6, 18 + 82 76 = 94 bytes

Requires (18 bytes):

using System.Linq;

Code (76 bytes):

bool a(string s)=>(s=s.ToLower()).Distinct().OrderBy(x=>x).SequenceEqual(s);

C# 6 supports lambdas to define a function, which is useful for golfing.

Non-C# 6 version:

bool a(string s){return (s=s.ToLower()).Distinct().OrderBy(x=>x).SequenceEqual(s);}

Ungolfed code:

bool IsInAlphabeticalOrder(string s)
{
    s = s.ToLower();
    return s.Distinct()
            .OrderBy(x => x)
            .SequenceEqual(s);
}
\$\endgroup\$
3
\$\begingroup\$

JavaScript (ES6) 54

Convert to uppercase, then to array and sort. If during sort two element are in the wrong order or equal, return 0 (falsy) else 1 (truthy)

Edit Shortened thx to @Optimizer (but still 2 more than the @Tamas solution implemented in ES6: F=s=>[...s.toUpperCase()].every(c=>(u=l,l=c)>u,l=''))

F=s=>[...s.toUpperCase(x=1)].sort((a,b)=>a<b?1:x=0)&&x

Test in Firefox / FireBug console

;['a','abcdefGHIjklmnopqrSTUVWXyz','aa','puz','puzz','puzZ','puZ','PuZ']
.map(w=>w+' '+F(w))

["a 1", "abcdefGHIjklmnopqrSTUVWXyz 1", "aa 0", "puz 1", "puzz 0", "puzZ 0", "puZ 1", "PuZ 1"]

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  • 1
    \$\begingroup\$ s= does not seem to be required... \$\endgroup\$ – Optimizer Mar 3 '15 at 17:26
  • \$\begingroup\$ @Optimizer right, it was a first try when at last i compared the original (uppercased) and the sorted \$\endgroup\$ – edc65 Mar 3 '15 at 18:26
3
\$\begingroup\$

C (44 bytes)

f(char*s){return(*s&=95)?f(s+1)>*s?*s:0:96;}

Test it here: http://ideone.com/q1LL3E

Posting this because I can't comment yet, otherwise it would be a suggestion to improve the existing C answer because I completely stole the case-insensitive idea from the existing C answer.

Returns 0 if the string is not ordered, and a non-zero value if ordered.

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3
\$\begingroup\$

Golang (65 bytes)

Go is not a golf friendly language, also, i suck at golf...

func a(s[]byte)(bool){return len(s)<2||s[0]|32<s[1]|32&&a(s[1:])}

Run it here: http://play.golang.org/p/xXJX8GjDvr

edit 106->102

edit 102->96

edit 96->91

edit 91->87

edit 87->65

I beat the java version, I can stop for today

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3
\$\begingroup\$

Java 8 - 90 89 87 85 chars

The idea here is to use a 'reduce' function that tracks the last char and "gives up" when it detects the sequence is not strictly ascending.

golfed:

int f(String s){return s.toLowerCase().chars().reduce(0,(v,c)->(v<0)?v:(c>v)?c:-1);}

ungolfed:

int f(String s){
    return s.toLowerCase()
            .chars()
            .reduce(0, (v,c) -> (v<0)? v : (c>v)?c:-1);
}

example:

System.out.println(new Quick().f("abc"));
System.out.println(new Quick().f("aa"));
System.out.println(new Quick().f("abcdefGHIjklmnopqrSTUVWXyz"));
System.out.println(new Quick().f("puZ"));
System.out.println(new Quick().f("Puz"));
System.out.println(new Quick().f("cba"));

output:

99
-1
122
122
122
-1
\$\endgroup\$
3
\$\begingroup\$

Perl 6, 35 bytes

{my@c=.uc.comb;@c eq@c.sort.unique}

This produces a callable block; if I could just assume that $_ is already set to the desired word, I could delete the surrounding curly braces and lose two more bytes, but probably the only reasonable way to make that assumption would be to run it with -n and feed the word as standard input, which would add the two bytes right back.

\$\endgroup\$
  • \$\begingroup\$ Sure it does. .uc.comb doesn't rearrange anything, so if the uppercased and combed array is equal to the sorted uppercased and combed array, that means it started out in sorted order. \$\endgroup\$ – Mark Reed Mar 3 at 22:26
  • \$\begingroup\$ right, it's checking the size of the intersection, which ignores order. Ok, updated. \$\endgroup\$ – Mark Reed Mar 3 at 22:33
3
\$\begingroup\$

R, 37 bytes

all(diff(utf8ToInt(scan(,''))%%32)>0)

Try it online!

Posting since this is substantially different and shorter than Michal's R answer.

Converts the letters to ASCII codepoints with utf8ToInt, then takes modulo 32 so that lower and upper letters are converted to the same numbers 1...26. Computes the pairwise differences, and checks that they are all positive.

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2
\$\begingroup\$

Perl, 27

@hsl's regexp dynamically build.

#!perl -p
$"="?";@x=a..z;$_=/^@x?$/i

Also we can do a reverse match: convert the input into a regexp: PuZ => .*p.*u.*z.* and then match this to a string of letters in alphabetical order. Result - also 27 characters.

#!perl -lp
$_=join(s//.*/g,a..z)=~lc
\$\endgroup\$
2
\$\begingroup\$

k (6 bytes)

&/>':_

& returns true if both args are true

/ modifies & to apply "over" a list, like a fold in functional languages

> greater than

': modifies > to apply "each-prior", so returns a vector of booleans stating which elements are greater than their predecessor

_ makes it argument lower case

  _"puzZ"
"puzz"
  >':_"puzZ"
1110b
  &/>':_"puzZ"
0b

(0b means boolean false)

q (13 bytes)

all(>':)lower

q is just syntactic sugar on k. all is defined as &/, and lower is _

\$\endgroup\$
  • 4
    \$\begingroup\$ Can you explain how this works? \$\endgroup\$ – flawr Mar 2 '15 at 16:11
  • \$\begingroup\$ This almost feels like cheating on other languages... Who needs function names, parentheses and semicolons? :) \$\endgroup\$ – Sanchises Mar 2 '15 at 22:59
  • \$\begingroup\$ @sanchises k has all of those things and they work pretty much the same way as in C style languages. It's just that this problem happens to be expressible as a single statement. \$\endgroup\$ – mollmerx Mar 4 '15 at 16:50
2
\$\begingroup\$

Python, 50 bytes

f=lambda x:sorted(set(x.lower()))==list(x.lower())

Try online here: http://repl.it/c5Y/2

\$\endgroup\$
2
\$\begingroup\$

VBA (161 bytes)

Function t(s As String)
t = 0
For i = 2 To Len(s)
a = Left(LCase(s), i)
    If Asc(Right(a, 1)) <= Asc(Right(a, 2)) Then Exit Function
Next
t = 1
End Function  

Compares ascii value with previous letter in lowercase, return 0 (false) when its value is smaller / equal and exit function

\$\endgroup\$
2
\$\begingroup\$

Python 2, 43 bytes

lambda s:eval('"%s"'%'"<"'.join(s.lower()))

Try it online!

Puts < symbols between all the letters (converted to lowercase), and then evals it. Python's chained comparison operators are perfectly happy to evaluate the whole thing as one big boolean expression.

\$\endgroup\$
1
\$\begingroup\$

Erlang, 51

f(S)->G=string:to_lower(S),ordsets:from_list(G)==G.

Uses an ordered set (analogous to java.util.TreeSet) to sort the characters and discard any duplicates. The new list is then compared with the input string.

Test Function:

test() ->
    [io:format("~p ~p~n", [S, f(S)]) || S <- ["a","abcdefGHIjklmnopqrSTUVWXyz","aa","puz","puzz","puzZ","puZ","PuZ"]].
\$\endgroup\$
1
\$\begingroup\$

Java, 96

boolean a(char[]a){int i=-1,l=0;for(;++i<a.length;l+=i>0&&a[i]<=a[i-1]?1:0)a[i]|=32;return l<1;}

Pretty straightforward here. Just convert all to lower and compare each to the previous character.

\$\endgroup\$

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