27
\$\begingroup\$

Your input will be an English sentence, phrase, or word. It will only contain a-zA-Z' -,.!?. Your task is to take the input, remove spaces, and then redistribute capitalization such that letters at indexes that were capitalized before (and only letters at indexes that were capitalized before) are capitalized.

For example, if the input is A Quick Brown Fox Jumped Over The Lazy Dog, the (0-based) indexes of the capital letters are 0, 2, 8, 14, 18, 25, 30, 34, 39. Next, remove spaces from the input: AQuickBrownFoxJumpedOverTheLazyDog. Next, lowercase all letters, but uppercase those at 0, 2, 8, 14, 18, 25, 30, 34, 39: AqUickbrOwnfoxJumpEdovertHelazYdog, which is your output.

Input

Your input will be an English sentence, phrase, or word. It can only contain lowercase letters, uppercase letters, hyphens, apostrophes, commas, periods, question marks, exclamation marks, and spaces.

Output

The input with spaces removed, lowercase-d, with letters at the index of capital letters in the input uppercase-d.

NOTE: Your program cannot crash (error such execution terminates) with an IndexOutOfRange or similar error.

Test Cases

Hi! Test!
Hi!tEst!

A Quick Brown Fox Jumped Over The Lazy Dog
AqUickbrOwnfoxJumpEdovertHelazYdog

testing TESTing TeStING testing testing TESTING
testingtESTIngteStInGTEstingtestingtestiNG

TESTING... ... ... success! EUREKA???!!! maybe, don't, NOOOOO
TESTING.........success!eureKA???!!!maybe,don't,nooooo

Enter        PASSWORD ---------
Enterpassword---------

A a B b C c D d E e F f G g H h I i J j K k L l M m N n O o P p Q q R r S s T t U u V v W w X x Z z
AabbCcddEeffGghhIijjKkllMmnnOoppQqrrSsttUuvvWwxxZz

  TEST
teST
\$\endgroup\$
  • \$\begingroup\$ Sandbox \$\endgroup\$ – Stephen Jun 28 '17 at 21:06
  • \$\begingroup\$ 'For example, if the input is "A Quick Brown Fox Jumped Over The Lazy Dog", the (0-based) indexes of the capital letters are 0, 2, 8, 14, 18, 23, 27, 32' They are 0, 2, 8, 14, 18, 25, 30, 34, 39 \$\endgroup\$ – Luke Sawczak Jun 28 '17 at 21:18
  • \$\begingroup\$ @LukeSawczak thank you, my bad \$\endgroup\$ – Stephen Jun 28 '17 at 21:23
  • \$\begingroup\$ Traling spaces not allowed, I assume? \$\endgroup\$ – Luis Mendo Jun 28 '17 at 21:38
  • \$\begingroup\$ @LuisMendo your assumption is correct. This is code-golf, right? :P \$\endgroup\$ – Stephen Jun 28 '17 at 21:40

33 Answers 33

7
\$\begingroup\$

Jelly, 14 13 bytes

nŒlTɓḲFŒlŒuṛ¦

Try it online!

How it works

nŒlTɓḲFŒlŒuṛ¦  Main link. Argument: s (string)

 Œl            Convert s to lowercase.
n              Perform character-wise "not equal" comparison.
   T           Get the indices of all truthy elements, i.e., the indices of all
               uppercase letters in s. Let's call the resulting array J.
    ɓ          Begin a dyadic chain with left argument s and right argument J.
     ḲF        Split s at spaces and flatten, removing the spaces.
       Œl      Convert s to lowercase.
            ¦  Sparse application:
         Œu        Convert s to uppercase.
           ṛ       Take the resulting items of the uppercased string at all indices
                   in J, the items of the lowercased string at all others.
\$\endgroup\$
14
\$\begingroup\$

C (gcc), 82 79 74 72 69 67 66 bytes

f(c){for(char*s=c,*p=c;c=*s++;c&&putchar(c^(*p++|~c/2)&32))c&=95;}

Try it online!

\$\endgroup\$
7
\$\begingroup\$

Python 2, 114 bytes

x=input()
X=x.replace(' ','')
print''.join([X[i].upper()if x[i].isupper()else X[i].lower()for i in range(len(X))])

Try it online!

Equivalently:

Python 2, 114 bytes

lambda x:''.join([[str.lower,str.upper][x[i].isupper()](x.replace(' ','')[i])for i in range(len(x)-x.count(' '))])

Try it online!

\$\endgroup\$
  • \$\begingroup\$ ''.join([(X[i].lower,X[i].upper)[x[i].isupper()]()for i in range(len(X))]) for -5 bytes. \$\endgroup\$ – ovs Jun 29 '17 at 12:08
5
\$\begingroup\$

Python 3, 78 75 72 bytes

s=input()
for c in s:s=s[c>' '!=print(end=(c+c).title()[s<'@'or'['<s]):]

Thanks to @xnor for golfing off 6 bytes!

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Can you just compare s instead of s[0]? \$\endgroup\$ – xnor Jun 29 '17 at 4:33
  • \$\begingroup\$ Yes, of course. Thanks! \$\endgroup\$ – Dennis Jun 29 '17 at 4:35
  • 1
    \$\begingroup\$ (c*2).title() can get you both cases, though switched. \$\endgroup\$ – xnor Jun 29 '17 at 4:52
  • \$\begingroup\$ Another 3 bytes. Thanks again! \$\endgroup\$ – Dennis Jun 29 '17 at 5:04
  • \$\begingroup\$ Tricky! Took me a while to figure out that c>' '!=f() is equivalent to (c>' ') and (' '!=f()). \$\endgroup\$ – Chas Brown Jun 29 '17 at 7:41
5
\$\begingroup\$

05AB1E, 15 14 bytes

-1 byte thanks to Emigna

ðKuvy¹Nè.lil}?

Try it online!

ðK             # Remove spaces
  u            # Convert to uppercase
   vy          # For each character...
     ¹Nè       #   Get the character at the same index from the original input
        .lil}  #   If it was a lowercase letter change this one to lowercase
             ? # Print without a newline
\$\endgroup\$
  • \$\begingroup\$ You save a byte if you uppercase the space-removed string and lowercase it in the condition. \$\endgroup\$ – Emigna Jun 29 '17 at 7:35
5
\$\begingroup\$

Haskell, 98 95 89 88 81 bytes

Thanks to @name, @nimi, @Zgarb, and @Laikoni for helping shave off 14 bytes total

import Data.Char
\s->zipWith(\p->last$toLower:[toUpper|isUpper p])s$filter(>' ')s

Ungolfed:

import Data.Char
\sentence -> zipWith (\oldChar newChar ->
                        if isUpper oldChar
                        then toUpper newChar
                        else toLower newChar)
                     sentence
                     (filter (/= ' ') sentence)
\$\endgroup\$
  • \$\begingroup\$ On mobile, but looks like you can save some bytes with filter(/=' ') \$\endgroup\$ – Henry Jun 29 '17 at 3:29
  • \$\begingroup\$ Yep, certainly can. Missed the part of the spec noting that spaces were the only whitespace that needed removing. \$\endgroup\$ – Julian Wolf Jun 29 '17 at 4:43
  • 1
    \$\begingroup\$ filter(>' ') for one byte less \$\endgroup\$ – nimi Jun 29 '17 at 5:33
  • 2
    \$\begingroup\$ I think the body of the lambda can be shortened to last(toLower:[toUpper|isUpper p])c \$\endgroup\$ – Zgarb Jun 29 '17 at 6:39
  • \$\begingroup\$ Switching the arguments of zipWith should save one more byte: f s=zipWith(\p->last$toLower:[toUpper|isUpper p])s$filter(>' ')s. \$\endgroup\$ – Laikoni Jun 29 '17 at 7:07
4
\$\begingroup\$

V, 24 bytes

ÄVuÓó
ejlDò/¥2lõ
vuk~òGd

Try it online!

These kind of challenges are exactly what V was made for. :)

Explanation:

Ä           " Duplicate this line
 Vu         " Convert it to lowercase
   Óó       " Remove all spaces
e           " Move to the end of this line
 j          " Move down a line (to the original)
  l         " Move one char to the right
   D        " And delete the end of this line
    ò       " Recursively:
     /      "   Search for:
         õ  "     An uppercase character
      ¥2l   "     On line 2
            "     (This will break the loop when there are no uppercase characters left)
vu          "   Convert it to lowercase
  k         "   Move up a line
   ~        "   Convert this to uppercase also
    ò       " Endwhile
     G      " Move to the last line
      d     " And delete it
\$\endgroup\$
  • \$\begingroup\$ @DLosc Good questions! The newlines signal the end of a regex command, such as a substitute (remove) or search command. More detail is on this page: github.com/DJMcMayhem/V/wiki/Regexes \$\endgroup\$ – DJMcMayhem Jun 29 '17 at 19:13
4
\$\begingroup\$

Python 2, 100 bytes

s=input()
print"".join([c.lower(),c.upper()][s[i].isupper()]for i,c in enumerate(s.replace(" ","")))
\$\endgroup\$
  • 3
    \$\begingroup\$ Welcome to PPCG, and very good first answer! \$\endgroup\$ – ETHproductions Jun 29 '17 at 19:15
3
\$\begingroup\$

Alice, 32 bytes

/..- ~l+u~mSloy
\ia''-y.'Qa.+a@/

Try it online!

Explanation

This is a standard template for programs that work entirely in ordinal mode. Unwrapped, the program is as follows:

i.' -l.uQm.lay.a-'~y+'~aS+o@

i       take input as string
.       duplicate
' -     remove spaces from copy
l.u     create all-lowercase and all-uppercase versions
Q       reverse stack, so original string is on top
m       truncate original string to length of spaces-removed string
.lay    convert everything except uppercase characters to \n
.a-'~y  convert everything except \n (i.e., convert uppercase characters) to ~
+       superimpose with lowercase string
        \n becomes the corresponding lowercase character, and ~ remains as is
'~aS    convert ~ to \n
+       superimpose with uppercase string
        lowercase in existing string stays as is because it has a higher code point
        \n becomes corresponding uppercase character
o       output
@       terminate
\$\endgroup\$
3
\$\begingroup\$

JavaScript (ES6), 94 91 85 bytes

s=>s.replace(/./g,c=>c==" "?"":c[`to${"@"<s[x]&s[x++]<"["?"Upp":"Low"}erCase`](),x=0)
  • 6 bytes saved with assistance from ETHproductions & Arnauld.

Try it

o.innerText=(f=

s=>s.replace(/./g,c=>c==" "?"":c[`to${"@"<s[x]&s[x++]<"["?"Upp":"Low"}erCase`](),x=0)

)(i.value="Hi! Test!");oninput=_=>o.innerText=f(i.value)
<input id=i><pre id=o>

\$\endgroup\$
  • \$\begingroup\$ Ninjad :P codegolf.stackexchange.com/a/128951/65836 \$\endgroup\$ – Stephen Jun 28 '17 at 22:22
  • \$\begingroup\$ Could you do '@'<s[i]&s[i]<'['?? \$\endgroup\$ – ETHproductions Jun 28 '17 at 23:42
  • \$\begingroup\$ @StepHen: Aw, man, didn't see that last night while I was working on this. \$\endgroup\$ – Shaggy Jun 29 '17 at 7:38
  • \$\begingroup\$ @ETHproductions: I was wondering if that might be shorter but I was too lazy to look up which characters I'd need to use :D Turns out it doea save a byte; thanks. \$\endgroup\$ – Shaggy Jun 29 '17 at 7:40
3
\$\begingroup\$

Retina, 77 71 bytes

.+
$&¶$&
T`L `l_`.+$
+`((.)*)[A-Z].*(¶(?<-2>.)*)
$1$3 
.+¶

T`l `L_` .?

Try it online! Link includes test suite. Explanation: The first stage duplicates the line while the second stage lowercases the duplicate and deletes its spaces. The third stage then loops through each uppercase letter from right to left and attempts to place a space before the corresponding character on the second line. The first line is deleted and the spaces are used to uppercase the relevant characters of the result. Edit: Saved 6 bytes thanks to @Kobi.

\$\endgroup\$
  • \$\begingroup\$ Small question: Are the (.?) and $4 parts needed? It looks like having an optional group at the end does nothing. \$\endgroup\$ – Kobi Jun 29 '17 at 11:38
  • \$\begingroup\$ @Kobi Nothing small about that question! It had originally been part of an attempt at using lookarounds to match the characters to be uppercased directly instead of having to translate them as a separate step. \$\endgroup\$ – Neil Jun 29 '17 at 12:26
3
\$\begingroup\$

Perl, 95 94 + 1 = 95 bytes

+1 byte penalty for -n

Save one byte by replace from s/\s//g to s/ //g

$s=$_;s/ //g;$_=lc($_);while(/(.)/gs){$p=$&;$p=uc($p)if(substr($s,$-[0],1)=~/[A-Z]/);print$p;}

Try it online!

Explanation:

  1. Make copy of input string.

  2. Remove all spaceses and transform string to lower case.

  3. Then start loop over each letter. Test letter in same position in saved string for upper case. If it upper - make current letter captitalized. Print letter.

Note that perl need to be run with "-n" command line switch

\$\endgroup\$
  • \$\begingroup\$ Welcome to PPCG! If you would like, you can add a link to Try It Online: tio.run/# (I would add it, but I don't know if this is Perl 5 or Perl 6) \$\endgroup\$ – Stephen Jul 3 '17 at 18:25
  • 1
    \$\begingroup\$ I think that you need to count +1 byte for the -n flag. Other than that, this looks good! Welcome to the site! :) \$\endgroup\$ – DJMcMayhem Jul 3 '17 at 18:25
  • \$\begingroup\$ @StepHen it's Perl 5, colud you add link? I failed to run my code there in propper way. \$\endgroup\$ – Veitcel Jul 4 '17 at 17:44
  • \$\begingroup\$ Happy to see a new Perl golfer! I've added the TIO link to your answer and improved the formatting. \$\endgroup\$ – Dada Jul 4 '17 at 18:06
2
\$\begingroup\$

MATL, 18 bytes

kXz"@GX@)tk<?Xk]&h

Same approach as Riley's 05AB1E answer.

Try it online!

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

Python 3, 117 bytes

s=input()
y=list(s.replace(' ','').lower())
i=0
for c in y:
 if s[i].isupper():y[i]=c.upper()
 i+=1
print(''.join(y))

Try It Online!

This is pretty much my first code golf, so it's likely to be bad, minus help from comments below!

P.S. Yes, it's dumb that defining and incrementing i saves bytes over range(len(y)). Oh well.

\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to PPCG! Nice first submission! However, complying to our site's I/O standards, your submission must either be a function of a string or take input; you cannot assume input is in a variable. Hope you enjoy your stay! :) \$\endgroup\$ – HyperNeutrino Jun 28 '17 at 21:29
  • \$\begingroup\$ Thanks; edited a function in, but also saved 5 bytes in the body :D \$\endgroup\$ – Luke Sawczak Jun 28 '17 at 21:35
  • 1
    \$\begingroup\$ @LukeSawczak save a ton of bytes by changing to one space for indentation, and maybe add a Try It Online! link if you want to \$\endgroup\$ – Stephen Jun 28 '17 at 21:37
  • 1
    \$\begingroup\$ You can remove the space after return. \$\endgroup\$ – CalculatorFeline Jun 28 '17 at 21:45
  • \$\begingroup\$ @LukeSawczak how's this? tio.run/… \$\endgroup\$ – Stephen Jun 28 '17 at 22:15
2
\$\begingroup\$

C# (.NET Core), 108 101 bytes

using System.Linq;s=>s.Replace(" ","").Select((c,i)=>s[i]>64&s[i]<91?char.ToUpper(c):char.ToLower(c))

Try it online!

  • 7 bytes saved after realizing that the char class has static ToUpper() and ToLower() methods.
\$\endgroup\$
2
\$\begingroup\$

Charcoal, 33 bytes

A⁰χFLθ¿⁼§θι A⁺¹χχ¿№α§θ⁻ιχ↥§θι↧§θι

Try it online!

As I still don't know how to pass a string with whitespaces as a single input parameter into Charcoal code, I just assign in the header the test string to the Charcoal variable that represents what would be the first input (θ):

AA Quick Brown Fox Jumped Over The Lazy Dogθ

So the code has the same number of bytes as if the string were passed as first input.

You can see here the verbose version of the code.

\$\endgroup\$
  • 1
    \$\begingroup\$ I said in another answer but just in case you forget, just input as a python array with one element \$\endgroup\$ – ASCII-only Jul 27 '17 at 2:38
  • \$\begingroup\$ I just require the input to have a trailing newline. \$\endgroup\$ – Neil Dec 1 '17 at 16:58
2
\$\begingroup\$

PHP, 181 bytes

I try get the minor numbers of bytes,this is my code:

<?php
$s=readline();
preg_match_all('/[A-Z]/',$s,$m,PREG_OFFSET_CAPTURE);
$s=strtolower(str_replace(' ','',$s));
while($d=each($m[0]))$s[$d[1][1]]=strtoupper($s[$d[1][1]]);
echo $s;

Try it online!

\$\endgroup\$
  • \$\begingroup\$ Instead of a Constant PREG_OFFSET_CAPTURE you can use the value 256 , $argn is a shorter variable as readline() for an input and I think ctype_upper and use of lcfirst and ucfirst will save lots of bytes with one loop and use of $$i and ternary operator \$\endgroup\$ – Jörg Hülsermann Jul 2 '17 at 22:24
2
\$\begingroup\$

Java 8, 184 177 161 bytes

s->{String r="";for(int i=0,j=i,t,u;i<s.length;){t=s[i++];if(t>32){u=s[j++];r+=(char)(t<65|t>90&t<97|t>122?t:u>64&u<91?t&~32:u>96&u<123|u<33?t|32:t);}}return r;}

Can definitely be golfed some more..
- 16 bytes thanks to @OlivierGrégoire by taking the input as char[] instead of String.

Explanation:

Try it here.

s->{                           // Method with char-array parameter and String return-type
  String r="";                 //  Result-String
  for(int i=0,j=i,t,u;         //  Some temp integers and indices
      i<s.length;){            //  Loop over the String
    t=s[i++];                  //   Take the next character and save it in `t` (as integer)
                               //   and raise index `i` by 1
    if(t>32){                  //   If `t` is not a space:
     u=s[j++];                 //   Take `u` and raise index `j` by 1
     r+=                       //   Append the result-String with:
      (char)                   //    Integer to char conversion of:
       (t<65|t>90&t<97|t>122?  //     If `t` is not a letter:
        t                      //      Simply use `t` as is
       :u>64&u<91?             //     Else if `u` is uppercase:
        t&~32                  //      Take `t` as uppercase
       :u>96&u<123|u<33?       //     Else if `u` is lowercase or a space:
        t|32                   //      Take `t` as lowercase
       :                       //     Else:
        t);                    //      Take `t` as is
    }
  }                            //  End of loop
  return r;                    //  Return result-String
}                              // End of method
\$\endgroup\$
  • 1
    \$\begingroup\$ Take a char[] instead of a String for this one, you'll save up plenty of bytes! \$\endgroup\$ – Olivier Grégoire Jun 30 '17 at 7:57
  • \$\begingroup\$ On the other hand, I answered too with another algorithm. And here, I take the opposite arguments: in = String, out = char[] :-) \$\endgroup\$ – Olivier Grégoire Jun 30 '17 at 8:39
2
\$\begingroup\$

Common Lisp, 104 bytes

(defun f(s)(map'string(lambda(x y)(if(upper-case-p x)(char-upcase y)(char-downcase y)))s(remove #\  s)))

Try it online!

Unusually short for the wordy Common Lisp!

Straightforward code:

(defun f (s)                     ; receive the string as parameter
  (map 'string                   ; map the following function of two arguments
       (lambda (x y)             ; x from the original string, y from the string with removed spaces
         (if (upper-case-p x)    ; if x is uppercase
             (char-upcase y)     ; get y uppercase
             (char-downcase y))) ; else get y lowercase
       s
       (remove #\  s)))
\$\endgroup\$
2
\$\begingroup\$

Java (OpenJDK 8), 150 117 113 97 bytes

s->{for(int i=0,j=0,c;i<s.length;)if((c=s[i++]&95)>0)System.out.printf("%c",c^(s[j++]|~c/2)&32);}

Try it online!

While golfing more, I came to 102 bytes:

s->{for(int i=0,j=0,c;i<s.length;)if((c=s[i++]&95)>0)System.out.printf("%c",c<64?c|32:c|s[j]&32,j++);}

Try it online!

But I remembered this was starting to look like Dennis' C answer so I simply ported his bit-twiddling and... magic happened. The big gain from the port is removing the branches and the repetitions inside them.

\$\endgroup\$
  • \$\begingroup\$ @ceilingcat that doesn't work: Hi! Test! should become Hi!tEst!, but with your solution it becomes Hi!Test. \$\endgroup\$ – Olivier Grégoire Nov 10 at 11:40
2
\$\begingroup\$

Google Sheets, 213 bytes

=ArrayFormula(JOIN("",IF(REGEXMATCH(MID(A1,ROW(OFFSET(A1,0,0,LEN(A1))),1),"[A-Z]"),MID(UPPER(SUBSTITUTE(A1," ","")),ROW(OFFSET(A1,0,0,LEN(A1))),1),MID(LOWER(SUBSTITUTE(A1," ","")),ROW(OFFSET(A1,0,0,LEN(A1))),1))))

Input is in cell A1 and the formula breaks down like this:

  • ArrayFormula() lets us evaluate each term of ROW() independently
  • JOIN() concatenates all those independent results into a single string
  • IF(REGEXMATCH(),UPPER(),LOWER() is what makes it alternate using upper or lower case depending on what the case was at that position in the input
  • ROW(OFFSET()) returns an array of values 1 to A1.length that can be fed into the MID() function so we can evaluate each character in turn

Results of test cases: (It's easier to read if you click though to the larger version.)

TestCases

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

Ruby, 80 bytes

->a{n=a.downcase.delete' '
n.size.times{|i|(?A..?Z)===a[i]&&n[i]=n[i].upcase}
n}

Try it online!

\$\endgroup\$
  • \$\begingroup\$ You can save a couple bytes by using n.gsub(/./){} instead of n.size.times{};n: n.gsub(/./){(?A..?Z)===a[i]?$&.upcase: $&}. \$\endgroup\$ – Jordan Dec 3 '17 at 2:06
2
\$\begingroup\$

Perl, 92 bytes

$p[$i++]=$-[0]while s/[A-Z]/lc($&)/e;s/\s//g;for$c(@p){substr($_,$c,1)=~tr[a-z][A-Z]};print;

Explanation:

$p[$i++]=$-[0]while s/[A-Z]/lc($&)/e;   #get locations of caps into an array at the same time converting letters to lowercase

s/\s//g;   #delete all spaces

for$c(@p){substr($_,$c,1)=~tr[a-z][A-Z]};   #convert lowercase letters to uppercase where uppercase letters were present

print;   # print (of course) :)
\$\endgroup\$
  • 1
    \$\begingroup\$ Welcome to PPCG! :) \$\endgroup\$ – Stephen Jul 3 '17 at 20:37
  • \$\begingroup\$ You need to add -n flag to make your answer valid. A few golfing things: s/ //g is enough (no need \s), y/a-z/A-Z/ is the same as tr[a-z][A-Z], you can use -p flag so you don't need the last print, you don't need the parenthesis in lc$&. \$\endgroup\$ – Dada Jul 4 '17 at 9:44
1
\$\begingroup\$

C, 103 bytes

i,j,c;f(char*s){for(i=j=0;c=tolower(s[j++]);)c-32&&putchar(c-32*(s[i]>64&&s[i]<91&&c>96&&c<123))&&++i;}

Try it online!

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

Python 3, 125, 124 bytes

lambda s:''.join(c.upper()if i in(s.find(q)for q in s if q.isupper())else c for i,c in enumerate(s.replace(' ','').lower()))

Try it online!

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

Python 2, 106 105 bytes

s=input()
print''.join(map(lambda(c,u):[c.lower,c.upper][u](),zip(s.replace(' ',''),map(str.isupper,s))))

Edit: save one byte via print ''.join => print''.join.

Lambda form, 99 bytes

lambda s:''.join(map(lambda(c,u):[c.lower,c.upper][u](),zip(s.replace(' ',''),map(str.isupper,s))))
\$\endgroup\$
1
\$\begingroup\$

SCALA, 128 chars, 128 bytes

var l=s.toLowerCase().filter(x=>x!=32)
for(i<-0 to l.size-1){if(s(i).isUpper)l=l.substring(0,i)+l(i).toUpper+l.substring(i+1)}
l

Thanks for this challenge. Try it online!

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

q/kdb+, 49 bytes

Solution:

{@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}

Examples:

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"Hi! Test!"
"Hi!tEst!"

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"A Quick Brown Fox Jumped Over The Lazy Dog"
"AqUickbrOwnfoxJumpEdovertHelazYdog"

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"testing TESTing TeStING testing testing TESTING"
"testingtESTIngteStInGTEstingtestingtestiNG"

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"TESTING... ... ... success! EUREKA???!!! maybe, don't, NOOOOO"
"TESTING.........success!eureKA???!!!maybe,don't,nooooo"

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"Enter        PASSWORD ---------"
"Enterpassword---------"

q){@[a;(&)(#:[a:lower x except" "]#x)in .Q.A;upper]}"A a B b C c D d E e F f G g H h I i J j K k L l M m N n O o P p Q q R r S s T t U u V v W w X x Z z"
"AabbCcddEeffGghhIijjKkllMmnnOoppQqrrSsttUuvvWwxxZz"

q){@[a;(&)#:[a:lower x except" "]#x in .Q.A;upper]}"  TEST"
"teST"

Explanation:

Find indices where input is uppercase and then apply function upper to those indices on a lowercase, space-removed version of the input string. Note that we cannot apply the function beyond the length of the string, so use take (#) to truncate the input string to length of the lowercase, space-removed version.

{@[a;where count[a:lower x except " "]#x in .Q.A;upper]} / ungolfed
{                                                      } / lambda function
 @[ ;                                           ;     ]  / apply FUNC to VAR at INDICES: @[VAR;INDICES;FUNC]
                                                 upper   / uppercase, upper["abc"] -> "ABC"
                                       x in .Q.A         / boolean list where input is in uppercase alphabet ABC..XYZ
                                      #                  / take this many elements from list on the right (ie truncate)
           count[                    ]                   / returns length of the stuff inside the brackets, count["ABC"] -> 3                                        
                         x except " "                    / remove " " from string
                   lower                                 / lowercase, lower["ABC"] -> "abc"
                 a:                                      / save in variable a
     where                                               / returns indices where true where[101b] -> 0 2
   a                                                     / our lowercased, space-stripped input

Bonus:

After reading the answers, thought I'd try a solution where I iterate over the input, so far I've only managed a 53 byte solution:

{a{$[y in .Q.A;upper x;x]}'#:[a:lower x except" "]#x}
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1
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Swift 3.0, 199 bytes

var s="AS Ff",i=[String](),p=[Int](),j=0;for c in s.characters{if c>="A"&&c<="Z"{p.append(j)};if c != " "{i.append(String(c).lowercased())};j=j+1};for c in p{i[c]=i[c].uppercased()};print(i.joined())

Try it Online!

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1
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Perl 5, 40 bytes

37 bytes of code + -F flag. (note that on old versions of Perl, you might need to add -an flags)

print$F[$i++]=~/[A-Z]/?uc:lc for/\S/g

Try it online!

Explanations:
Thanks to -F, @F contains a list of every characters of the input.
for/\S/g iterates over every non-space character of the input. We use $i to count at which iteration we are. If $F[$i++] is an uppercase character (/[A-Z]/), then we print the uppercase current character (uc), otherwise, we print it lowercase (lc). Note that uc and lc return their argument unchanged if it isn't a letter.


Previous version (less golfed: 47 bytes):

 s/ //g;s%.%$_=$&;$F[$i++]=~/[A-Z]/?uc:lc%ge

Try it online!

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