Remove Spaces, Maintaining Capitalization

Question

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

Answer 1

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;}

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Answer 2

Jelly, 14 13 bytes

nŒlTɓḲFŒlŒuṛ¦

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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.

Answer 3

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))])

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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(' '))])

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Answer 4

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!

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Answer 5

05AB1E, 15 14 bytes

-1 byte thanks to Emigna

ðKuvy¹Nè.lil}?

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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

Answer 6

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)

Answer 7

Python 2, 100 bytes

s=input()
print"".join([c.lower(),c.upper()][s[i].isupper()]for i,c in enumerate(s.replace(" ","")))

Answer 8

V, 24 bytes

ÄVuÓó
ejlDò/¥2lõ
vuk~òGd

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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

Answer 9

JavaScript (ES6), 94 91 85 84 81 bytes

s=>s.replace(r=/./g,c=>1/c?"":c[`to${r<s[x]&s[x++]<{}?"Upp":"Low"}erCase`](),x=0)

• 6 bytes saved with assistance from ETHproductions & Arnauld.
• 3 bytes saved thanks to l4m2

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Answer 10

Alice, 32 bytes

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

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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

Answer 11

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.

Answer 12

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

Answer 13

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;}

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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

Answer 14

Python 3, 117 114 98 bytes

s=input()
i,*y=0,*s.replace(' ','').lower()
for c in y:print(end=[c,c.upper()]['@'<s[i]<'[']);i+=1

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Edit: 3 bytes saved by UndoneStudios.

Edit: 14 bytes saved by Unrelated String.

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.

---

Python 3, 85 bytes

lambda s:''.join([c,c.upper()]['@'<b<'[']for b,c in zip(s,s.replace(' ','').lower()))

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This solution is entirely due to att; I'm keeping it separate because the strategy seems to me to have changed enough.

Answer 15

Java (JDK), 84 bytes

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

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Credits :

• Dennis for the bit twiddling as it's a good way to shorten the answer
• Fhuvi for trimming 12 bytes.

Answer 16

MATL, 18 bytes

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

Same approach as Riley's 05AB1E answer.

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Answer 17

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))

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• 7 bytes saved after realizing that the char class has static ToUpper() and ToLower() methods.

Answer 18

Charcoal, 33 bytes

Ａ⁰χＦＬθ¿⁼§θι Ａ⁺¹χχ¿№α§θ⁻ιχ↥§θι↧§θι

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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.

Answer 19

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;

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Answer 20

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)))

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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)))

Answer 21

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.)

Answer 22

Ruby, 80 bytes

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

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Answer 23

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) :)

Answer 24

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

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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

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Answer 25

Charcoal, 18 bytes

⪫Ｅ⪫⪪θ ω⎇№α§θκ↥ι↧ιω

Try it online! Link is to verbose version of code. Explanation:

    θ               Input string
   ⪪                Split on spaces
  ⪫   ω             Join together
 Ｅ                  Loop over each character
          §θκ       Get character from original string
        №α          Search uppercase letters
       ⎇            If true (nonzero)
             ↥ι     Uppercase the current character
               ↧ι   Else lowercase the current character
⪫                ω  Join together
                    Implicitly print

The recent addition of StringMap shaves off a couple of bytes:

⭆⪫⪪θ ω⎇№α§θκ↥ι↧ι

⭆ is roughly equivalent to ⪫Ｅ...ω.

Answer 26

GNU sed, 67 + 1 = 68 bytes

+1 byte for -r flag.

:A
s/$/;/
:
s/([A-Z]);(.?)/;\l\1\u\2/
t
s/(\S);/;\1/
t
s/ ;//
/ /bA

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Explanation

The algorithm is pretty simple:

1. Insert a cursor (;) at the end of the line:

   :A
   s/$/;/
   :
   

2. If the character before the cursor is an uppercase letter, lowercase it (\l) and uppercase the character after the cursor (\u), moving the cursor to the left, and branch back to :

   s/([A-Z]);(.?)/;\l\1\u\2/
   t
   

3. Otherwise, if the character before the cursor is not a space, move the cursor to the left and branch back to ;:

   s/(\S);/;\1/
   t
   

4. Otherwise, if the cursor is preceded by a space, delete both:

   s/ ;//
   

5. Finally, if there are any spaces left, branch back to :A:

   / /bA
   

Answer 27

Rust, 143 bytes

|s|str::to_lowercase(s).replace(' ',"").chars().zip(s.chars()).for_each(|(c,d)|print!("{}",if d.is_uppercase(){c.to_ascii_uppercase()}else{c}))

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Ungolfed:

  |input: &str|
    input.to_lowercase()          // create lowercase string
         .replace(' ', "")        // remove all spaces
         .chars()                 // iterate over characters
         .zip(input.chars())      // zip with input character iterator
         .for_each(|(c, input_c)| // for each character in both strings:
           print!("{}",                  // print the following:
             if input_c.is_uppercase() { // if the input character is uppercase
               c.to_ascii_uppercase()    // `c` in uppercase
             }
             else { c }                  // else `c`
           )
         )

Answer 28

Go, 184 bytes

import(."strings";u"unicode")
func f(s string)string{S:=[]rune(ReplaceAll(ToLower(s)," ",""))
for i:=range s{if i<len(S)&&u.IsUpper(rune(s[i])){S[i]=u.ToUpper(S[i])}}
return string(S)}

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Answer 29

Noulith, 88 bytes

\s->(for(p<<-s.lower filter (!=' '))yield if(s[p[0]].is_upper)p[1].upper else p[1])join""

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Answer 30

Vyxal, 4 bytes

ȧ⇩$•

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Helps to have a "with capitalisation of" built-in

Explained

ȧ⇩$•­⁡​‎‎⁡⁠⁡‏‏​⁡⁠⁡‌⁢​‎‎⁡⁠⁢‏‏​⁡⁠⁡‌⁣​‎‎⁡⁠⁣‏⁠‎⁡⁠⁤‏‏​⁡⁠⁡‌⁤​‎‏​⁢⁠⁡‌­
ȧ     # ‎⁡Remove whitespace from the input
 ⇩    # ‎⁢and lowercase it
  $•  # ‎⁣Transfer the capitalisation of the original input to the stripped input.
# ‎⁤The reason for `•` doing this is because on lists, `•` is mould. Therefore, the idea was that it "moulds" one string to the capitalization of the other
💎

Created with the help of Luminespire.