The SBEVE Challenge

Question

She said s(he) be(lie)ve(d), he said sbeve.

Input

• A non-empty string, s. It's guaranteed that s will have only printable ASCII characters and at least one word (defined as [A-Za-z0-9]+) in parentheses, and all parentheses will be closed respectively.

Output

• A string containing all non-whitespace characters (whitespaces are defined as spaces, tabs, carriage returns, new lines, vertical tabs and form feeds characters) that are not in parentheses.

Test cases

Input -> Output
s(he) be(lie)ve(d) -> sbeve
s(h3) (1s) br(0k)3n -> sbr3n
(I) (K)now (Ill) Be
(My) Best (Self) -> nowBeBest
sho(u)lder (should)er
s(ho)u(ld)er s(h)ould(er) -> sholderersuersould
p(er)f(ection) -> pf
(hello) (world) ->               

The last output is an empty string.

This is code-golf so shortest code in bytes wins.

Answer 1

Jelly, 11 bytes

>Ƈ⁶Ø(yṣ”)m2

A full program printing the result.

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How?

Whitespace printable ASCII characters are "\t\r\n\f\v " and non-whitespace printable ASCII are all greater than ' ', so:

>Ƈ⁶Ø(yṣ”)m2 - Link: list of printable ASCII characters, s
 Ƈ          - filter keep those (characters) for which:
> ⁶         -   greater than literal space character
   Ø(       - literal list of characters ['(', ')']
     y      - translate (replace all '(' with ')')
       ”)   - literal ')' character
      ṣ     - split at
         m2 - modulo-two-slice (every other entry)
            - implicit, smashing print

Answer 2

05AB1E, 16 bytes

ηʒ„()S¢Æ_}€θJžKÃ

Try it online!, or Verify all test cases

---

Explanation

η                         - Prefixes of the string 
 ʒ       }                - filter these when...
  „()S¢                   - the counts of ( and ) characters 
       Æ_                 - are the same
          €θ              - get the last character from each of these prefixes
            J             - Join all these last characters
             žKÃ          - and remove any that aren't in [a-zA-Z0-9]

Answer 3

Retina, 11 bytes

\(.*?\)|\s

(Note the trailing newline in the code).

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How it works

                 Replace either
  .*             a sequence of characters
\(   \)          in parentheses,
    ?            matched non-greedily,
       |         or
        \s       any whitespace character
                 by nothing

Answer 4

Bash + GNU utilities, 36 32 28 bytes

xargs|sed 's/([^)]*)\|\s//g'

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Thanks to @user41805 for 4 bytes!

Answer 5

SNOBOL4 (CSNOBOL4), 119 bytes

 I =INPUT
S I ARB . L '(' ARB ')' REM . I :F(W)
 O =O L :(S)
W O NOTANY(&UCASE &LCASE 84 ** 9) ='' :S(W)
 OUTPUT =O
END

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Answer 6

Jelly, 12 bytes

e€Ø(œpm2FfØB

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A monadic link taking and returning a Jelly string. Now handles other whitespace characters.

Alternative 12 bytes

Ø(yṣ”)m2FfØB

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Answer 7

APL+WIN, 25 bytes or 30 bytes

Prompts for string.

((~x∨≠\x←s∊'()')/s←⎕)~' '

Try it online! Courtesy of Dyalog Classic

If I now have to handle none alphanumeric characters this works for 5 extra bytes in APL+WIN but will not work in Dyalog Classic so no TIO

((~x∨≠\x←s∊'()')/s←⎕)~⎕av[⍳33]

Answer 8

05AB1E, 12 bytes

ŒʒÁ„)(å}KžKÃ

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Π              # substrings 
 ʒ     }        # filter, keep each substring if:
  Á             #  after being rotated right
   „)(å         #  it contains ")("
        K       # remove those substrings from the input
         žKÃ    # keep only characters in [a-zA-Z0-9]

Answer 9

Stax, 11 bytes

û┼╢╕jN&╪º╛╘

Run and debug it

Unpacked:

"\(.*?\)| "zR

this is just a simple regex replace. It replaces all strings that match /\(.*?\)| / with z (an empty string)

Answer 10

Perl 5 + -lF/\(\w+\)|\s+/ -M5.10.0, 5 bytes

say@F

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Answer 11

05AB1E, 10 11 bytes

„()S¡ιнJʒð›

+1 byte as bug-fix (žu to „() and žKÃ to ʒð›), since apparently all printable ASCII + whitespaces are valid input-characters, instead of only alphanumeric, whitespaces, and parenthesis..

Try it online or verify all test cases.

Explanation:

„()S         # Push string "()", and split it to a list of characters: ["(",")"]
    ¡        # Split the (implicit) input-string by that
     ι       # Uninterleave this list
      н      # Only keep the first inner list
       J     # Join it together to a single string
        ʒ    # Filter the characters in this string by:
         ð›  #  Check if the character is larger than a space " " (by ASCII codepoint)
             # (after which the result is output implicitly)

Answer 12

Japt, 12 bytes

r"%s|%(.*?%)

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Answer 13

Python 3, ⁴⁷ 45 bytes

Saved 2 bytes thanks to FryAmTheEggman!!!

lambda s:re.sub(r"\(\w+\)|\s","",s)
import re

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Answer 14

Charcoal, 18 bytes

Φθ∧›ι ⁼№…θ⊕κ(№…θκ)

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

 θ                  Input string
Φ                   Filtered where
    ι               Current character
   ›                Is not whitespace
  ∧                 Logical And
       №            Count of
            (       Open parentheses
        …θ⊕κ        In input string so far (inclusive)
      ⁼             Equals
             №      Count of
                 )  Close parentheses
              …θκ   In input string so far (exclusive)

Answer 15

Perl 5 -p0, 15 bytes

s/\s|\(.*?\)//g

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Answer 16

sed, 32 31 bytes

:l;s/([^)]*)\|\s//g;N;s/\n//;tl

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Shaved off one byte by removing the -E option and just using basic regular expressions, surprisingly :-) .

sed is almost perfect for this challenge. The only issue is that sed is a stream editor, processing one line at a time: the trailing newline on each line is not processed, which doesn't make it straightforward to delete any newline characters.

So here's how this script does it:

:l
   Label we can jump to later.

s/([^)]*)\|\s//g
   Delete all parenthetical expressions and whitespace before the first newline.

N
   If we're not at the last line, append the next line to the pattern space.  (The next line is appended immediately _after_ the \n at the end of the previous line.  That \n is still there, now in the middle of the pattern space, where it's available for processing by sed.)

s/\n//
    Delete a \n in the middle of the pattern space, if any.  (There will be one if we weren't at the last line in the previous step already.)

tl
    If the last command found a \n to delete, jump back to label l, and do it again!

Answer 17

C (gcc), 70 bytes

f(char*s){for(;*s;)s=*s-40?isalnum(*s)&&putchar(*s),s+1:strchr(s,41);}

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Answer 18

Red, 52 bytes

func[s][parse trim/all s[any[to"("remove thru")"]]s]

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Answer 19

JavaScript (Node.js), 30 bytes

s=>s.replace(/\(.*?\)|\s/g,"")

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Answer 20

APL (Dyalog Extended), ²¹ 19 bytes

∊⊢⊆⍨<>(≠\∨⊢)⍤∊∘'()'

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How it works

∊⊢⊆⍨<>(≠\∨⊢)⍤∊∘'()'  ⍝ Input: string s
             ∊∘'()'  ⍝ Boolean vector (1 if a member of '()')
      (≠\  )⍤        ⍝ Scan by boolean XOR
                     ⍝ (gives 1 between '()'s, including '(' but not ')')
         ∨⊢          ⍝ Include back ')'
     >               ⍝ Bitmask indicating negation of the above,
    <                ⍝ plus each char not being whitespace
∊⊢⊆⍨                 ⍝ Filter s by the bitmask above

Boolean filtering uses this APL golfing tip.

Answer 21

Gema, 7 characters

(*)=
 =

Sample run:

bash-5.0$ gema '(*)=; =' <<< 's(he) be(lie)ve(d)'
sbeve

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Answer 22

Kotlin, 75 bytes

{fold(""){a,v->when(v){')'->a.takeWhile{it!='('}
in " \r\n"->a
else->a+v}}}

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Answer 23

Pip, 14 bytes

aRMw,`\(.+?\)`

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Unweave works really well in this solution.

-2 bytes from DLosc, with the first documented use of , for Regex alternation!