Make Gimbap cutter

Question

Gimbap(김밥) is Korean food, somewhat looks like sushi roll.

Here is Korean emoticon represent Gimbap : @))))))))))

Your quest is make ASCII Gimbap cutter.

Rule

Input is string made with only @ and ).

Output cuts every valid Gimbap with ), and then add @ between them. So for example @)) to @) @).

Valid Gimbap starts with @ and followed by any amount of ).

If there is no valid Gimbap, output is blank.

Input and output

Input | Output
@))))) | @) @) @) @) @)
@))))))) | @) @) @) @) @) @) @)
@))@))) | @) @) @) @) @)
@) | @)
)) | 
@ | 
@@)@@@))) | @) @) @) @)
@)@)@)) | @) @) @) @)
@@@)) | @) @)
))@) | @)

Winning condition

This is code-golf, so shortest code wins.

from sandbox

Is the rule equivalent to "output one copy of @) for every ) in the input not counting those before any @? May our output include a trailing space, like "@) @) "? — xnor, Jun 9, 2019 at 4:54
Do we have to output a space delimited string or can we output an array of slices? Also, are we limited to those 2 characters or can we use any 2 characters of our choosing? — Shaggy, Jun 9, 2019 at 13:15
My first time trying to read the title, "Gimp-bat clutter? What?" — Draco18s no longer trusts SE, Jun 9, 2019 at 17:13

xnor · Accepted Answer · 2019-06-09 05:53:15Z

10

Haskell, 32 bytes

f(')':s)=f s
f s=do ')'<-s;"@) "

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Recursion beats using span for removing the initial ('s.

Haskell, 33 bytes

f s=do ')'<-snd$span(<'@')s;"@) "

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edited Jun 9, 2019 at 5:53

answered Jun 9, 2019 at 5:38

xnor

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

JavaScript, 42 41 bytes

s=>s.split(/(?<=@.*)\)/).fill``.join`@) `

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 9, 2019 at 16:44

Shaggy

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att · Accepted Answer · 2019-06-09 10:53:02Z

5

C (gcc), 53 bytes

i;f(char*_){for(i=1;*_;!i&*_++&&printf("@) "))i&=*_;}

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i;f(char*_){for(    *_;                      )      }   //loop over the string:
                i=1;   !i&                    i&=*_;    //skip leading `)`s
                       !i&*_++&&printf("@) ")           //and print "@) "for each `)` thereafter

answered Jun 9, 2019 at 10:53

att

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

JavaScript (Node.js), ⁴⁸ 47 bytes

x=>"@) ".repeat(x.split(/(?<=@.*)\)/).length-1)

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JavaScript (Node.js), ⁵³ 49 bytes

x=>[...x].map(y=>y<"0"?m+=s&&"@) ":s=1,s=m="")&&m

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JavaScript (Node.js), 60 bytes

x=>x.replace(/(@?)(\)*)/g,(_,a,b)=>a&&b.replace(/./g,"@) "))

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 9, 2019 at 14:15

Shieru Asakoto

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dzaima · Accepted Answer · 2019-06-11 18:16:05Z

5

Canvas, 10 9 bytes

@ｓｊ∑Ｌ@) ×

Try it here!

edited Jun 11, 2019 at 18:16

answered Jun 9, 2019 at 7:50

dzaima

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Neil · Accepted Answer · 2019-06-09 09:02:48Z

4

Retina 0.8.2, 14 bytes

^\)+|@

\)
@)

Try it online! Link includes test cases. Explanation:

^\)+|@

Delete leading )s and all @s.

\)
@)

Replace all )s with @)s. (Note: trailing space.)

answered Jun 9, 2019 at 9:02

Neil

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xnor · Accepted Answer · 2019-06-09 05:04:15Z

3

Python 2, 39 bytes

lambda s:'@) '*s.lstrip(')').count(')')

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answered Jun 9, 2019 at 5:04

xnor

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Erik the Outgolfer · Accepted Answer · 2019-06-09 21:46:16Z

3

Jelly, 8 bytes

ṣṀḊẎ”@pK

Try it online!

-1 thanks to Jonathan Allan.

edited Jun 9, 2019 at 21:46

answered Jun 9, 2019 at 19:32

Erik the Outgolfer

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Xcali · Accepted Answer · 2019-06-09 05:56:43Z

2

Perl 5 `-p`, 26 bytes

$_=s/^.*?@//&&'@) 'x y/)//

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answered Jun 9, 2019 at 5:56

Xcali

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\$\begingroup\$ 24 bytes \$\endgroup\$
– wastl
Jun 15, 2019 at 10:57

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Emigna · Accepted Answer · 2019-06-09 09:59:26Z

2

05AB1E, 12 bytes

')Û'@KS'@ìðý

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Explanation

')Û            # trim leading ")"
   '@K         # remove all "@"
      S        # split to list of characters
       '@ì     # prepend "@" to each
          ðý   # join on spaces

answered Jun 9, 2019 at 9:59

Emigna

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Neil · Accepted Answer · 2019-06-09 10:01:30Z

Batch, 58 bytes

@set s=%1@
@set s=%s:*@=(%
@set s=%s:@=%
@echo%s:)=@) %

Takes input as a command-line parameter. Explanation:

@set s=%1@

Suffix an @ in case the input doesn't contain any.

@set s=%s:*@=(%

Delete up to the first @, replacing it with a ( to ensure that the string is not empty (because %:% doesn't work on empty strings). The ( also makes the echo work if the rest of the string is empty.

@set s=%s:@=%

Delete any remaining @s.

@echo%s:)=@) %

Expand any remaining )s.

Erik the Outgolfer · Accepted Answer · 2019-06-09 23:21:20Z

2

05AB1E (legacy), 10 bytes

Z¡¦JS'@ìðý

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This bug forces me to use the legacy version. This is the code for the current version of 05AB1E (11 bytes):

'@¡¦JS'@ìðý

Try it online!

Port of my Jelly answer.

answered Jun 9, 2019 at 23:21

Erik the Outgolfer

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Shaggy · Accepted Answer · 2019-06-10 08:56:00Z

Japt v2.0a0 `-S`, 15 bytes

r/^\)+|@/ ¬mi'@

Try it

r/^\)+|@/ ¬mi'@     :Implicit input of string
r                   :Remove
 /^\)+|@/           :  "@"s and leading ")"s
          ¬         :Split
           m        :Map
            i'@     :  Prepend "@"
                    :Implicit output, joined with spaces

Alternative

e/^\)/ è\) Æ"@)

Try it

e/^\)/ è\) Æ"@)     :Implicit input of string
e                   :Recursively remove
 /^\)/              :  Leading ")"
       è\)          :Count remaining ")"s
           Æ        :Map the range [0,Count)
            "@)     :  Literal string
                    :Implicit output, joined with spaces

Nitrodon · Accepted Answer · 2019-06-10 22:11:11Z

2

brainfuck, 49 bytes

,[[-<+>>++++<]>[[-]<<<[[.>]<---------.[-]]>[-]],]

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answered Jun 10, 2019 at 22:11

Nitrodon

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Japt v2.0a0 `-P`, 15 bytes

f/@\)+/ ËÅç"@)

Saved 2 bytes thanks to @Shaggy.

Try it

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 9, 2019 at 5:10

Gymhgy

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\$\begingroup\$ 15 bytes \$\endgroup\$
– Shaggy
Jun 9, 2019 at 8:56
\$\begingroup\$ @Shaggy Thanks, I haven't golfed in Japt recently \$\endgroup\$
– Gymhgy
Jun 9, 2019 at 17:07

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Nick Kennedy · Accepted Answer · 2019-06-09 18:08:08Z

1

Jelly, 12 bytes

œl”)ḟ”@⁾@ jⱮ

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answered Jun 9, 2019 at 18:08

Nick Kennedy

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bb94 · Accepted Answer · 2019-06-10 09:09:14Z

1

Perl 6, 34 bytes

{~map {'@)'xx.chars-1},m:g/\@\)+/}

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answered Jun 10, 2019 at 9:09

bb94

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Value Ink · Accepted Answer · 2019-06-10 23:25:34Z

Ruby `-p`, 28 bytes

$_= ~/@/&&'@) '*$'.count(?))

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Explanation

                                # -p gets a line of STDIN
$_=                             # Set output to
    ~/@/                        # Find first '@' in input
                                # nil (falsey) if not found
        &&                      # If found, set output to
          '@) '                 # Sliced gimbap
               *                # Repeat
                $'              # In the string after the first '@',
                  .count(?))    # ... count the number of ')'
                                # -p outputs the contents of $_
                                # nil outputs as a blank string

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Java 10, 49 bytes

s->s.replaceAll("^\\)+|@+","").replace(")","@) ")

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 11, 2019 at 7:11

Kevin Cruijssen

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Vicente Bolea · Accepted Answer · 2019-06-11 13:24:35Z

1

sed, 30 bytes

s/)\?@\()\?\)/\1/g; s/)/@) /gp

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edited Jun 11, 2019 at 13:24

answered Jun 11, 2019 at 12:55

Vicente Bolea

1113 bronze badges

\$\begingroup\$ Welcome to PPCG. Unfortunately, your code can't handle leading )s and multiple @ properly. And, how about using Try it online? \$\endgroup\$
– LegenDUST
Jun 11, 2019 at 13:00
1

\$\begingroup\$ As you can see in 5th or last example, leading )s have to be ignored. \$\endgroup\$
– LegenDUST
Jun 11, 2019 at 13:08
\$\begingroup\$ @LegenDUST, you are right! it was not that easy. I guess the working version is much uglier \$\endgroup\$
– Vicente Bolea
Jun 11, 2019 at 13:25
\$\begingroup\$ 28 chars: s/^)*//;s/[^)]//g;s/./@) /gp \$\endgroup\$
– jnfnt
Jun 13, 2019 at 2:22

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Roman · Accepted Answer · 2019-06-11 15:46:31Z

1

Wolfram Language (Mathematica), 91 90 85 71 70 59 57 bytes

StringReplace@{(StartOfString~~")"..)|"@"->"",")"->"@) "}

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edited Jun 11, 2019 at 15:46

answered Jun 11, 2019 at 12:36

Roman

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randomdude999 · Accepted Answer · 2019-06-11 18:31:53Z

Pyth, 20 bytes

*?}\@z/>zxz\@\)0"@)

Try it online! Note that there is a trailing space at the end of the program. This one is (or rather, started out as) a rather direct translation of the Python 2 answer (though the lstrip part was surprisingly difficult).

Explanation:

*            # repeat string
  ?          # repeat count: ternary
    }\@z     # condition: check whether input contains @
    /        # if condition is true: count occurrences of one string in another
      >      # array slice: all elements of array (or string) from a specific index and upwards
        z    # the thing to slice (input)
        xz\@ # the index first occurrence of \@ in z
      \)     # string to count occurrences of (\x is shorthand for "x")
    0        # value when ternary condition is false
  "@) "      # the string to be repeated (automatically terminated by end-of-line)

Jonathan Frech · Accepted Answer · 2019-06-13 01:14:58Z

krrp, 63 bytes

^":\L,^*':?#?E'E!-@1#!r'?=#!f'$64.-?*L$64.L$41.L$32.-@0#!r'.0".

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Explanation

^":                   ~ take the string as a parameter named `"`
 \L                   ~ import the list module
 ,^*':                ~ apply a binary function
  ?#?E'               ~  if the string is empty,
   E                  ~   return the empty string; else
   !-@1#!r'           ~   define `-` as the cut Gimbap
   ?=#!f'$64.         ~    if an at sign is seen,
    -                 ~    return the cut Gimbap; else
    ?*                ~    if an at sign has been seen,
     L$64.L$41.L$32.- ~     return a Gimbap piece together
                      ~     with freshly cut Gimbap; else
     @0#!r'           ~     proceed to cut
 .0".                 ~ to zero and the above taken string

Try it online!

mazzy · Accepted Answer · 2019-07-16 05:24:52Z

1

PowerShell, 42 bytes

''+($args|sls '(?<=@.*)\)'-a|% m*|%{'@)'})

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Unrolled:

$arrayOfCuttedGimbaps = $args|select-string '(?<=@.*)\)' -AllMatches|% Matches|%{'@)'}
''+($arrayOfCuttedGimbaps)    # toString and output

answered Jul 16, 2019 at 5:24

mazzy

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Stack Exchange Network

Make Gimbap cutter

Rule

Input and output

Winning condition

24 Answers 24

Haskell, 32 bytes

JavaScript, 42 41 bytes

C (gcc), 53 bytes

JavaScript (Node.js), ⁴⁸ 47 bytes

JavaScript (Node.js), ⁵³ 49 bytes

JavaScript (Node.js), 60 bytes

Canvas, 10 9 bytes

Retina 0.8.2, 14 bytes

Python 2, 39 bytes

Jelly, 8 bytes

Perl 5 `-p`, 26 bytes

05AB1E, 12 bytes

Batch, 58 bytes

05AB1E (legacy), 10 bytes

Japt v2.0a0 `-S`, 15 bytes

Alternative

brainfuck, 49 bytes

Japt v2.0a0 `-P`, 15 bytes

Jelly, 12 bytes

Perl 6, 34 bytes

Ruby `-p`, 28 bytes

Java 10, 49 bytes

sed, 30 bytes

Wolfram Language (Mathematica), 91 90 85 71 70 59 57 bytes

Pyth, 20 bytes

krrp, 63 bytes

Explanation

PowerShell, 42 bytes

Your Answer

Not the answer you're looking for? Browse other questions tagged
code-golf
ascii-art
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Hot Network Questions

Make Gimbap cutter

Rule

Input and output

Winning condition

24 Answers 24

Haskell, 32 bytes

JavaScript, 42 41 bytes

C (gcc), 53 bytes

JavaScript (Node.js), 48 47 bytes

JavaScript (Node.js), 53 49 bytes

JavaScript (Node.js), 60 bytes

Canvas, 10 9 bytes

Retina 0.8.2, 14 bytes

Python 2, 39 bytes

Jelly, 8 bytes

Perl 5 -p, 26 bytes

05AB1E, 12 bytes

Batch, 58 bytes

05AB1E (legacy), 10 bytes

Japt v2.0a0 -S, 15 bytes

Alternative

brainfuck, 49 bytes

Japt v2.0a0 -P, 15 bytes

Jelly, 12 bytes

Perl 6, 34 bytes

Ruby -p, 28 bytes

Java 10, 49 bytes

sed, 30 bytes

Wolfram Language (Mathematica), 91 90 85 71 70 59 57 bytes

Pyth, 20 bytes

krrp, 63 bytes

Explanation

PowerShell, 42 bytes

Your Answer

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Related

Hot Network Questions

JavaScript (Node.js), ⁴⁸ 47 bytes

JavaScript (Node.js), ⁵³ 49 bytes

Perl 5 `-p`, 26 bytes

Japt v2.0a0 `-S`, 15 bytes

Japt v2.0a0 `-P`, 15 bytes

Ruby `-p`, 28 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
ascii-art
or ask your own question.