Parentheses Clusters

Question

Write a program that groups a string into parentheses cluster. Each cluster should be balanced.

Examples :

split("((())d)") ➞ ["((()))"]

split("(h(e(l)l)o)(w(o)r)l(d)(w)h(a(t)(s)u)p") ➞ ["((()))", "(())", "()", "()", "(()())"]

split("((())())(()(()()))") ➞ ["((())())", "(()(()()))"]

Input may contain letters other than parentheses but you are not sorting them. (The possible characters in the input are uppercase and lowercase letters, numbers, and ())

Given parentheses will always be complete (i.e : ( will be matched with ) )

Cluster is when outermost ( ends with ). In short when your parentheses are all balanced at that moment so :

(((    ---> this will be balanced when 3 ) are added
)))    ---> now its balanced this is cluster 1
()     ---> this is cluster 2
(((((((((((((((((((((((((((((( ---> this is start of cluster 3. will need equal amount of ) to balance 

This is code-golf, shortest code wins

(Link to challenge on Code Wars : link for those that wish to check it)

Answer 1

Vim, 20 bytes

:s/\w//g
qq%a
<Esc>@qq@q

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Programmer's text editor builtins ftw! \o/

Explanation:

:s/\w//g    # Remove all letters and numbers
qq          # Start macro 'q'
  %a        #  Jump to matching parenthesis and insert newline
<Esc>
     @q     #  Call macro 'q'
       q    #  End macro 'q'
        @q  # Call macro 'q'

Answer 2

J, 35 bytes

(](]<;.2~0=[:+/\_1+2*i.)[-.-.)&'()'

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• Remove all parens from the input.
• Turn ( into -1 and ) into 1, and take the scan sum, then convert that into a mask which is 1 wherever the sum is 0.
• Finally, use that mask to cut the cleaned input into chunks.

Answer 3

Jelly, 11 bytes

fØ(µO-*ĬṖk

Try it online!

The Footer just runs all the test cases, joins each cluster by a single newline and each test case by 2.

How it works

fØ(µO-*ĬṖk - Main link. Takes a string S on the left
fØ(         - Remove all non-"()" characters
   µ        - Use this string of parentheses P as the argument from here:
    O       - Convert to ords ("(" -> 40, ")" -> 41)
     -*     - Raise -1 to that power, yielding "(" -> 1, ")" -> -1
       Ä    - Cumulative sum
        ¬   - Logical NOT, converting 0 -> 1, everything else -> 0
         Ṗ  - Remove the trailing 1
          k - Split P at the indices of the 1s

Answer 4

APL(Dyalog Unicode), 26 bytes ^SBCS

{⍵⊂⍨¯1⌽0=+\¯1*')'=⍵}∩∘'()'

Try it on APLgolf! or Try it with step by step output

∩∘'()' intersect right argument with (). This removes all other characters.
{ ... } call the dfn with the cleaned string as its right argument ⍵.

')'=⍵ for each character in the string, is it equal to )?
¯1* this maps ( to 1 and ) to ¯1.
+\ cumulative sums. This is the nesting level for each character.
0= equal to 0?
¯1⌽ rotate the trailing 1 to the front.
⍵⊂⍨ partition the string at characters with nesting level 0.

Answer 5

Python 3, 75 bytes

def f(s,x=1):
 for i in s:y=(i==")")-(i=="(");x+=y;y and print(i,end=" "*x)

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prints the clusters separated by spaces

Python 3, 88 bytes

def f(s,x=1,w=""):
 for i in s:y=(i==")")-(i=="(");x+=y;w+=y*y*i+" "*x
 return w.split()

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returns a list of clusters

Answer 6

05AB1E, 18 17 14 bytes

žuÃDÇÈ·<.¥_Å¡¦

-1 byte because of the rule change from all printable ASCII to just parenthesis and alphanumeric characters in the input
-3 bytes and improved performance thanks to @cairdCoinheringaahing

Output as a list of list of characters.

Try it online or verify all test cases.

Explanation:

žu              # Push constant string "()<>[]{}"
  Ã             # Only keep those characters from the (implicit) input-string
   D            # Duplicate this string
    Ç           # Convert each character in the copy to its unicode integer:
                #  "("→0; ")"→1
     É          #  Check for each whether it's odd: "("→0; ")"→1
      ·         #  Double each: "("→0; ")"→2
       <        #  Decrease each by 1: "("→-1; ")"→1
        .¥      #  Undelta this list
          _     #  Check for each whether it's equal to 0 (1 if 0; 0 otherwise)
           Å¡   # Split the string we've duplicated at the truthy indices,
                # which implicitly converts the parts to character-lists
             ¦  # And remove the leading empty list
                # (after which the result is output implicitly)

Answer 7

Ruby, 37 bytes

->s{s.tr('^()','').scan /\(\g<0>*\)/}

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Returns an array of clusters. The numbered subpattern \g<0> nests the balanced-parentheses-matching regex within itself.

Answer 8

Japt v2.0a0, 15 13 12 bytes

r\w ó@T±JpXc

Try it

r\w ó@T±JpXc     :Implicit input of string U
r                :Replace
 \w              :  RegEx /[a-z0-9]/gi
    ó            :Partition after each character X
     @           :That returns falsey (0) when passed through the following function
      T±         :  Increment T (initially 0) by
        J        :  -1
         p       :  Raised to the power of
          Xc     :  Charcode of X

Answer 9

JavaScript (ES6),  61 60  59 bytes

Saved 1 byte thanks to @tsh

Outputs a string with one cluster per line.

s=>s.replace(/./g,c=>c=='('?(s=-~s,c):c==')'?--s?c:`)
`:'')

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Answer 10

Pip, 28 25 24 23 bytes

Fca@X^pI$==_NyPBcMpY/Py

Try it here! Or, with a variable definition in the header, you can also Try it online!

Explanation

Pushes one parenthesis at a time onto y, checks whether y is balanced, and if so, outputs and resets it.

Fca@X^pI$==_NyPBcMpY/Py
                         a is cmdline arg; p is "()"; y is "" (implicit)
Fc                       For each character c in
  a@                     each regex match in a of
    X                    a regex matching
     ^                   either of the characters in
      p                  "()":
       I                  If
             yPBc         we push c onto the end of y
           _N             and then the number of occurrences in y of
                 Mp       each character in "()"
        $==               is equal:
                     Py    Print y
                    /      and invert (resulting in nil because y isn't a number)
                   Y       and yank that as the new value of y

This works because pushing a string onto a variable that is nil sets the variable to the pushed string, the same as if the variable's value had been "".

Answer 11

Java 8, 89 82 81 bytes

s->{int t=0;for(var c:s)System.out.print(c==40?++t>0?c:c:c==41?--t<1?") ":c:"");}

-7 bytes thanks to @ZaelinGoodman

Input as character-array; output as a space-delimited string.

Try it online.

Explanation:

s->{                   // Method with character-array as parameter and no return
  int t=0;             //  Create a temp integer, starting at 0
  for(var c:s)         //  Loop over the characters of the input-array:
    System.out.print(  //   Print:
      c==40?           //    If the current character is '(':
        ++t            //     Increase `t` by 1
        >0?c:c         //     And print the '('
      :c==41?          //    Else-if the current character is ')':
        --t            //     Decrease `t` by 1
           <1?         //     If it is now 0:
              ") "     //      Print the ')', plus a space
             :         //     Else (it's not 0):
              c        //      Print just the ')'
      :                //    Else (it's a different character):
       "");}           //     Print nothing

Answer 12

JavaScript (Node.js), 83 bytes

A=>[...A.replace(/[^()]/g,(o=0,t=''))].map(l=>(t+=l,!(o+=l<")"||-1)&&t+(t="")||""))

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If it accepts array having , in between parentheses otherwise :

JavaScript (Node.js), 101 100 95 bytes

A=>[...A.replace(/[^()]/g,(o=0,a=[],t=''))].map(l=>(o+=l<")"||-1,t+=l,!o&&a.push(t+(t=''))))&&a

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Answer 13

Java (JDK), 86 80 79 bytes

n->{int o=0;for(var c:n)if(o!=(o+=c>41?0:81-2*c))System.out.print(o<1?") ":c);}

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-7 thanks to Kevin Kruijssn and Zaelin Goodman!

Answer 14

JavaScript, 55 53 bytes

Outputs a space delimited string with a trailing space.

s=>s.replace(/./g,x=>x>{}?``:(n+=x>s||-1)?x:`) `,n=0)

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Answer 15

Perl 5, 26 bytes

y/()//cd;s/\((?R)*\)/$&
/g

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y/()//cd;           #delete all chars from input line except ( and )
s/\((?R)*\)/$&      #print clusters separated by newline
/g                  #found by recursive regexp with (?R)

Answer 16

PowerShell Core, 89 80 bytes

-9 bytes thanks to @mazzy!!!

$args-replace"\w"|sls "(\((?=\(*(?<S>\)))|\k<S>(?<-S>))+?(?(S)(?!))"-a|% M*|% V*

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Answer 17

PowerShell, 62 bytes

switch($args){'('{$r+=$_;++$l}')'{$r+=$_;if(!--$l){$r;$r=''}}}

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Answer 18

C (gcc), 70 bytes

I think at this point I have to admit I'm on a ternary operator abuse spree.

Big thanks to ErikF for -14 bytes.

And thanks the ceilingcat for -4 more!

c;f(d){for(d=0;read(0,&c,1);c>47||(d+=2*putchar(c)-81)||putchar(32));}

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Answer 19

R, 99 92 bytes

Or R>=4.1, 85 bytes by replacing the word function with \.

-7 bytes thanks to @Dominic van Essen.

function(s,x=utf8ToInt(s))Map(intToUtf8,split(a<-x[x<42],head(diffinv(!cumsum(a*2-81)),-1)))

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As always, R is terrible in string challenges...

Answer 20

Lua, 46 bytes

print(arg[1]:gsub("%w",""):gsub("%b()","%1;"))

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Looks like it happened working. Though I don't know much about Lua.

Answer 21

Retina 0.8.2, 25 bytes

\w

!`\(((\()|(?<-2>.))*.

Try it online! Link includes test cases. Explanation: The first two lines simply delete letters and digits. The final line matches groups of (assumed) parentheses. .NET balancing groups keep track of the number of unmatched (s and don't allow more )s than (s. This means that the inner loop stops when it reaches the matching ) for the outer (. The ! causes the groups themselves to be output on separate lines.

Answer 22

Charcoal, 19 bytes

ＦＳ≡ι(«ι⊞υυ»)«ι¿¬⊟υ⸿

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

ＦＳ≡ι

Switch over each character of the input string.

(«ι⊞υυ»

If it's a ( then print it and push the predefined empty list to itself. The list therefore contains itself as many times as there are unbalanced (s.

)«ι

If it's a ) then print it, and...

¿¬⊟υ⸿

... if popping the list leaves it empty again then move to the next line.