N(e(s(t))) a string

Question

To "function nest" a string, you must:

• Treat the first character as a function, and the following characters as the arguments to that function. For example, if the input string was Hello, then the first step would be:

  H(ello)
  

• Then, repeat this same step for every substring. So we get:

  H(ello)
  H(e(llo))
  H(e(l(lo)))
  H(e(l(l(o))))
  

Your task is to write a program or function that "function nests" a string. For example, if the input string was Hello world!, then you should output:

H(e(l(l(o( (w(o(r(l(d(!)))))))))))

The input will only ever contain printable ASCII, and you may take the input and the output in any reasonable format. For example, STDIN/STDOUT, function arguments and return value, reading and writing to a file, etc.

For simplicity's sake, you may also assume the input will not contain parentheses, and will not be empty.

Input:
Nest a string
Output:
N(e(s(t( (a( (s(t(r(i(n(g))))))))))))

Input:
foobar
Output:
f(o(o(b(a(r)))))

Input:
1234567890
Output:
1(2(3(4(5(6(7(8(9(0)))))))))

Input:
code-golf
Output:
c(o(d(e(-(g(o(l(f))))))))

Input:
a
Output:
a

Input:
42
Output:
4(2)

As usual, all of our default rules and loopholes apply, and the shortest answer scored in bytes wins!

Answer 1

Kotlin, 83 63 bytes

-20 bytes using toList; thanks to 12Me21 tipping me off this could be shorter.

{s:String->s.toList().joinToString("(")+")".repeat(s.length-1)}

Try it online!

Answer 2

Forth (gforth), 66 58 bytes

: f over 1 type 1 -1 d+ dup 0> if ." ("recurse ." )"then ;

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-8 bytes by switching to a recursive solution

Explanation

Prints the first character of the string. If it's not the last character in the string, print an open parentheses, recursively call on the remainder of the string, and print a close parentheses.

Code Explanation

: f             \ start a new word definition
  over 1 type   \ print the first character of the string
  1 -1 d+       \ remove the first character of the string
  dup           \ duplicate the string length value
  0> if         \ check if remaining string length is greater than 0
    ." ("       \ print an open parentheses
    recurse     \ call recursively on remainder of string
    ." )"       \ print a close parentheses
  then          \ end the if block
;               \ end the word definition            

Answer 3

Vyxal, 7 bytes

L‹k(*sY

Try it Online!

Get gamed on @emanresuA

Explained

L‹k(*sY
L‹      # Push the length of the input, but decremented
  k(    # Push "()"
    *   # and repeat it ↑↑ times
     s  # sort that
      Y # and interleave it with the input

Answer 4

JavaScript (Node.js), 35 ⁴⁰ bytes

Only 2 bytes above the best JS answer! (using array of characters as input, as OP apparently allowed in his comments)

s=>s.reduceRight((a,e)=>e+`(${a})`)

The reduceRight combines the advantage of the reduce (n-1 iterations for n items, because in the output there is always one less couple of parenthesis than characters) with a reverse (necessary in this approach for accumulating the nesting by starting with the end)

Try it online!

Answer 5

Perl 5, 27 bytes (25 + 1 for -l + 1 for -n)

This is a translation of Wheat Wizard's approach to Perl.

say$_.")"x s/.(?!$)/$&(/g

Run it like this:

perl -nlE 'say$_.")"x s/.(?!$)/$&(/g' <<< 'Hello'

Explanation:

say$_.")"x s/.(?!$)/$&(/g
           s/.(?!$)/   /g # replace every char not followed by end of string ...
                    $&    # ... with the full match (that's the one char)
                       (  # and an open parenthesis
           s/.(?!$)/$&(/g # this operates on and changes $_ and ...
                          # ... returns the number of substitutes
      ")"x                # repeat closing paren number of substitues times
     .                    # append
   $_                     # to the string that has already been changed to f(o(o
say                       # print with newline

Answer 6

Pyke, 9 bytes

\(JQl\)*+

Try it here!

Answer 7

Python 3, 40 chars

s=input()
print("(".join(s)+len(s)*")")

Answer 8

Gema, 31 13 characters

Shamelessly borrowing Titus's idea from his recursive PHP solution.

\B?=?
?#=(?#)

Sample run:

bash-4.3$ echo -n 'Hello world!' | gema '\B?=?;?#=(?#)'
H(e(l(l(o( (w(o(r(l(d(!)))))))))))

Answer 9

Scala, 46 bytes

(s:String)=>(s.init:\(""+s.last))(_+"("+_+")")

Explanation:

(s:String)=>    //define a function
  (s.init       //take everything but the last char
   :\           //foldRight
   (""+s.last)    //with the last char as a string as a start
  )(              //combine the chars right to left with this function:
    _+"("+_+")"   //take the char, append "(", append everything we've got so far, append ")"
  )

Answer 10

Groovy, 38 36 Bytes thanks to @manatwork

{it.reverse().inject{r,i->i+"($r)"}}

Yeah, the C&P messed up the first time.

Answer 11

Python 2, 60 57 51 Bytes

def c(s):if(len(s)==1):return s;return s[0]+"("+c(s[1:])+")"

After some clarification on the rules from @manatwork on white space,

def c(s):return s if(len(s)==1)else s[0]+"("+c(s[1:])+")"

Thanks again @manatwork

def c(s):return s[0]+"("+c(s[1:])+")"if s[1:]else s

Ungolfed:

def c(s):
    if(len(s)==1):
        return s;
    return s[0]+"("+c(s[1:])+")"

Recursively calls itself and adds to the string.

c("CodeGolf") = 'C(o(d(e(G(o(l(f)))))))'

Answer 12

Python 50 49 bytes

saved 1 byte (actually 3) thanks to @ETHproductions

This is my recursive version written in Python. Would probably be shorter the iterative aproach.. but I simply like recursion :)

F=lambda s:s[0]+'('+F(s[1:])+')'if len(s)>1else s

Answer 13

Dart 44 bytes

p(s)=>s.split("").join("(")+")"*~-s.length;

I tried to be clever, but nothing beat this simple version. Notable mention:

r(s,[x=0])=>s[x++]+(x<s.length?"(${r(s,x)})":"");
q(s)=>s[0]+((s=s.substring(1))==""?s:"(${q(s)})");

but they drowned in necessary parentheses.

Answer 14

Java, 72 bytes

(a,s,l)->{l=a.length;s=""+a[--l];for(;l>0;)s=a[--l]+"("+s+")";return s;}

Ungolfed

public class Main {

  interface X {
    String f(char[]a,String s,int l);
  }

  static X x = (a,s,l) -> {
    l = a.length;
    s = "" + a[--l]; // start with the last character
    for (;l>0;)
      s = a[--l] + "(" + s +")"; // wrap in parentheses and prepend with the previous letter.
    return s;
  };

  public static void main(String[] args) {
    System.out.println(x.f("Hello World!".toCharArray(),"",0));
  }
}

Answer 15

C, 84 82 76 70 68 66 Bytes

i;f(char*s){for(i=1;*s+~i;putchar(*s?i&1?*s++:40:41),i+=*s?1:-2);}

Now using just one for loop and one putchar...

Test is like this

main(c,v)char**v;
{
    f(v[1]);puts("");
    f("foobar");puts("");     
    f("code-golf");puts("");     
}

output

H(e(l(l(o( (W(o(r(l(d))))))))))
f(o(o(b(a(r)))))
c(o(d(e(-(g(o(l(f))))))))

Answer 16

Clojure, 94 60 59 48 bytes

-34 by making it a actual recursive solution. The biggest saving here was getting rid of the repeat part to generate the end brackets.

-1 by rearranging it, eliminating a conditional.

-11 bytes thanks to NikoNyrh. Now deconstructs the parameter directly.

(defn n[[f & r]](if(str f(if r(str\((n r)\))))))

Recursive. Basically (str head "(" (recur tail) ")"), with the brackets being added only if a tail exists.

Uses unoptimized recursion. Can handle strings up to around 5235 characters long.

Ungolfed:

(defn nest [[f & r]]
    (if f ; When it exists, construct string and recur, else, base-case
      (str f (if r (str \( (nest r) \))))))

Answer 17

Python 3, 43 bytes

lambda x:print(*x,sep='(',end=')'*~-len(x))

Longer than shortest Python submission, but it is fairly different, so I thought I should post it. It is a function, but prints to STDOUT.

Answer 18

Haskell, 38 30 bytes

8 bytes saved thanks to @JanDvorak

f[x]=[x]
f(x:s)=x:'(':f s++")"

This is my first attempt at a haskell golf, probably not optimal yet.

Explanation

We define a function f. If this function receives input that matches the pattern [x], that is a length 1 string, we return the input. If we receive anything else as input we return the x:'(':f s++")", or the first character plus the rest result of f on the rest of the string all enclosed in parentheses.

Answer 19

Groovy, 46 bytes

{it.reverse().inject{r,i->")$r($i"}.reverse()}

Answer 20

Swift 3.1,  85  83 bytes

This is an anonymous function accepting an Array of Strings and returning a String.

{$0.joined(separator:"(")+String(repeating:")",count:$0.count-1)}as([String])->String

Try it here!

Swift 3.1,  88 87  86 bytes

This is a named function accepting an Array of Strings and printing a String.

func f(n:[String]){print(n.joined(separator:"(")+(1..<n.count).map{_ in")"}.joined())}

Try it here!

Answer 21

Ly, 50 bytes

iyspr1[1[=!["("o1$]p1$]1[=[pp2$]p1$]o]l1-[")"o1-];

Try it online!

This is ridiculous. Ly is lacking severely in the string manipulation department.

Answer 22

Java 8 (63 82 bytes)

"s" is a char[]

s->{String r="",p=r;for(char c:s){r+="("+c;p+=")";}return r+p;}

s->{String r="",p=r;for(char c:s){if(r==p)r+=c;else{r+="("+c;p+=")";}}return r+p;}

Edit: I didn't realise there aren't parentheses around the whole string.

Answer 23

Implicit, 24 23 bytes

~@~.(0-1@40@~.);;(0@41;

Try it online!

Ungolfed/explained:

~@~.     # read first character, print, read second, increment
(0       # do-while top of stack truthy
 -1      #  decrement top of stack
 @40     #  print open parenthesis (ASCII 40)
 @       #  print top of stack
 ~.      #  read char and increment
)        # while top of stack truthy
;;       # pop top two values (EOF, last char)

(0       # do-while top of stack truthy
 @41     #  print close parenthesis (ASCII 41)
 ;       #  pop top of stack
¶        # (implicit) forever

Test cases:

Nest a string
N(e(s(t( (a( (s(t(r(i(n(g))))))))))))

foobar
f(o(o(b(a(r)))))

1234567890
1(2(3(4(5(6(7(8(9(0)))))))))

code-golf
c(o(d(e(-(g(o(l(f))))))))

a
a

42
4(2)

Answer 24

SNOBOL4 (CSNOBOL4), 111 bytes

	INPUT LEN(1) . O REM . N
	W =SIZE(N)
N	N LEN(1) . X REM . N	:F(O)
	O =O '(' X	:(N)
O	OUTPUT =O DUPL(')',W)
END

Try it online!

Answer 25

Canvas, 8 bytes

(＊；Ｌ╷)×＋

Try it here!

Answer 26

Python 3.6+, 39 bytes

z=lambda x,*y:x+f'({z(*y)})'if y else x

Takes each character as an individual argument (can be used like z(*'asdf')). This uses recursion with iterable unpacking and f-string interpolation.

---

Note: This is almost non-competing. Python 3.6 was released after this challenge was posted, but was feature-frozen in beta on 2016-09-12, so this feature was available before the challenge was posted.

---

Though it turns out you can do it in the same number of bytes with %-interpolation (this works on more Python versions):

z=lambda x,*y:x+'(%s)'%z(*y)if y else x

Answer 27

Whitespace, 132 bytes

[S S S N
_Push_0][N
S S N
_Create_Label_LOOP][S N
S _Duplicate][S N
S _Duplicate][T N
T   S _Read_characters_as_STDIN][T  T   T   _Retrieve][S S S T  S T S S T   N
_Push_41_)][T   S S T   _Subtract][N
T   S S N
_If_0_Jump_to_Label_PRINT_TRAILING][S N
S _Duplicate][N
T   S T N
_If_0_Jump_to_Label_SKIP][S S S T   S T S S S N
_Push_40_(][T   N
S S _Print_as_character][N
S S T   N
_Create_Label_SKIP][S N
S _Duplicate][T T   T   _Retrieve][T    N
S S _Print_as_character][S S S T    N
_Push_1][T  S S S _Add][N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT_TRAILING][S S S T   N
_Push_1][T  S S T   _Subtract][S N
S _Duplicate][N
T   S S S N
_If_0_Jump_to_Label_EXIT][S S S T   S T S S T   N
_Push_41_)][T   N
S S _Print_as_character][N
S N
S N
_Jump_to_Label_PRINT_TRAILING]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Since Whitespace's STDIN don't know when the input is done since it can only read a single character or number at a time, the input will need a trailing ) to indicate we're done inputting (which is possible since the challenge rules state: "For simplicity's sake, you may also assume the input will not contain parentheses").

Explanation in pseudo-code:

Integer n = 0
Start LOOP:
  Character c = read character from STDIN
  If(c == ')'):
    Jump to function PRINT_TRAILING
  If(n != 0)
    Print '(' to STDOUT
  Print c to STDOUT
  n = n + 1
  Jump to next iteration of LOOP

function PRINT_TRAILING:
  n = n - 1
  If(n == 0)
    Stop program
  Print ')' to STDOUT
  Jump to function PRINT_TRAILING

Answer 28

J, 29 bytes

f=:(')'#~<:@#),~}.@,@('(',.])

Try it online!

Answer 29

C++, 66 bytes

[&](string s){return s[1]?s.substr(0,1)+"("+f(s.substr(1))+")":s;}

Try it Online

Matlab, 45 bytes

@(s)fold(@(x,y)strcat(y,'(',x,')'),fliplr(s))

Try it Online!

The problem with this is that octave in try it online doesn't really support the fold function which of course exists in Matlab. Let me know if I should delete this solution since we can't test it on tio.

Answer 30

Racket 195 bytes

(let p((r "")(l(reverse(string->list s)))(i 0))(cond[(= i(length l))r][(= i 0)(set! r(string(list-ref l i)))
(p r l(+ 1 i))][(set! r(string-append(string(list-ref l i))"("r")" ))(p r l(+ 1 i))]))

Ungolfed:

(define (f s)
  (let loop ((r "")
             (l (reverse (string->list s)))
             (i 0))
    (cond
      [(>= i (length l)) r] 
      [(= i 0) (set! r (string (list-ref l i)))
               (loop r l (add1 i))]
      [else (set! r (string-append (string (list-ref l i)) "(" r ")" ))
            (loop r l (add1 i))]
      )))

Testing:

(f "Hello")

Output:

"H(e(l(l(o))))"

Edit: 2 bytes saved following suggestion by @JonathanFrech (add1 to + 1)