Parenthetical Names

Question

Parenthetical names

Your challenge is to build a program that puts parentheses around the 'title' part of names. For example, Sonic the Hedgehog is converted to Sonic (the Hedgehog).

Rules:

• Around every the that passes the following rules, an opening parentheses will be added before the the and a corresponding closing parentheses will be added at the end of the string.
• Inputs will only contain words (uppercase/lowercase letters and spaces).
• Trailing thes are ignored.
• With more than one thes in a row, every other the is parenthesized. (So pig the the the hog hog hog turns into pig (the the (the hog hog hog)).)
• Remaining thes are skipped unless surrounded by spaces. For example, the the in their would not be parenthesized.

I know this may seem lacking in detail. However, it's hard to get one set of really specific guidelines working- just follow the above rules and you're good.

Inspired by this xkcd comic: https://xkcd.com/2252/

This is code-golf, so least number of bytes wins.

Answer 1

QuadR, 27 bytes

\b(the )+
 (the .*)
&
 (\1)

Try it online!

\b(the )+ replace multiple thes with:
…& themselves

(the .*) replace a space followed by the and everything until the end with:
… (\1) a space followed by the parenthesised text

Answer 2

Retina 0.8.2, 27 bytes

+`(?<!\bthe) (the .+)
 ($1)

Try it online! Link includes test cases. Explanation: Parenthesises from all the s to the end of the line, unless the the follows a previous the.

Answer 3

Charcoal, 40 bytes

≔⪪Ｓ θ≔⌕Ａθtheη≔⁻η⁺⊕η⟦⁰⊖Ｌθ⟧η⪫Ｅθ⁺×№ηκ(ι ⭆η)

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

≔⪪Ｓ θ

Split the input on spaces.

≔⌕Ａθtheη

Find all the indices of the.

≔⁻η⁺⊕η⟦⁰⊖Ｌθ⟧η

Remove any index that's 1 more than another index, and 0 and the last index (because those thes weren't surrounded by spaces).

⪫Ｅθ⁺×№ηκ(ι 

Prefix ( to the remaining indices and join with spaces.

⭆η)

Output the matching )s at the end.

A port of @Steffan's Python answer is also 40 bytes:

⪫⪪⪫⪪Ｓ the ¦ (the ¦the (the¦the the×)№ＫＡ(

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

⪫⪪⪫⪪Ｓ the ¦ (the ¦the (the¦the the

Substitute the with (the and the (the with the the.

×)№ＫＡ(

Output enough )s to match the number of (s.

Answer 4

05AB1E, 31 bytes

„ €€ÀDć'(©«ì:…€€ (8∍D®K:D®¢')׫

Port of @Steffan's Python answer.

Try it online or verify all test cases.

Explanation:

            :  # Replace all
„ €€À          # " the ":
„ €€           #  Push dictionary string "  the"
    À          #  Rotate it once towards the right: " the "
     Dć'(©«ì  '# with " (the ":
     D         #  Duplicate the " the "
      ć        #  Extract the head
       '(     '#  Push "("
         ©     #  Store it in variable `®` (without popping)
          «    #  Append it to the extracted space
           ì   #  Merge it in front of the remaining string: " (the "

          :    # Then replace all
…€€ (8∍        # "the (the":
…€€ (          #  Push dictionary string "the ("
     8∍        #  Extend it to size 8: "the (the"
       D®K     # with "the the":
       D       #  Duplicate the "the (the"
        ®K     #  Remove the `®` ("("): "the the"

D              # Duplicate the resulting string
 ®¢            # Pop the copy, and count the amount of `®` ("(") in it
   ')×        '# Pop and push a string with that many ")"
      «        # Append it at the end of the string
               # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to use the dictionary?) to understand why „ €€ is " the" and …€€ ( is "the (".

Answer 5

Python, 93 bytes

lambda s,t='the',r=' (',w=' ':(p:=s.replace(w+t+w,r+t+w).replace(t+r+t,t+w+t))+p.count(r)*')'

Attempt This Online!

Vyxal, 31 bytes

⌈D‛λλÞIf$L‹o:›Fꜝλ\(p;VṄ:\(O\)*J

Try it Online!

Port of Charcoal answer.

Answer 6

Python3, 116 bytes:

import re
f=lambda x,c=0:re.sub(f'(?<{"!="[c]}e) the {["","[^t]"][c]}[\w\W]+',lambda x:f' ({f(x.group()[1:],1)})',x)