freeCodeCamp/curriculum/challenges/english/99-dev-playground/daily-coding-challenges-python/python-challenge-16.md

id, title, challengeType, dashedName

id	title	challengeType	dashedName
6821eca2237de8297eaee79e	Python Challenge 16: Reverse Parenthesis	29	python-challenge-16

--description--

Given a string that contains properly nested parentheses, return the decoded version of the string using the following rules:

• All characters inside each pair of parentheses should be reversed.
• Parentheses should be removed from the final result.
• If parentheses are nested, the innermost pair should be reversed first, and then its result should be included in the reversal of the outer pair.
• Assume all parentheses are evenly balanced and correctly nested.

--hints--

decode("(f(b(dc)e)a)") should return "abcdef".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(decode("(f(b(dc)e)a)"), "abcdef")`)
}})

decode("((is?)(a(t d)h)e(n y( uo)r)aC)") should return "Can you read this?".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(decode("((is?)(a(t d)h)e(n y( uo)r)aC)"), "Can you read this?")`)
}})

decode("f(Ce(re))o((e(aC)m)d)p") should return "freeCodeCamp".

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(decode("f(Ce(re))o((e(aC)m)d)p"), "freeCodeCamp")`)
}})

--seed--

--seed-contents--

def decode(s):

    return s

--solutions--

def decode(s):
    while ')' in s:
        close_index = s.index(')')
        open_index = s.rindex('(', 0, close_index)
        inner = s[open_index + 1:close_index][::-1]
        s = s[:open_index] + inner + s[close_index + 1:]
    return s