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67 lines
1.6 KiB
Markdown
67 lines
1.6 KiB
Markdown
---
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id: 6821ebda237de8297eaee792
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title: "Challenge 16: Reverse Parenthesis"
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challengeType: 29
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dashedName: challenge-16
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---
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# --description--
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Given a string that contains properly nested parentheses, return the decoded version of the string using the following rules:
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- All characters inside each pair of parentheses should be reversed.
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- Parentheses should be removed from the final result.
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- 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.
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- Assume all parentheses are evenly balanced and correctly nested.
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# --hints--
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`decode("(f(b(dc)e)a)")` should return `"abcdef"`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(decode("(f(b(dc)e)a)"), "abcdef")`)
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}})
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```
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`decode("((is?)(a(t d)h)e(n y( uo)r)aC)")` should return `"Can you read this?"`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(decode("((is?)(a(t d)h)e(n y( uo)r)aC)"), "Can you read this?")`)
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}})
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```
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`decode("f(Ce(re))o((e(aC)m)d)p")` should return `"freeCodeCamp"`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(decode("f(Ce(re))o((e(aC)m)d)p"), "freeCodeCamp")`)
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}})
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```
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# --seed--
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## --seed-contents--
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```py
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def decode(s):
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return s
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```
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# --solutions--
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```py
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def decode(s):
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while ')' in s:
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close_index = s.index(')')
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open_index = s.rindex('(', 0, close_index)
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inner = s[open_index + 1:close_index][::-1]
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s = s[:open_index] + inner + s[close_index + 1:]
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return s
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```
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