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81 lines
1.6 KiB
Markdown
81 lines
1.6 KiB
Markdown
---
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id: 69b58ce40693f140c84c8559
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title: "Challenge 240: Palindrome Characters"
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challengeType: 29
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dashedName: challenge-240
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---
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# --description--
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Given a string, determine if it's a palindrome and return the middle character (if it's odd length) or middle two characters (if it's even).
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- A palindrome is a string that is the same forward and backward.
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- If it's not a palindrome, return `"none"`.
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# --hints--
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`palindrome_locator("racecar")` should return `"e"`.
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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(palindrome_locator("racecar"), "e")`)
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}})
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```
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`palindrome_locator("level")` should return `"v"`.
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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(palindrome_locator("level"), "v")`)
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}})
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```
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`palindrome_locator("freecodecamp")` should return `"none"`.
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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(palindrome_locator("freecodecamp"), "none")`)
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}})
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```
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`palindrome_locator("noon")` should return `"oo"`.
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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(palindrome_locator("noon"), "oo")`)
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}})
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```
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`palindrome_locator("11100111")` should return `"00"`.
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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(palindrome_locator("11100111"), "00")`)
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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 palindrome_locator(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 palindrome_locator(s):
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if s != s[::-1]:
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return "none"
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mid = len(s) // 2
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return s[mid] if len(s) % 2 == 1 else s[mid - 1] + s[mid]
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```
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