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99 lines
2.2 KiB
Markdown
99 lines
2.2 KiB
Markdown
---
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id: 698a1a863194f1f4e63f645e
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title: "Challenge 207: Smallest Gap"
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challengeType: 29
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dashedName: challenge-207
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---
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# --description--
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Given a string, return the substring between the two identical characters that have the smallest number of characters between them (smallest gap).
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- There will always be at least one pair of matching characters.
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- The returned substring should exclude the matching characters.
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- If two or more gaps are the same length, return the characters from the first one.
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For example, given `"ABCDAC"`, return `"DA"` because:
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- Only `"A"` and `"C"` repeat in the string.
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- The number of characters between the two `"A"` characters is 3, and between the `"C"` characters is 2.
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- So return the string between the two `"C"` characters.
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# --hints--
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`smallest_gap("ABCDAC")` should return `"DA"`.
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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(smallest_gap("ABCDAC"), "DA")`)
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}})
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```
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`smallest_gap("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(smallest_gap("racecar"), "e")`)
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}})
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```
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`smallest_gap("A{5e^SD*F4i!o#q6e&rkf(po8|we9+kr-2!3}=4")` should return `"#q6e&rkf(p"`.
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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(smallest_gap("A{5e^SD*F4i!o#q6e&rkf(po8|we9+kr-2!3}=4"), "#q6e&rkf(p")`)
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}})
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```
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`smallest_gap("Hello World")` should return `""`.
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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(smallest_gap("Hello World"), "")`)
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}})
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```
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`smallest_gap("The quick brown fox jumps over the lazy dog.")` should return `"fox"`.
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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(smallest_gap("The quick brown fox jumps over the lazy dog."), "fox")`)
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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 smallest_gap(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 smallest_gap(s):
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min_gap = float("inf")
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result = ""
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for i in range(len(s)):
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for j in range(i + 1, len(s)):
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if s[i] == s[j]:
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gap = j - i - 1
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if gap < min_gap:
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min_gap = gap
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result = s[i + 1:j]
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break
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return result
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```
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