42 lines
735 B
Markdown
42 lines
735 B
Markdown
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## Compress String Solution
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1. Write a function that gets a string and compresses it
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- 'aaaabbccc' -> 'a4b2c3'
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- 'abbbc' -> 'a1b3c1'
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```
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def compress_str(mystr: str) -> str:
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result = ''
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if mystr:
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prevchar = mystr[0]
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else:
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return result
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count = 1
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for nextchar in mystr[1:]:
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if nextchar == prevchar:
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count += 1
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else:
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result += prevchar + str(count)
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count = 1
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prevchar = nextchar
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result += prevchar + str(count)
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return result
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```
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2. Write a function that decompresses a given string
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- 'a4b2c3' -> 'aaaabbccc'
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- 'a1b3c1' -> 'abbbc'
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```
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def decompress_str(mystr: str) -> str:
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result = ''
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for index in range(0, len(mystr), 2):
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result += mystr[index] * int(mystr[index + 1])
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return result
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```
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