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" This notebook was prepared by [Donne Martin](http://donnemartin.com). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges). "
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" # Solution Notebook "
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" ## Problem: Compress a string such that ' AAABCCDDDD ' becomes ' A3BC2D4 ' . Only compress the string if it saves space. \n " ,
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" \n " ,
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" * [Constraints](#Constraints) \n " ,
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" * [Test Cases](#Test-Cases) \n " ,
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" * [Algorithm](#Algorithm) \n " ,
" * [Code](#Code) \n " ,
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" * [Unit Test](#Unit-Test) "
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" ## Constraints \n " ,
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" \n " ,
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" * Can we assume the string is ASCII? \n " ,
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" * Yes \n " ,
" * Note: Unicode strings could require special handling depending on your language \n " ,
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" * Is this case sensitive? \n " ,
" * Yes \n " ,
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" * Can we use additional data structures? \n " ,
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" * Yes \n " ,
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" * Can we assume this fits in memory? \n " ,
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" * Yes "
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" ## Test Cases \n " ,
" \n " ,
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" * None -> None \n " ,
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" * ' ' -> ' ' \n " ,
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" * ' AABBCC ' -> ' AABBCC ' \n " ,
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" * ' AAABCCDDDD ' -> ' A3BC2D4 ' "
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" ## Algorithm \n " ,
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" \n " ,
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" * For each char in string \n " ,
" * If char is the same as last_char, increment count \n " ,
" * Else \n " ,
" * Append last_char and count to compressed_string \n " ,
" * last_char = char \n " ,
" * count = 1 \n " ,
" * Append last_char and count to compressed_string \n " ,
" * If the compressed string size is < string size \n " ,
" * Return compressed string \n " ,
" * Else \n " ,
" * Return string \n " ,
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" \n " ,
" Complexity: \n " ,
" * Time: O(n) \n " ,
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" * Space: O(n) \n " ,
" \n " ,
" Complexity Note: \n " ,
" * Although strings are immutable in Python, appending to strings is optimized in CPython so that it now runs in O(n) and extends the string in-place. Refer to this [Stack Overflow post](http://stackoverflow.com/a/4435752). "
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" ## Code "
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" class CompressString(object): \n " ,
" \n " ,
" def compress(self, string): \n " ,
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" if string is None or not string: \n " ,
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" return string \n " ,
" result = ' ' \n " ,
" prev_char = string[0] \n " ,
" count = 0 \n " ,
" for char in string: \n " ,
" if char == prev_char: \n " ,
" count += 1 \n " ,
" else: \n " ,
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" result += self._calc_partial_result(prev_char, count) \n " ,
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" prev_char = char \n " ,
" count = 1 \n " ,
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" result += self._calc_partial_result(prev_char, count) \n " ,
" return result if len(result) < len(string) else string \n " ,
" \n " ,
" def _calc_partial_result(self, prev_char, count): \n " ,
" return prev_char + (str(count) if count > 1 else ' ' ) "
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" ## Unit Test "
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" text " : [
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" Overwriting test_compress.py \n "
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]
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" source " : [
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" %% writefile test_compress.py \n " ,
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" from nose.tools import assert_equal \n " ,
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" \n " ,
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" \n " ,
" class TestCompress(object): \n " ,
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" \n " ,
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" def test_compress(self, func): \n " ,
" assert_equal(func(None), None) \n " ,
" assert_equal(func( ' ' ), ' ' ) \n " ,
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" assert_equal(func( ' AABBCC ' ), ' AABBCC ' ) \n " ,
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" assert_equal(func( ' AAABCCDDDDE ' ), ' A3BC2D4E ' ) \n " ,
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" assert_equal(func( ' BAAACCDDDD ' ), ' BA3C2D4 ' ) \n " ,
" assert_equal(func( ' AAABAACCDDDD ' ), ' A3BA2C2D4 ' ) \n " ,
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" print( ' Success: test_compress ' ) \n " ,
" \n " ,
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" \n " ,
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" def main(): \n " ,
" test = TestCompress() \n " ,
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" compress_string = CompressString() \n " ,
" test.test_compress(compress_string.compress) \n " ,
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" \n " ,
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" \n " ,
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" if __name__ == ' __main__ ' : \n " ,
" main() "
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" Success: test_compress \n "
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" %r un -i test_compress.py "
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