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" This notebook was prepared by [hashhar](https://github.com/hashhar), second solution added by [janhak] (https://github.com/janhak). Source and license info is on [GitHub](https://github.com/donnemartin/interactive-coding-challenges). "
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]
} ,
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" # Solution Notebook "
]
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" ## Problem: Compress a string such that ' AAABCCDDDD ' becomes ' A3BCCD4 ' . Only compress the string if it saves space. \n " ,
" \n " ,
" * [Constraints](#Constraints) \n " ,
" * [Test Cases](#Test-Cases) \n " ,
" * [Algorithm](#Algorithm) \n " ,
" * [Code](#Code) \n " ,
" * [Unit Test](#Unit-Test) "
]
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" ## Constraints \n " ,
" \n " ,
" * Can we assume the string is ASCII? \n " ,
" * Yes \n " ,
" * Note: Unicode strings could require special handling depending on your language \n " ,
" * Is this case sensitive? \n " ,
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" * Yes \n " ,
" * Can we use additional data structures? \n " ,
" * Yes \n " ,
" * Can we assume this fits in memory? \n " ,
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" * Yes "
]
} ,
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" ## Test Cases \n " ,
" \n " ,
" * None -> None \n " ,
" * ' ' -> ' ' \n " ,
" * ' AABBCC ' -> ' AABBCC ' \n " ,
" * ' AAABCCDDDD ' -> ' A3BCCD4 ' "
]
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" ## Algorithm \n " ,
" \n " ,
" Since Python strings are immutable, we ' ll use a list of characters to build the compressed string representation. We ' ll then convert the list to a string. \n " ,
" \n " ,
" * Calculate the size of the compressed string \n " ,
" * Note the constraint about compressing only if it saves space \n " ,
" * If the compressed string size is >= string size, return string \n " ,
" * Create compressed_string \n " ,
" * For each char in string \n " ,
" * If char is the same as last_char, increment count \n " ,
" * Else \n " ,
" * If the count is more than 2 \n " ,
" * Append last_char to compressed_string \n " ,
" * append count to compressed_string \n " ,
" * count = 1 \n " ,
" * last_char = char \n " ,
" * If count is 1 \n " ,
" * Append last_char to compressed_string \n " ,
" * count = 1 \n " ,
" * last_char = char \n " ,
" * If count is 2 \n " ,
" * Append last_char to compressed_string \n " ,
" * Append last_char to compressed_string once more \n " ,
" * count = 1 \n " ,
" * last_char = char \n " ,
" * Append last_char to compressed_string \n " ,
" * Append count to compressed_string \n " ,
" * Return compressed_string \n " ,
" \n " ,
" Complexity: \n " ,
" * Time: O(n) \n " ,
" * Space: O(n) "
]
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" ## Code "
]
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" execution_count " : null ,
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" def compress_string(string): \n " ,
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" if string is None or len(string) == 0: \n " ,
" return string \n " ,
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" \n " ,
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" # Calculate the size of the compressed string \n " ,
" size = 0 \n " ,
" last_char = string[0] \n " ,
" for char in string: \n " ,
" if char != last_char: \n " ,
" size += 2 \n " ,
" last_char = char \n " ,
" size += 2 \n " ,
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" \n " ,
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" # If the compressed string size is greater than \n " ,
" # or equal to string size, return original string \n " ,
" if size >= len(string): \n " ,
" return string \n " ,
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" \n " ,
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" # Create compressed_string \n " ,
" # New objective: \n " ,
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" # Single characters are to be left as is \n " ,
" # Double characters are to be left as are \n " ,
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" compressed_string = list() \n " ,
" count = 0 \n " ,
" last_char = string[0] \n " ,
" for char in string: \n " ,
" if char == last_char: \n " ,
" count += 1 \n " ,
" else: \n " ,
" # Do the old compression tricks only if count exceeds two \n " ,
" if count > 2: \n " ,
" compressed_string.append(last_char) \n " ,
" compressed_string.append(str(count)) \n " ,
" count = 1 \n " ,
" last_char = char \n " ,
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" # If count is either 1 or 2 \n " ,
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" else: \n " ,
" # If count is 1, leave the char as is \n " ,
" if count == 1: \n " ,
" compressed_string.append(last_char) \n " ,
" count = 1 \n " ,
" last_char = char \n " ,
" # If count is 2, append the character twice \n " ,
" else: \n " ,
" compressed_string.append(last_char) \n " ,
" compressed_string.append(last_char) \n " ,
" count = 1 \n " ,
" last_char = char \n " ,
" compressed_string.append(last_char) \n " ,
" compressed_string.append(str(count)) \n " ,
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" \n " ,
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" # Convert the characters in the list to a string \n " ,
" return \" \" .join(compressed_string) "
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]
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" ## Algorithm: Split to blocks and compress \n " ,
" \n " ,
" Let us split the string first into blocks of identical characters and then compress it block by block. \n " ,
" \n " ,
" * Split the string to blocks \n " ,
" * For each character in string \n " ,
" * Add this character to block \n " ,
" * If the next character is different \n " ,
" * Return block \n " ,
" * Erase the content of block \n " ,
" \n " ,
" \n " ,
" * Compress block \n " ,
" * If block consists of two or fewer characters \n " ,
" * Return block \n " ,
" * Else \n " ,
" * Append length of the block to the first character and return \n " ,
" \n " ,
" \n " ,
" * Compress string \n " ,
" * Split the string to blocks \n " ,
" * Compress blocks \n " ,
" * Join compressed blocks \n " ,
" * Return result if it is shorter than original string \n " ,
" \n " ,
" Complexity: \n " ,
" * Time: O(n) \n " ,
" * Space: O(n) "
]
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" def split_to_blocks(string): \n " ,
" block = ' ' \n " ,
" for char, next_char in zip(string, string[1:] + ' ' ): \n " ,
" block += char \n " ,
" if char is not next_char: \n " ,
" yield block \n " ,
" block = ' ' \n " ,
" \n " ,
" \n " ,
" def compress_block(block): \n " ,
" if len(block) <= 2: \n " ,
" return block \n " ,
" else: \n " ,
" return block[0] + str(len(block)) \n " ,
" \n " ,
" \n " ,
" def compress_string(string): \n " ,
" if string is None or not string: \n " ,
" return string \n " ,
" compressed = (compress_block(block) for block in split_to_blocks(string)) \n " ,
" result = ' ' .join(compressed) \n " ,
" return result if len(result) < len(string) else string "
]
} ,
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" ## Unit Test "
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" cell_type " : " code " ,
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" source " : [
" %% writefile test_compress.py \n " ,
" from nose.tools import assert_equal \n " ,
" \n " ,
" \n " ,
" class TestCompress(object): \n " ,
" \n " ,
" def test_compress(self, func): \n " ,
" assert_equal(func(None), None) \n " ,
" assert_equal(func( ' ' ), ' ' ) \n " ,
" assert_equal(func( ' AABBCC ' ), ' AABBCC ' ) \n " ,
" assert_equal(func( ' AAABCCDDDD ' ), ' A3BCCD4 ' ) \n " ,
" assert_equal(func( ' aaBCCEFFFFKKMMMMMMP taaammanlaarrrr seeeeeeeeek tooo ' ), ' aaBCCEF4KKM6P ta3mmanlaar4 se9k to3 ' ) \n " ,
" print( ' Success: test_compress ' ) \n " ,
" \n " ,
" \n " ,
" def main(): \n " ,
" test = TestCompress() \n " ,
" test.test_compress(compress_string) \n " ,
" \n " ,
" \n " ,
" if __name__ == ' __main__ ' : \n " ,
" main() "
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" source " : [
" %r un -i test_compress.py "
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