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Tweaked compress challenge.
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@ -66,9 +66,10 @@
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"source": [
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"## Algorithm: List\n",
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"\n",
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"Since Python strings are immutable, we'll use a list of characters instead to exercise string manipulation as you would get with a C string. We'll convert the list to a string at the end of the algorithm.\n",
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"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",
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"\n",
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"* Calculate the size of the compressed string\n",
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" * Note the constraint about compressing only if it saves space\n",
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"* If the compressed string size is >= string size, return string\n",
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"* Create compressed_string\n",
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" * For each char in string\n",
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@ -84,7 +85,7 @@
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: O(n) additional space for the list"
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"* Space: O(n)"
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]
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},
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{
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@ -134,6 +135,8 @@
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" last_char = char\n",
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" compressed_string.append(last_char)\n",
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" 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",
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" return \"\".join(compressed_string)"
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]
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},
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@ -143,13 +146,20 @@
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"source": [
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"## Algorithm: Byte Array\n",
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"\n",
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"As a bonus solution, we can solve this problem with a byte array.\n",
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"As a bonus solution, we can also solve this problem with a byte array.\n",
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"\n",
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"The byte array algorithm similar when using a list, except we will need to work with the bytearray's character codes (using the function ord) instead of the characters as we did above when we implemented this solution with a list.\n",
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"The byte array algorithm similar when using a list, except we will need to work with the bytearray's character codes (using the function ord) instead of the characters as when we implemented this solution with a list.\n",
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"\n",
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"Complexity:\n",
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"* Time: O(n)\n",
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"* Space: O(m) where m is the size of the compressed bytearray"
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"* Space: O(n)"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code: Byte Array"
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]
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},
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{
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@ -194,14 +204,7 @@
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" last_char_code = char_code\n",
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" compressed_string[pos] = last_char_code\n",
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" compressed_string[pos+1] = ord(str(count))\n",
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" return compressed_string"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Code: Byte Array"
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" return str(compressed_string)"
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]
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},
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{
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