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\n<\/p>\n
I would follow this steps:<\/p>\n
collections.Counter<\/code><\/li>\n<\/ol>\nAbout your code, the_string[i] == the_string[i]<\/code> will always be true.<\/p>\n
\nSPOILER: solution<\/h2>\nfrom collections import Counter\n\n\ndef repeating_letters(the_string):\n partials = []\n initial = 0\n for i, character in enumerate(the_string):\n if character == the_string[initial]:\n continue\n partials.append(the_string[initial:i])\n initial = i\n partials.append(the_string[initial:]) # Needed for the last partial string\n return Counter(partials)\n<\/code><\/pre>\nAs @prahantrana mentions in a comment, getting the partials can be done in a one-liner with the groupby<\/code> method from the itertools<\/code> library.<\/p>\nfrom collections import Counter\nfrom itertools import groupby\n\n\ndef repeating_letters(the_string):\n return Counter(''.join(group) for _, group in groupby(the_string))\n<\/code><\/pre>\nOr<\/p>\n
from collections import Counter\nfrom itertools import groupby\n\n\ndef repeating_letters(the_string):\n return Counter(char*len(list(group)) for char, group in groupby(the_string))\n<\/code><\/pre>\nI’m not sure which of them is faster.<\/p>\n<\/p><\/div>\n