Python3, 94% faster with explanation
in python3 a string is unmutable, so we need to build a new string.\n\nI was a little lazy, and just built a while loop which checks if i is greater than s, eve
in python3 a string is unmutable, so we need to build a new string.\n\nI was a little lazy, and just built a while loop which checks if i is greater than s, even though it should never be, really the existence of \'s\' is our end case.\n\nfrom here we just check if i is divisible by two, if it is, we reverse k amount of chars and add them to our string, if it isnt we simply add the current amount of chars. then we delete k many letters from s, when s no longer exists, we can return our new string. \n\n```\nclass Solution:\n    def reverseStr(self, s: str, k: int) -> str:\n        news, ogLen = \'\', len(s)\n        i = 0\n        while (i < ogLen) and s:\n            if i % 2 == 0:\n                news += s[:k][::-1]\n            else:\n                news += s[:k]\n            s = s[k:]\n            i += 1\n        return news
['Python', 'Python3']