Python 3 Beats 100.0% Simple
This solution beats 100% both experimentally and should a lot run faster than the average solution theoretically too.\n\nHere\'s the "average" solution:\n\n    
This solution beats 100% both experimentally and should a lot run faster than the average solution theoretically too.\n\nHere\'s the "average" solution:\n```\n    def fizzBuzz(self, n: int) -> List[str]:\n        arr = []\n\t\t\n        for x in range(1, n + 1):\n            if x % 15 == 0:\n                arr.append(\'FizzBuzz\')\n            elif x % 3 == 0:\n                arr.append(\'Fizz\')\n            elif x % 5 == 0:\n                arr.append(\'Buzz\')\n            else:\n                arr.append(str(x))\n\t\t\t\n        return arr\n```\nHowever, this is slow for many reasons. First, the % operator is costly. It\'s a lot slower than most people would expect it to be. Second, the "average" case where it\'s just a normal integer that you add to the list, you have to go through 4 % operators. This is extremely unnecessary.\n\nNow, here\'s an improved code:\n```\n    def fizzBuzz(self, n: int) -> List[str]:\n        f, b, fb = \'Fizz\', \'Buzz\', \'FizzBuzz\'\n        arr = [str(x + 1) for x in range(n)]\n        \n        for i in range(2, n, 3):\n            arr[i] = f\n        \n        for i in range(4, n, 5):\n            arr[i] = b\n        \n        for i in range(14, n, 15):\n            arr[i] = fb\n        \n        return arr\n```\n\nNow, instead of the average case having to go through 4 % operators, it just has to be initiated once. There are no % operators *or* if else\'s, making it extremely fast.
['Python', 'Python3']