Fizz Buzz without conditionals or booleans

Summary

Fizz Buzz without conditionals or booleansI recently learned about the Feeling of Computing podcast and listened to the latest episode. One of the hosts challenged listeners to “write Fizz Buzz with no booleans, no conditionals, no pattern matching, or other things that are like disguised booleans.”Here’s my Python solution:from itertools import cycle def string_mask(a, b): return b + a[len(b) :] def main(): fizz = cycle(["", "", "Fizz"]) buzz = cycle(["", "", "", "", "Buzz"]) numbers = range(1, 101) for f, b, n in zip(fizz, buzz, numbers): print(string_mask(str(n), f + b)) main() This solution is basically three things put together:Create endless cycling sequences of "", "", "fizz", "", "", "fizz", "", "", "fizz", ... and the same idea for buzz.Combine those two sequences with the numbers 1 through 100 using zip, to get the following sequence:... ("", "", 7) ("", "", 8) ("Fizz", "", 9) ("", "Buzz", 10) ("", "", 11) ("Fizz", "", 12) ("", "", 13) ("", "", 14) ("Fizz", "Buzz", 15) ("", "", 16) ... Convert the number to a string, then “mask” it with any “Fizz”, “Buzz”, or “FizzBuzz” string. For example, string_mask("3", "Fizz") returns "Fizz", and string_mask("10015", "Buzz") returns "Buzz5".Because of this, my code breaks once you reach 10,000 because the digits start “leaking out” the end of the string. You’ll start seeing results like Buzz0 and Fizz2.I’m sure there are better ways to do this, but that was my quick solution. How would you solve this problem?