First re-pass at FizzBuzz after an interview

fizz_buzz_1.py

cache = {}


def fizz_buzz(n: int, factor_1: int = 3, factor_2: int = 5) -> None:
    """
    Given an integer n, for a list of integers 1-n, the task is to print,

    "FizzBuzz" if i is divisible by `factor_1` and `factor_2`,
    "Fizz" if i is divisible by `factor_1`,
    "Buzz" if i is divisible by `factor_2`
    "i" as a string, if none of the conditions are true.

    :param n: The upper limit, inclusive, of the range to evaluate
    :param factor_1: The first factor
    :param factor_2: The second factor
    """

    common_factor = factor_1 * factor_2

    # Iterate over 1 to n, inclusive
    for full_size_num in range(1, n + 1):
        output = ""

        # Calculate `constrained_num` as the remainder of `full_size_num` divided by `common_factor`
        # i.e. 0 <= `constrained_num` < `common_factor`
        if full_size_num > common_factor:
            # Constrain `full_size_num`
            constrained_num = full_size_num % common_factor
        else:
            # Do nothing
            constrained_num = full_size_num

        if full_size_num not in cache:
            # `constrained_num` is NOT in `cache`
            if constrained_num % factor_1 == 0:
                output += "Fizz"

            if constrained_num % factor_2 == 0:
                output += "Buzz"

            if len(output) < 1:
                output = str(full_size_num)

            # Add `output` to `cache`
            cache[constrained_num] = output
        else:
            # `constrained_num` is already in `cache`
            pass

        print(cache[constrained_num])


fizz_buzz(61)

## Output
# 1
# 2
# Fizz
# 4
# Buzz
# Fizz
# 7
# 8
# Fizz
# Buzz
# 11
# Fizz
# 13
# 14
# FizzBuzz
# 16
# 17
# Fizz
# 19
# Buzz
# Fizz
# 22
# 23
# Fizz
# Buzz
# 26
# Fizz
# 28
# 29
# FizzBuzz
# 31
# 32
# Fizz
# 34
# Buzz
# Fizz
# 37
# 38
# Fizz
# Buzz
# 41
# Fizz
# 43
# 44
# FizzBuzz
# 46
# 47
# Fizz
# 49
# Buzz
# Fizz
# 52
# 53
# Fizz
# Buzz
# 56
# Fizz
# 58
# 59
# FizzBuzz
# 61