Nikhilam Navatascaramam Dastah Part 3.py

def main(num1=int, num2=int):
    # First priority conditions:
    # 1. If any number is equal to 0, return 0
    # 2. If any number is equal to 1, return the other number
    # 3. If both numbers are less than equal to 5, use the multiple table
    if num1 == 0 or num2 == 0:
        return 0

    if num1 == 1:
        return num2
    elif num2 == 1:
        return num1

    if num1 <= 5 and num2 <= 5:
        return multiple_table[num1][num2]

    if num2 < num1:
        # Swap num1 and num2 so that num1 will always be less than num2
        num1, num2 = num2, num1

    # Determine bases for both numbers
    # Using the minus method
    base1 = int('1' + ('0' * len(str(num1))))
    # Using the plus method
    base2 = int('1' + ('0' * (len(str(num2)) - 1)))

    if not base1 == base2:
        return 'Formula Not Applicable'

    # Numbers cannot be a multiple of 10
    if num1 % 10 == 0 or num2 % 10 == 0:
        return 'Formula Not Applicable'

    zeros = len(str(base1)) - 1

    # First step is to figure out the right values
    right1 = base1 - num1
    right2 = num2 - base2

    # Getting the left and right portions of the final answer
    left_ans = num1 + right2
    right_ans = main(right1, right2)

    # Becuase of recursion, right_ans can end up being 'Formula Not Applicable'
    if right_ans == 'Formula Not Applicable':
        return 'Formula Not Applicable'

    right_ans = base1 - int(right_ans)
    left_ans -= 1

    if right_ans < 0:
        return 'Formula Not Applicable'

    while len(str(right_ans)) < zeros:
        right_ans = '0' + str(right_ans)

    return int(str(left_ans) + str(right_ans))


# Hard-coded 5x5 multiplication table
multiple_table = [[0, 0, 0, 0, 0, 0],
                  [0, 1, 2, 3, 4, 5],
                  [0, 2, 4, 6, 8, 10],
                  [0, 3, 6, 9, 12, 15],
                  [0, 4, 8, 12, 16, 20],
                  [0, 5, 10, 15, 20, 25],
                  ]