#!/usr/bin/env python3

# We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234,
# is 1 through 5 pandigital.
#
# The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
#
# Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
#
# HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

from timeit import default_timer

import numpy as np
from numpy import zeros

from projecteuler import is_pandigital


def main():
    start = default_timer()

    n = 0
#   Initially I used a bigger array, but printing the resulting products
#   shows that 10 values are sufficient.
    products = zeros(10, int)

#   To get a 1 to 9 pandigital concatenation of the two factors and product,
#   we need to multiply a 1 digit number times a 4 digit numbers (the biggest
#   one digit number 9 times the biggest 3 digit number 999 multiplied give
#   8991 and the total digit count is 8, which is not enough), or a 2 digit
#   number times a 3 digit number (the smallest two different 3 digits number,
#   100 and 101, multiplied give 10100, and the total digit count is 11, which
#   is too many). The outer loop starts at 2 because 1 times any number gives
#   the same number, so its digit will be repeated and the result can't be
#   pandigital. The nested loop starts from 1234 because it's the smallest
#   4-digit number with no repeated digits, and it ends at 4987 because it's
#   the biggest number without repeated digits that multiplied by 2 gives a
#   4 digit number.
    for i in range(2, 9):
        for j in range(1234, 4987):
            p = i * j
            num_s = str(i) + str(j) + str(p)

            if len(num_s) > 9:
                break

            num = int(num_s)

            if is_pandigital(num, 9):
                products[n] = p
                n = n + 1

#   The outer loop starts at 12 because 10 has a 0 and 11 has two 1s, so
#   the result can't be pandigital. The nested loop starts at 123 because
#   it's the smallest 3-digit number with no digit repetitions and ends at
#   833, because 834*12 has 5 digits.
    for i in range(12, 99):
        for j in range(123, 834):
            p = i * j

            num_s = str(i) + str(j) + str(p)

            if len(num_s) > 9:
                break

            num = int(num_s)

            if is_pandigital(num, 9):
                products[n] = p
                n = n + 1

#   Sort the found products to easily see if there are duplicates.
    products = np.sort(products[:n])

    sum_ = products[0]

    for i in range(1, n):
        if products[i] != products[i-1]:
            sum_ = sum_ + products[i]

    end = default_timer()

    print('Project Euler, Problem 32')
    print(f'Answer: {sum_}')

    print(f'Elapsed time: {end - start:.9f} seconds')


if __name__ == '__main__':
    main()