# 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.

# So, we only need inputs where the summands are of size 5 and the output of size 4

from itertools import combinations
from itertools import permutations

def splits(input, left, right):
    assert len(input) == left + right

    result = []

    combis = combinations(input, left)
    for c in combis:
        r = list(set(input) - set(c))
        result.append((list(c), r))

    return result

def check(series):
    result = set()

    for i in range(4):
        # print("Testing", toInt(series[0:i+1]) , toInt(series[i+1:5]) , toInt(series[5:]))
        if toInt(series[0:i+1]) * toInt(series[i+1:5]) == toInt(series[5:]):
            result.add(toInt(series[5:]))

    return result

def toInt(l):
    return int("".join(list(map(str,l))))

def main():
    print("Hello, this is Patrick")

    digits = range(1, 10)

    perms = permutations(digits)

    products = set()
    for perm in perms:
        products.update(check(perm))
    print(products)

    print(sum(products))



if __name__ == "__main__":
    main()