Solve project Euler problem #38 http://projecteuler.net/problem=38

euler_38.py

# Analysis:
# 
# Let's try to find the optimal range. We know that
# 1. 918273645 is a 1-9 pandigital & is a concatenated product, which means that this number is either the largest or
#    below it, so we search from this number up
# 2. Then we find that the number we're trying to search for should start with 9, e.g. 9, 91, 945, etc
# 3. len(concat_product(x=90, n=3)) = 8, len(concat_product(x=90, n=4)) = 11. This range won't work.
# 4. len(concat_product(x=9000, n=2)) = 9. This is where we should start.
# 5. Taking #1 in consideration, we will start from 9182 till 9999
# 6. To get the pandigital number from a this range, we can multiply by n * 100000 + n * 2 => n * 100002
# 7. We get the largest x fitting our definition by walking backwards

from time import time
before = time()
concatenated_products = (x * 100002 for x in xrange(9999, 9181, -1))
digits = set('123456789')
pandigitals = (p for p in concatenated_products if set(str(p)) == digits)
print next(pandigitals)
print time() - before