Skip to content

Commit

Permalink
Add problem 32 solution (TheAlgorithms#1257)
Browse files Browse the repository at this point in the history
  • Loading branch information
Hyftar authored and cclauss committed Oct 3, 2019
1 parent d28fc71 commit 6e69208
Showing 1 changed file with 62 additions and 0 deletions.
62 changes: 62 additions & 0 deletions project_euler/problem_32/solution.py
Original file line number Diff line number Diff line change
@@ -0,0 +1,62 @@
"""
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.
"""
import itertools


def isCombinationValid(combination):
"""
Checks if a combination (a tuple of 9 digits)
is a valid product equation.
>>> isCombinationValid(('3', '9', '1', '8', '6', '7', '2', '5', '4'))
True
>>> isCombinationValid(('1', '2', '3', '4', '5', '6', '7', '8', '9'))
False
"""
return (
int(''.join(combination[0:2])) *
int(''.join(combination[2:5])) ==
int(''.join(combination[5:9]))
) or (
int(''.join(combination[0])) *
int(''.join(combination[1:5])) ==
int(''.join(combination[5:9]))
)


def solution():
"""
Finds the sum of all products whose multiplicand/multiplier/product identity
can be written as a 1 through 9 pandigital
>>> solution()
45228
"""

return sum(
set(
[
int(''.join(pandigital[5:9]))
for pandigital
in itertools.permutations('123456789')
if isCombinationValid(pandigital)
]
)
)

if __name__ == "__main__":
print(solution())

0 comments on commit 6e69208

Please sign in to comment.