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Add problem 32 solution (TheAlgorithms#1257)
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""" | ||
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 | ||
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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])) | ||
) | ||
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def solution(): | ||
""" | ||
Finds the sum of all products whose multiplicand/multiplier/product identity | ||
can be written as a 1 through 9 pandigital | ||
>>> solution() | ||
45228 | ||
""" | ||
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||
return sum( | ||
set( | ||
[ | ||
int(''.join(pandigital[5:9])) | ||
for pandigital | ||
in itertools.permutations('123456789') | ||
if isCombinationValid(pandigital) | ||
] | ||
) | ||
) | ||
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if __name__ == "__main__": | ||
print(solution()) |