Python/project_euler/problem_04/sol1.py
Bruno Simas Hadlich 267b5eff40 Added doctest and more explanation about Dijkstra execution. (#1014)
* Added doctest and more explanation about Dijkstra execution.

* tests were not passing with python2 due to missing __init__.py file at number_theory folder

* Removed the dot at the beginning of the imported modules names because 'python3 -m doctest -v data_structures/hashing/*.py' and 'python3 -m doctest -v data_structures/stacks/*.py' were failing not finding hash_table.py and stack.py modules.

* Moved global code to main scope and added doctest for project euler problems 1 to 14.

* Added test case for negative input.

* Changed N variable to do not use end of line scape because in case there is a space after it the script will break making it much more error prone.

* Added problems description and doctests to the ones that were missing. Limited line length to 79 and executed python black over all scripts.

* Changed the way files are loaded to support pytest call.

* Added __init__.py to problems to make them modules and allow pytest execution.

* Added project_euler folder to test units execution

* Changed 'os.path.split(os.path.realpath(__file__))' to 'os.path.dirname()'
2019-07-17 01:09:53 +02:00

51 lines
1.3 KiB
Python

"""
Problem:
A palindromic number reads the same both ways. The largest palindrome made from
the product of two 2-digit numbers is 9009 = 91 x 99.
Find the largest palindrome made from the product of two 3-digit numbers which
is less than N.
"""
from __future__ import print_function
try:
raw_input # Python 2
except NameError:
raw_input = input # Python 3
def solution(n):
"""Returns the largest palindrome made from the product of two 3-digit
numbers which is less than n.
>>> solution(20000)
19591
>>> solution(30000)
29992
>>> solution(40000)
39893
"""
# fetchs the next number
for number in range(n - 1, 10000, -1):
# converts number into string.
strNumber = str(number)
# checks whether 'strNumber' is a palindrome.
if strNumber == strNumber[::-1]:
divisor = 999
# if 'number' is a product of two 3-digit numbers
# then number is the answer otherwise fetch next number.
while divisor != 99:
if (number % divisor == 0) and (
len(str(int(number / divisor))) == 3
):
return number
divisor -= 1
if __name__ == "__main__":
print(solution(int(raw_input().strip())))