mirror of
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ebc2d5d79f
Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>
70 lines
2.1 KiB
Python
70 lines
2.1 KiB
Python
"""
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Project Euler Problem 79: https://projecteuler.net/problem=79
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Passcode derivation
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A common security method used for online banking is to ask the user for three
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random characters from a passcode. For example, if the passcode was 531278,
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they may ask for the 2nd, 3rd, and 5th characters; the expected reply would
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be: 317.
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The text file, keylog.txt, contains fifty successful login attempts.
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Given that the three characters are always asked for in order, analyse the file
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so as to determine the shortest possible secret passcode of unknown length.
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"""
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import itertools
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from pathlib import Path
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def find_secret_passcode(logins: list[str]) -> int:
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"""
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Returns the shortest possible secret passcode of unknown length.
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>>> find_secret_passcode(["135", "259", "235", "189", "690", "168", "120",
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... "136", "289", "589", "160", "165", "580", "369", "250", "280"])
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12365890
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>>> find_secret_passcode(["426", "281", "061", "819" "268", "406", "420",
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... "428", "209", "689", "019", "421", "469", "261", "681", "201"])
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4206819
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"""
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# Split each login by character e.g. '319' -> ('3', '1', '9')
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split_logins = [tuple(login) for login in logins]
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unique_chars = {char for login in split_logins for char in login}
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for permutation in itertools.permutations(unique_chars):
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satisfied = True
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for login in logins:
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if not (
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permutation.index(login[0])
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< permutation.index(login[1])
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< permutation.index(login[2])
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):
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satisfied = False
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break
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if satisfied:
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return int("".join(permutation))
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raise Exception("Unable to find the secret passcode")
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def solution(input_file: str = "keylog.txt") -> int:
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"""
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Returns the shortest possible secret passcode of unknown length
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for successful login attempts given by `input_file` text file.
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>>> solution("keylog_test.txt")
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6312980
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"""
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logins = Path(__file__).parent.joinpath(input_file).read_text().splitlines()
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return find_secret_passcode(logins)
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if __name__ == "__main__":
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print(f"{solution() = }")
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