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Improve solution (~30 times - from 900+ seconds to ~30 seconds)
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
97 lines
2.6 KiB
Python
97 lines
2.6 KiB
Python
"""
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Project Euler problem 145: https://projecteuler.net/problem=145
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Author: Vineet Rao, Maxim Smolskiy
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Problem statement:
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Some positive integers n have the property that the sum [ n + reverse(n) ]
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consists entirely of odd (decimal) digits.
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For instance, 36 + 63 = 99 and 409 + 904 = 1313.
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We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
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Leading zeroes are not allowed in either n or reverse(n).
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There are 120 reversible numbers below one-thousand.
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How many reversible numbers are there below one-billion (10^9)?
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"""
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EVEN_DIGITS = [0, 2, 4, 6, 8]
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ODD_DIGITS = [1, 3, 5, 7, 9]
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def reversible_numbers(
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remaining_length: int, remainder: int, digits: list[int], length: int
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) -> int:
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"""
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Count the number of reversible numbers of given length.
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Iterate over possible digits considering parity of current sum remainder.
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>>> reversible_numbers(1, 0, [0], 1)
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0
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>>> reversible_numbers(2, 0, [0] * 2, 2)
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20
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>>> reversible_numbers(3, 0, [0] * 3, 3)
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100
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"""
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if remaining_length == 0:
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if digits[0] == 0 or digits[-1] == 0:
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return 0
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for i in range(length // 2 - 1, -1, -1):
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remainder += digits[i] + digits[length - i - 1]
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if remainder % 2 == 0:
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return 0
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remainder //= 10
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return 1
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if remaining_length == 1:
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if remainder % 2 == 0:
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return 0
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result = 0
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for digit in range(10):
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digits[length // 2] = digit
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result += reversible_numbers(
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0, (remainder + 2 * digit) // 10, digits, length
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)
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return result
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result = 0
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for digit1 in range(10):
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digits[(length + remaining_length) // 2 - 1] = digit1
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if (remainder + digit1) % 2 == 0:
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other_parity_digits = ODD_DIGITS
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else:
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other_parity_digits = EVEN_DIGITS
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for digit2 in other_parity_digits:
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digits[(length - remaining_length) // 2] = digit2
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result += reversible_numbers(
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remaining_length - 2,
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(remainder + digit1 + digit2) // 10,
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digits,
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length,
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)
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return result
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def solution(max_power: int = 9) -> int:
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"""
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To evaluate the solution, use solution()
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>>> solution(3)
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120
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>>> solution(6)
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18720
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>>> solution(7)
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68720
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"""
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result = 0
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for length in range(1, max_power + 1):
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result += reversible_numbers(length, 0, [0] * length, length)
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return result
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if __name__ == "__main__":
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print(f"{solution() = }")
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