Python/project_euler/problem_051/sol1.py
Joyce bcfca67faa
[mypy] fix type annotations for all Project Euler problems (#4747)
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3

* [mypy] fix type annotations for project euler problem007/sol2

* [mypy] fix type annotations for project euler problem008/sol2

* [mypy] fix type annotations for project euler problem009/sol1

* [mypy] fix type annotations for project euler problem014/sol1

* [mypy] fix type annotations for project euler problem 025/sol2

* [mypy] fix type annotations for project euler problem026/sol1.py

* [mypy] fix type annotations for project euler problem037/sol1

* [mypy] fix type annotations for project euler problem044/sol1

* [mypy] fix type annotations for project euler problem046/sol1

* [mypy] fix type annotations for project euler problem051/sol1

* [mypy] fix type annotations for project euler problem074/sol2

* [mypy] fix type annotations for project euler problem080/sol1

* [mypy] fix type annotations for project euler problem099/sol1

* [mypy] fix type annotations for project euler problem101/sol1

* [mypy] fix type annotations for project euler problem188/sol1

* [mypy] fix type annotations for project euler problem191/sol1

* [mypy] fix type annotations for project euler problem207/sol1

* [mypy] fix type annotations for project euler problem551/sol1
2021-10-12 00:33:44 +08:00

114 lines
3.0 KiB
Python

"""
https://projecteuler.net/problem=51
Prime digit replacements
Problem 51
By replacing the 1st digit of the 2-digit number *3, it turns out that six of
the nine possible values: 13, 23, 43, 53, 73, and 83, are all prime.
By replacing the 3rd and 4th digits of 56**3 with the same digit, this 5-digit
number is the first example having seven primes among the ten generated numbers,
yielding the family: 56003, 56113, 56333, 56443, 56663, 56773, and 56993.
Consequently 56003, being the first member of this family, is the smallest prime
with this property.
Find the smallest prime which, by replacing part of the number (not necessarily
adjacent digits) with the same digit, is part of an eight prime value family.
"""
from __future__ import annotations
from collections import Counter
def prime_sieve(n: int) -> list[int]:
"""
Sieve of Erotosthenes
Function to return all the prime numbers up to a certain number
https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
>>> prime_sieve(3)
[2]
>>> prime_sieve(50)
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]
"""
is_prime = [True] * n
is_prime[0] = False
is_prime[1] = False
is_prime[2] = True
for i in range(3, int(n ** 0.5 + 1), 2):
index = i * 2
while index < n:
is_prime[index] = False
index = index + i
primes = [2]
for i in range(3, n, 2):
if is_prime[i]:
primes.append(i)
return primes
def digit_replacements(number: int) -> list[list[int]]:
"""
Returns all the possible families of digit replacements in a number which
contains at least one repeating digit
>>> digit_replacements(544)
[[500, 511, 522, 533, 544, 555, 566, 577, 588, 599]]
>>> digit_replacements(3112)
[[3002, 3112, 3222, 3332, 3442, 3552, 3662, 3772, 3882, 3992]]
"""
number_str = str(number)
replacements = []
digits = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
for duplicate in Counter(number_str) - Counter(set(number_str)):
family = [int(number_str.replace(duplicate, digit)) for digit in digits]
replacements.append(family)
return replacements
def solution(family_length: int = 8) -> int:
"""
Returns the solution of the problem
>>> solution(2)
229399
>>> solution(3)
221311
"""
numbers_checked = set()
# Filter primes with less than 3 replaceable digits
primes = {
x for x in set(prime_sieve(1_000_000)) if len(str(x)) - len(set(str(x))) >= 3
}
for prime in primes:
if prime in numbers_checked:
continue
replacements = digit_replacements(prime)
for family in replacements:
numbers_checked.update(family)
primes_in_family = primes.intersection(family)
if len(primes_in_family) != family_length:
continue
return min(primes_in_family)
return -1
if __name__ == "__main__":
print(solution())