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Hacktoberfest: Add a solution for Project Euler 50 #2703

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2 changes: 2 additions & 0 deletions DIRECTORY.md
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* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_48/sol1.py)
* Problem 49
* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_49/sol1.py)
* Problem 50
* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_50/sol1.py)
* Problem 52
* [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_52/sol1.py)
* Problem 53
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102 changes: 102 additions & 0 deletions project_euler/problem_50/sol1.py
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"""
Consecutive prime sum

Problem 50

The prime 41, can be written as the sum of six consecutive primes:

41 = 2 + 3 + 5 + 7 + 11 + 13

This is the longest sum of consecutive primes that adds to
a prime below one-hundred.

The longest sum of consecutive primes below one-thousand that adds to a prime,
contains 21 terms, and is equal to 953.

Which prime, below one-million, can be written as the sum of
the most consecutive primes?

Solution:

First of all, we need to generate all prime numbers
from 2 to the closest prime number with 1000000.
Then, use sliding window to get the answer.
"""

from math import floor, sqrt


def is_prime(number: int) -> bool:
"""
function to check whether a number is a prime or not.
>>> is_prime(2)
True
>>> is_prime(6)
False
>>> is_prime(1)
False
>>> is_prime(-7000)
False
>>> is_prime(104729)
True
"""

if number < 2:
return False

for n in range(2, floor(sqrt(number)) + 1):
if number % n == 0:
return False

return True


def solution(constraint=10 ** 6):
"""
Return the problem solution.
>>> solution(1000)
953
"""

prime_list = [2] + [x for x in range(3, constraint, 2) if is_prime(x)]

cumulative_sum = []
tmp = 0
for prime in prime_list:
tmp += prime
if tmp < constraint:
cumulative_sum.append(tmp)
else:
break

upper_limit_idx = 0
for i in range(len(prime_list)):
if prime_list[i] < constraint:
upper_limit_idx = i
else:
break

max_count = -1
answer = 0
for number in reversed(cumulative_sum):
count_prime = cumulative_sum.index(number) + 1

if not is_prime(number):
tmp = number

for i in range(upper_limit_idx):
count_prime -= 1
tmp -= prime_list[i]

if is_prime(tmp) or tmp < 0:
break

if max_count < count_prime:
max_count = count_prime
answer = tmp

return answer


if __name__ == "__main__":
print(solution())