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Add new solution for the euler project problem 9 (TheAlgorithms#12771)
* Add new solution for the euler project problem 9 - precompute the squares.

* Update sol4.py

* updating DIRECTORY.md

* Update sol4.py

* Update sol4.py

* Update sol4.py

---------

Co-authored-by: Maxim Smolskiy <mithridatus@mail.ru>
Co-authored-by: MaximSmolskiy <MaximSmolskiy@users.noreply.github.com>
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3 people authored Aug 26, 2025
commit 55db5a1b8d463a514522dd759053d30357538b62
1 change: 1 addition & 0 deletions DIRECTORY.md
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* [Sol1](project_euler/problem_009/sol1.py)
* [Sol2](project_euler/problem_009/sol2.py)
* [Sol3](project_euler/problem_009/sol3.py)
* [Sol4](project_euler/problem_009/sol4.py)
* Problem 010
* [Sol1](project_euler/problem_010/sol1.py)
* [Sol2](project_euler/problem_010/sol2.py)
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60 changes: 60 additions & 0 deletions project_euler/problem_009/sol4.py
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"""
Project Euler Problem 9: https://projecteuler.net/problem=9
Special Pythagorean triplet
A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a^2 + b^2 = c^2.
For example, 3^2 + 4^2 = 9 + 16 = 25 = 5^2.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.
References:
- https://en.wikipedia.org/wiki/Pythagorean_triple
"""


def get_squares(n: int) -> list[int]:
"""
>>> get_squares(0)
[]
>>> get_squares(1)
[0]
>>> get_squares(2)
[0, 1]
>>> get_squares(3)
[0, 1, 4]
>>> get_squares(4)
[0, 1, 4, 9]
"""
return [number * number for number in range(n)]


def solution(n: int = 1000) -> int:
"""
Precomputing squares and checking if a^2 + b^2 is the square by set look-up.
>>> solution(12)
60
>>> solution(36)
1620
"""

squares = get_squares(n)
squares_set = set(squares)
for a in range(1, n // 3):
for b in range(a + 1, (n - a) // 2 + 1):
if (
squares[a] + squares[b] in squares_set
and squares[n - a - b] == squares[a] + squares[b]
):
return a * b * (n - a - b)

return -1


if __name__ == "__main__":
print(f"{solution() = }")