From ef53bbdf5a6648f6035f48c9bcbf0cf7499fba22 Mon Sep 17 00:00:00 2001
From: Nelson Stoik <nelsonstoik@gmail.com>
Date: Wed, 7 Oct 2020 21:22:24 -0600
Subject: [PATCH] Style Improvements for project_euler/problem_26 (#2958)

* add typehints and docstrings

* add typehint and default value

* add typehint and default value. Removed unused variable.

* do not modifiy the given solution

* add doctests

* update sol1 after running black

* add typehint, docstring, and doctest

* update sol2 after running black

* add full problem statement and solution function with typehint and doctest

* renamed original function instead of adding new one

* don't alter original solution
---
 project_euler/problem_26/sol1.py | 26 ++++++++++++++++++++++----
 1 file changed, 22 insertions(+), 4 deletions(-)

diff --git a/project_euler/problem_26/sol1.py b/project_euler/problem_26/sol1.py
index cab8e0eb5..64e0bbfef 100644
--- a/project_euler/problem_26/sol1.py
+++ b/project_euler/problem_26/sol1.py
@@ -1,20 +1,38 @@
 """
 Euler Problem 26
 https://projecteuler.net/problem=26
+
+Problem Statement:
+
+A unit fraction contains 1 in the numerator. The decimal representation of the
+unit fractions with denominators 2 to 10 are given:
+
+1/2	= 	0.5
+1/3	= 	0.(3)
+1/4	= 	0.25
+1/5	= 	0.2
+1/6	= 	0.1(6)
+1/7	= 	0.(142857)
+1/8	= 	0.125
+1/9	= 	0.(1)
+1/10	= 	0.1
+Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be
+seen that 1/7 has a 6-digit recurring cycle.
+
 Find the value of d < 1000 for which 1/d contains the longest recurring cycle
 in its decimal fraction part.
 """
 
 
-def find_digit(numerator: int, digit: int) -> int:
+def solution(numerator: int = 1, digit: int = 1000) -> int:
     """
     Considering any range can be provided,
     because as per the problem, the digit d < 1000
-    >>> find_digit(1, 10)
+    >>> solution(1, 10)
     7
-    >>> find_digit(10, 100)
+    >>> solution(10, 100)
     97
-    >>> find_digit(10, 1000)
+    >>> solution(10, 1000)
     983
     """
     the_digit = 1