Fix coding style for Project Euler problem 39 (#3023)

* improvements for project euler task 39

* add tests for solution()

* fixed a typo

* Update sol1.py

Co-authored-by: Dhruv <dhruvmanila@gmail.com>

project_euler/problem_39/sol1.py

@@ -1,4 +1,6 @@
 """
+Problem 39: https://projecteuler.net/problem=39
+
 If p is the perimeter of a right angle triangle with integral length sides,
 {a,b,c}, there are exactly three solutions for p = 120.
 {20,48,52}, {24,45,51}, {30,40,50}
@@ -8,10 +10,11 @@ For which value of p ≤ 1000, is the number of solutions maximised?
 
 from __future__ import annotations
 
+import typing
 from collections import Counter
 
 
-def pythagorean_triple(max_perimeter: int) -> dict:
+def pythagorean_triple(max_perimeter: int) -> typing.Counter[int]:
     """
     Returns a dictionary with keys as the perimeter of a right angled triangle
     and value as the number of corresponding triplets.
@@ -22,19 +25,31 @@ def pythagorean_triple(max_perimeter: int) -> dict:
     >>> pythagorean_triple(50)
     Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1, 48: 1})
     """
-    triplets = Counter()
+    triplets: typing.Counter[int] = Counter()
     for base in range(1, max_perimeter + 1):
         for perpendicular in range(base, max_perimeter + 1):
             hypotenuse = (base * base + perpendicular * perpendicular) ** 0.5
-            if hypotenuse == int((hypotenuse)):
+            if hypotenuse == int(hypotenuse):
                 perimeter = int(base + perpendicular + hypotenuse)
                 if perimeter > max_perimeter:
                     continue
-                else:
                 triplets[perimeter] += 1
     return triplets
 
 
+def solution(n: int = 1000) -> int:
+    """
+    Returns perimeter with maximum solutions.
+    >>> solution(100)
+    90
+    >>> solution(200)
+    180
+    >>> solution(1000)
+    840
+    """
+    triplets = pythagorean_triple(n)
+    return triplets.most_common(1)[0][0]
+
+
 if __name__ == "__main__":
-    triplets = pythagorean_triple(1000)
-    print(f"{triplets.most_common()[0][0] = }")
+    print(f"Perimeter {solution()} has maximum solutions")