improvements for project euler task 39

Author: fa1l

Diff for: project_euler/problem_39/sol1.py

@@ -1,4 +1,6 @@
 """
+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 @@
 
 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.
@@ -22,7 +25,7 @@ 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
@@ -35,6 +38,13 @@ def pythagorean_triple(max_perimeter: int) -> dict:
     return triplets
 
 
+def solution(n: int = 1000) -> int:
+    """Returns perimeter with maximum solutions."""
+    triplets = pythagorean_triple(n)
+    perimeter, _ = triplets.most_common()[0]
+    return perimeter
+
+
 if __name__ == "__main__":
-    triplets = pythagorean_triple(1000)
-    print(f"{triplets.most_common()[0][0] = }")
+    result = solution(1000)
+    print(f"Perimiter {result} has maximum solutions")