From 9a32f0b46cbbfca72e72537c1ba96318819ddb8e Mon Sep 17 00:00:00 2001
From: Kushagra Bansal <kushagrabansalajmer@gmail.com>
Date: Tue, 18 Aug 2020 16:19:02 +0530
Subject: [PATCH] Created problem_39 in project_euler (#2330)

* Create __init__.py

* Add files via upload

* Update sol1.py

* Update sol1.py

* Update project_euler/problem_39/sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>

* Update sol1.py

Co-authored-by: Christian Clauss <cclauss@me.com>
---
 project_euler/problem_39/__init__.py |  1 +
 project_euler/problem_39/sol1.py     | 39 ++++++++++++++++++++++++++++
 2 files changed, 40 insertions(+)
 create mode 100644 project_euler/problem_39/__init__.py
 create mode 100644 project_euler/problem_39/sol1.py

diff --git a/project_euler/problem_39/__init__.py b/project_euler/problem_39/__init__.py
new file mode 100644
index 000000000..792d60054
--- /dev/null
+++ b/project_euler/problem_39/__init__.py
@@ -0,0 +1 @@
+#
diff --git a/project_euler/problem_39/sol1.py b/project_euler/problem_39/sol1.py
new file mode 100644
index 000000000..5c21d4bec
--- /dev/null
+++ b/project_euler/problem_39/sol1.py
@@ -0,0 +1,39 @@
+"""
+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}
+
+For which value of p ≤ 1000, is the number of solutions maximised?
+"""
+
+from typing import Dict
+from collections import Counter
+
+
+def pythagorean_triple(max_perimeter: int) -> Dict:
+    """
+    Returns a dictionary with keys as the perimeter of a right angled triangle
+    and value as the number of corresponding triplets.
+    >>> pythagorean_triple(15)
+    Counter({12: 1})
+    >>> pythagorean_triple(40)
+    Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1})
+    >>> pythagorean_triple(50)
+    Counter({12: 1, 30: 1, 24: 1, 40: 1, 36: 1, 48: 1})
+    """
+    triplets = 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)):
+                perimeter = int(base + perpendicular + hypotenuse)
+                if perimeter > max_perimeter:
+                    continue
+                else:
+                    triplets[perimeter] += 1
+    return triplets
+
+
+if __name__ == "__main__":
+    triplets = pythagorean_triple(1000)
+    print(f"{triplets.most_common()[0][0] = }")