From d35117e4170b75f5ab8dc2ab3e8bcf5c4ac25800 Mon Sep 17 00:00:00 2001
From: daviddoji <daviddoji@pm.me>
Date: Thu, 9 Sep 2021 15:58:18 +0200
Subject: [PATCH] Solution to problem 39

---
 src/Python/Problem039.py | 44 ++++++++++++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 src/Python/Problem039.py

diff --git a/src/Python/Problem039.py b/src/Python/Problem039.py
new file mode 100644
index 0000000..f2d979d
--- /dev/null
+++ b/src/Python/Problem039.py
@@ -0,0 +1,44 @@
+#!/usr/bin/env python3
+"""
+Created on 05 Jun 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 39 of Project Euler
+https://projecteuler.net/problem=39
+"""
+
+from utils import timeit
+
+
+@timeit("Problem 39")
+def compute():
+    """
+    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?
+    """
+    ans, val = 0, 0
+    for p in range(2, 1001, 2):
+        sol = 0
+        for a in range(1, p):
+            for b in range(a+1, p-2*a):     
+                c = p - (a + b)
+                if a**2 + b**2 == c**2:
+                    sol += 1 
+                elif a**2 + b**2 > c**2:
+                    # As we continue our innermost loop, the left side 
+                    # gets bigger, right gets smaller, so we're done here
+                    break
+        if sol > ans:
+            ans, val = sol, p
+
+    return val
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 39: {compute()}")