From ae4b708689655f555f0216303d34633a454f1435 Mon Sep 17 00:00:00 2001
From: Kalpana Jangra <kalpana.2k3@gmail.com>
Date: Mon, 5 Oct 2020 06:34:16 +0530
Subject: [PATCH] Added Project Euler Problem 71: Fixes #2695

---
 project_euler/problem_71/__init__.py |  0
 project_euler/problem_71/sol1.py     | 48 ++++++++++++++++++++++++++++
 2 files changed, 48 insertions(+)
 create mode 100644 project_euler/problem_71/__init__.py
 create mode 100644 project_euler/problem_71/sol1.py

diff --git a/project_euler/problem_71/__init__.py b/project_euler/problem_71/__init__.py
new file mode 100644
index 000000000000..e69de29bb2d1
diff --git a/project_euler/problem_71/sol1.py b/project_euler/problem_71/sol1.py
new file mode 100644
index 000000000000..415b127e5374
--- /dev/null
+++ b/project_euler/problem_71/sol1.py
@@ -0,0 +1,48 @@
+"""
+Ordered fractions
+Problem 71
+https://projecteuler.net/problem=71
+
+Consider the fraction n/d, where n and d are positive
+integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
+
+If we list the set of reduced proper fractions for d ≤ 8
+in ascending order of size, we get:
+    1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7,
+    1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
+
+It can be seen that 2/5 is the fraction immediately to the left of 3/7.
+
+By listing the set of reduced proper fractions for d ≤ 1,000,000
+in ascending order of size, find the numerator of the fraction
+immediately to the left of 3/7.
+"""
+
+
+def solution(numerator: int = 3, denominator: int = 7, limit: int = 1000000) -> int:
+    """
+    Returns the closest numerator of the fraction immediately to the
+    left of given fraction (numerator/denominator) from a list of reduced
+    proper fractions.
+    >>> solution()
+    428570
+    >>> solution(3, 7, 8)
+    2
+    >>> solution(6, 7, 60)
+    47
+    """
+    max_numerator = 0
+    max_denominator = 1
+
+    for current_denominator in range(1, limit + 1):
+        current_numerator = current_denominator * numerator // denominator
+        if current_denominator % denominator == 0:
+            current_numerator -= 1
+        if current_numerator * max_denominator > current_denominator * max_numerator:
+            max_numerator = current_numerator
+            max_denominator = current_denominator
+    return max_numerator
+
+
+if __name__ == "__main__":
+    print(solution(numerator=3, denominator=7, limit=1000000))