From f10605a4c97982721e9c0af124b7090d415b10f0 Mon Sep 17 00:00:00 2001
From: Vishal Pankaj Chandratreya <19171016+tfpf@users.noreply.github.com>
Date: Mon, 25 Mar 2024 16:01:19 +0530
Subject: [PATCH] Optimised solution. Runs in 100 ms.

---
 src/solutions/counting_fractions.rs | 24 ++++++++++++++++++++++--
 1 file changed, 22 insertions(+), 2 deletions(-)

diff --git a/src/solutions/counting_fractions.rs b/src/solutions/counting_fractions.rs
index baa8a90..c29f2cd 100644
--- a/src/solutions/counting_fractions.rs
+++ b/src/solutions/counting_fractions.rs
@@ -1,4 +1,24 @@
 pub fn solve() -> i64 {
-    assert_eq!(303963552391i64, 303963552391);
-    303963552391
+    // Calculate Euler's totient of every number up to the limit.
+    let mut totients = (0..=1000000).collect::<Vec<i64>>();
+    for num in 2..totients.len() {
+        if totients[num] == num as i64 {
+            // This number is prime. The totient of each of its multiples will
+            // contain the term
+            //     1/(1 - num)
+            // in its product expression. Instead of multiply-assigning that
+            // term, we use an equivalent subtract-assign.
+            for multiple in (num..totients.len()).step_by(num) {
+                totients[multiple] -= totients[multiple] / num as i64;
+            }
+        }
+    }
+
+    // The number of reduced fractions with a particular denominator is the
+    // totient of the denominator. Hence, the total number of fractions is the
+    // sum.
+    let result = totients[2..].iter().sum();
+
+    assert_eq!(result, 303963552391);
+    result
 }