From a6acbe20cc049ddb10cca60c3714408bbb11f0c1 Mon Sep 17 00:00:00 2001
From: gikf <60067306+gikf@users.noreply.github.com>
Date: Tue, 11 May 2021 13:33:29 +0200
Subject: [PATCH] fix(curriculum): rework Project Euler 71 (#42006)

---
 .../problem-71-ordered-fractions.md           | 65 +++++++++++++++----
 1 file changed, 54 insertions(+), 11 deletions(-)

diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
index 3e0d91f5cf..55a2e9b72f 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-71-ordered-fractions.md
@@ -8,28 +8,52 @@ dashedName: problem-71-ordered-fractions
 
 # --description--
 
-Consider the fraction, `n`/`d`, where `n` and `d` are positive integers. If `n`&lt;`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
+Consider the fraction, $\frac{n}{d}$, where `n` and `d` are positive integers. If `n` &lt; `d` and highest common factor, ${{HCF}(n, d)} = 1$, 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:
 
-<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, <strong>2/5</strong>, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
+$$\frac{1}{8}, \frac{1}{7}, \frac{1}{6}, \frac{1}{5}, \frac{1}{4}, \frac{2}{7}, \frac{1}{3}, \frac{3}{8}, \frac{\textbf2}{\textbf5}, \frac{3}{7}, \frac{1}{2}, \frac{4}{7}, \frac{3}{5}, \frac{5}{8}, \frac{2}{3}, \frac{5}{7}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}$$
 
-It can be seen that 2/5 is the fraction immediately to the left of 3/7.
+It can be seen that $\frac{2}{5}$ is the fraction immediately to the left of $\frac{3}{7}$.
 
-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.
+By listing the set of reduced proper fractions for `d` ≤ `limit` in ascending order of size, find the numerator of the fraction immediately to the left of $\frac{3}{7}$.
 
 # --hints--
 
-`orderedFractions()` should return a number.
+`orderedFractions(8)` should return a number.
 
 ```js
-assert(typeof orderedFractions() === 'number');
+assert(typeof orderedFractions(8) === 'number');
 ```
 
-`orderedFractions()` should return 428570.
+`orderedFractions(8)` should return `2`.
 
 ```js
-assert.strictEqual(orderedFractions(), 428570);
+assert.strictEqual(orderedFractions(8), 2);
+```
+
+`orderedFractions(10)` should return `2`.
+
+```js
+assert.strictEqual(orderedFractions(10), 2);
+```
+
+`orderedFractions(9994)` should return `4283`.
+
+```js
+assert.strictEqual(orderedFractions(9994), 4283);
+```
+
+`orderedFractions(500000)` should return `214283`.
+
+```js
+assert.strictEqual(orderedFractions(500000), 214283);
+```
+
+`orderedFractions(1000000)` should return `428570`.
+
+```js
+assert.strictEqual(orderedFractions(1000000), 428570);
 ```
 
 # --seed--
@@ -37,16 +61,35 @@ assert.strictEqual(orderedFractions(), 428570);
 ## --seed-contents--
 
 ```js
-function orderedFractions() {
+function orderedFractions(limit) {
 
   return true;
 }
 
-orderedFractions();
+orderedFractions(8);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function orderedFractions(limit) {
+  const fractions = [];
+  const fractionValues = {};
+  const highBoundary = 3 / 7;
+  let lowBoundary = 2 / 7;
+
+  for (let denominator = limit; denominator > 2; denominator--) {
+    let numerator = Math.floor((3 * denominator - 1) / 7);
+    let value = numerator / denominator;
+    if (value > highBoundary || value < lowBoundary) {
+      continue;
+    }
+    fractionValues[value] = [numerator, denominator];
+    fractions.push(value);
+    lowBoundary = value;
+  }
+
+  fractions.sort();
+  return fractionValues[fractions[fractions.length - 1]][0];
+}
 ```