From 4260416ac3f39e7bb15eade6a764786ec31c6f55 Mon Sep 17 00:00:00 2001
From: gikf <60067306+gikf@users.noreply.github.com>
Date: Tue, 11 May 2021 14:56:58 +0200
Subject: [PATCH] fix(curriculum): rework Project Euler 75 (#42066)

* fix: rework challenge to use argument in function

* fix: add solution

* fix: position block evenly between paragraphs
---
 ...lem-75-singular-integer-right-triangles.md | 76 ++++++++++++++++---
 1 file changed, 66 insertions(+), 10 deletions(-)

diff --git a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
index 4a71f87c72..a20c0a95ea 100644
--- a/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
+++ b/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-75-singular-integer-right-triangles.md
@@ -16,29 +16,47 @@ It turns out that 12 cm is the smallest length of wire that can be bent to form
   <strong>30 cm:</strong> (5,12,13)<br>
   <strong>36 cm:</strong> (9,12,15)<br>
   <strong>40 cm:</strong> (8,15,17)<br>
-  <strong>48 cm:</strong> (12,16,20)<br>
+  <strong>48 cm:</strong> (12,16,20)<br><br>
 </div>
 
 In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.
 
 <div style='margin-left: 4em;'>
-  <strong>120 cm:</strong> (30,40,50), (20,48,52), (24,45,51)
+  <strong>120 cm:</strong> (30,40,50), (20,48,52), (24,45,51)<br><br>
 </div>
 
-Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
+Given that L is the length of the wire, for how many values of L ≤ `n` can exactly one, integer sided right angle, triangle be formed?
 
 # --hints--
 
-`singularIntRightTriangles()` should return a number.
+`singularIntRightTriangles(48)` should return a number.
 
 ```js
-assert(typeof singularIntRightTriangles() === 'number');
+assert(typeof singularIntRightTriangles(48) === 'number');
 ```
 
-`singularIntRightTriangles()` should return 161667.
+`singularIntRightTriangles(48)` should return `6`.
 
 ```js
-assert.strictEqual(singularIntRightTriangles(), 161667);
+assert.strictEqual(singularIntRightTriangles(48), 6);
+```
+
+`singularIntRightTriangles(700000)` should return `75783`.
+
+```js
+assert.strictEqual(singularIntRightTriangles(700000), 75783);
+```
+
+`singularIntRightTriangles(1000000)` should return `107876`.
+
+```js
+assert.strictEqual(singularIntRightTriangles(1000000), 107876);
+```
+
+`singularIntRightTriangles(1500000)` should return `161667`.
+
+```js
+assert.strictEqual(singularIntRightTriangles(1500000), 161667);
 ```
 
 # --seed--
@@ -46,16 +64,54 @@ assert.strictEqual(singularIntRightTriangles(), 161667);
 ## --seed-contents--
 
 ```js
-function singularIntRightTriangles() {
+function singularIntRightTriangles(n) {
 
   return true;
 }
 
-singularIntRightTriangles();
+singularIntRightTriangles(48);
 ```
 
 # --solutions--
 
 ```js
-// solution required
+function singularIntRightTriangles(limit) {
+  function euclidFormula(m, n) {
+    return [m ** 2 - n ** 2, 2 * m * n, m ** 2 + n ** 2];
+  }
+
+  function gcd(numberA, numberB) {
+    if (numberB === 0) {
+      return numberA;
+    }
+    return gcd(numberB, numberA % numberB);
+  }
+
+  function notBothOdd(numberA, numberB) {
+    return (numberA + numberB) % 2 === 1;
+  }
+
+  function areCoprime(numberA, numberB) {
+    return gcd(numberA, numberB) === 1;
+  }
+
+  const trianglesWithPerimeter = new Array(limit + 1).fill(0);
+  const mLimit = Math.sqrt(limit / 2);
+
+  for (let m = 2; m < mLimit; m++) {
+    for (let n = 1; n < m; n++) {
+      if (notBothOdd(m, n) && areCoprime(m, n)) {
+        const [sideA, sideB, sideC] = euclidFormula(m, n);
+        const perimeter = sideA + sideB + sideC;
+        let curPerimeter = perimeter;
+        while (curPerimeter <= limit) {
+          trianglesWithPerimeter[curPerimeter]++;
+          curPerimeter += perimeter;
+        }
+      }
+    }
+  }
+  return trianglesWithPerimeter.filter(trianglesCount => trianglesCount === 1)
+    .length;
+}
 ```