6cfd0fc503
* fix: improve Project Euler descriptions and test case Improve formatting of Project Euler test descriptions. Also add poker hands array and new test case for problem 54 * feat: add typeof tests and gave functions proper names for first 100 challenges * fix: continue fixing test descriptions and adding "before test" sections * fix: address review comments * fix: adjust grids in 18 and 67 and fix some text that reference files rather than the given arrays * fix: implement bug fixes and improvements from review * fix: remove console.log statements from seed and solution
2.1 KiB
2.1 KiB
id, challengeType, title, forumTopicId
| id | challengeType | title | forumTopicId |
|---|---|---|---|
| 5900f3931000cf542c50fea6 | 5 | Problem 39: Integer right triangles | 302054 |
Description
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 ≤ n, is the number of solutions maximized?
Instructions
Tests
tests:
- text: <code>intRightTriangles(500)</code> should return a number.
testString: assert(typeof intRightTriangles(500) === 'number');
- text: <code>intRightTriangles(500)</code> should return 420.
testString: assert(intRightTriangles(500) == 420);
- text: <code>intRightTriangles(800)</code> should return 720.
testString: assert(intRightTriangles(800) == 720);
- text: <code>intRightTriangles(900)</code> should return 840.
testString: assert(intRightTriangles(900) == 840);
- text: <code>intRightTriangles(1000)</code> should return 840.
testString: assert(intRightTriangles(1000) == 840);
Challenge Seed
function intRightTriangles(n) {
// Good luck!
return n;
}
intRightTriangles(500);
Solution
// Original idea for this solution came from
// https://www.xarg.org/puzzle/project-euler/problem-39/
function intRightTriangles(n) {
// store the number of triangles with a given perimeter
let triangles = {};
// a is the shortest side
for (let a = 3; a < n / 3; a++)
// o is the opposite side and is at least as long as a
for (let o = a; o < n / 2; o++) {
let h = Math.sqrt(a * a + o * o); // hypotenuse
let p = a + o + h; // perimeter
if ((h % 1) === 0 && p <= n) {
triangles[p] = (triangles[p] || 0) + 1;
}
}
let max = 0, maxp = null;
for (let p in triangles) {
if (max < triangles[p]) {
max = triangles[p];
maxp = parseInt(p);
}
}
return maxp;
}