freeCodeCamp/problem-39-integer-right-triangles.english.md at 3fd336d3e9f2662605b4ef1dba5aa16e3fb0781d

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Oliver Eyton-Williams a7075a579c Fix: Problem 39: Integer right triangles (#38145 )

* fix: correct test and add solution

I also changed the seed to report the results of an easier example to
the user, since just evaluating the function mostly wastes time.

* fix: use a better solution

* fix: credit original author

2020-02-08 21:09:32 +05:30

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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 maximised?

Instructions

Tests

tests:
  - 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;
}

console.log(intRightTriangles(500)); // 420

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 = p;
    }
  }
  return maxp;
}

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