freeCodeCamp/curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-58-spiral-primes.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f3a61000cf542c50feb9	Problem 58: Spiral primes	5	302169	problem-58-spiral-primes

--description--

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

37 36 35 34 33 32 31
38 17 16 15 14 13 30
39 18  5  4  3 12 29
40 19  6  1  2 11 28
41 20  7  8  9 10 27
42 21 22 23 24 25 26
43 44 45 46 47 48 49

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ≈ 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the percent of primes along both diagonals first falls below percent?

--hints--

spiralPrimes(50) should return a number.

assert(typeof spiralPrimes(50) === 'number');

spiralPrimes(50) should return 11.

assert.strictEqual(spiralPrimes(50), 11);

spiralPrimes(15) should return 981.

assert.strictEqual(spiralPrimes(15), 981);

spiralPrimes(10) should return 26241.

assert.strictEqual(spiralPrimes(10), 26241);

--seed--

--seed-contents--

function spiralPrimes(percent) {

  return true;
}

spiralPrimes(50);

--solutions--

function spiralPrimes(percent) {
  function isPrime(n) {
    if (n <= 3) {
      return n > 1;
    } else if (n % 2 === 0 || n % 3 === 0) {
      return false;
    }

    for (let i = 5; i * i <= n; i += 6) {
      if (n % i === 0 || n % (i + 2) === 0) {
        return false;
      }
    }
    return true;
  }

  let totalCount = 1;
  let primesCount = 0;
  let curNumber = 1;
  let curSideLength = 1;
  let ratio = 1;
  const wantedRatio = percent / 100;

  while (ratio >= wantedRatio) {
    curSideLength += 2;
    for (let i = 0; i < 4; i++) {
      curNumber += curSideLength - 1;
      totalCount++;
      if (i !== 3 && isPrime(curNumber)) {
        primesCount++;
      }
    }
    ratio = primesCount / totalCount;
  }
  return curSideLength;
}