freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/abundant-deficient-and-perfect-number-classifications.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
594810f028c0303b75339acd	Abundant, deficient and perfect number classifications	5	302221	abundant-deficient-and-perfect-number-classifications

--description--

These define three classifications of positive integers based on their proper divisors.

Let P(n) be the sum of the proper divisors of n where proper divisors are all positive integers n other than n itself.

If P(n) < n then n is classed as deficient

If P(n) === n then n is classed as perfect

If P(n) > n then n is classed as abundant

Example: 6 has proper divisors of 1, 2, and 3. 1 + 2 + 3 = 6, so 6 is classed as a perfect number.

--instructions--

Implement a function that calculates how many of the integers from 1 to 20,000 (inclusive) are in each of the three classes. Output the result as an array in the following format [deficient, perfect, abundant].

--hints--

getDPA should be a function.

assert(typeof getDPA === 'function');

getDPA should return an array.

assert(Array.isArray(getDPA(100)));

getDPA return value should have a length of 3.

assert(getDPA(100).length === 3);

getDPA(20000) should equal [15043, 4, 4953]

assert.deepEqual(getDPA(20000), solution);

--seed--

--after-user-code--

const solution = [15043, 4, 4953];

--seed-contents--

function getDPA(num) {

}

--solutions--

function getDPA(num) {
  const dpa = [1, 0, 0];
  for (let n = 2; n <= num; n += 1) {
    let ds = 1;
    const e = Math.sqrt(n);
    for (let d = 2; d < e; d += 1) {
      if (n % d === 0) {
        ds += d + (n / d);
      }
    }
    if (n % e === 0) {
      ds += e;
    }
    dpa[ds < n ? 0 : ds === n ? 1 : 2] += 1;
  }
  return dpa;
}