freeCodeCamp/curriculum/challenges/english/08-coding-interview-prep/rosetta-code/abundant-deficient-and-perf...

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title id challengeType
Abundant, deficient and perfect number classifications 594810f028c0303b75339acd 5

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.

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].

Instructions

Tests

tests:
  - text: <code>getDPA</code> is a function.
    testString: assert(typeof getDPA === 'function', '<code>getDPA</code> is a function.');
  - text: <code>getDPA</code> should return an array.
    testString: assert(Array.isArray(getDPA(100)), '<code>getDPA</code> should return an array.');
  - text: <code>getDPA</code> return value should have a length of 3.
    testString: assert(getDPA(100).length === 3, '<code>getDPA</code> return value should have a length of 3.');
  - text: <code>getDPA(20000)</code> should equal [15043, 4, 4953]
    testString: assert.deepEqual(getDPA(20000), solution, '<code>getDPA(20000)</code> should equal [15043, 4, 4953]');

Challenge Seed

function getDPA (num) {
  // Good luck!
}

After Test

const solution = [15043, 4, 4953];

Solution

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;
}