freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/rosetta-code/9-billion-names-of-god-the-integer.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5949b579404977fbaefcd736	9 billion names of God the integer	5	302219	9-billion-names-of-god-the-integer

--description--

This task is a variation of the short story by Arthur C. Clarke.

(Solvers should be aware of the consequences of completing this task.)

In detail, to specify what is meant by a "name":

• The integer 1 has 1 name "1".
• The integer 2 has 2 names "1+1" and "2".
• The integer 3 has 3 names "1+1+1", "2+1", and "3".
• The integer 4 has 5 names "1+1+1+1", "2+1+1", "2+2", "3+1", "4".
• The integer 5 has 7 names "1+1+1+1+1", "2+1+1+1", "2+2+1", "3+1+1", "3+2", "4+1", "5".

This can be visualized in the following form:

          1
        1   1
      1   1   1
    1   2   1   1
  1   2   2   1   1
1   3   3   2   1   1

Where row n corresponds to integer n, and each column C in row m from left to right corresponds to the number of names beginning with C.

Optionally note that the sum of the $n$-th row P(n) is the integer partition function.

--instructions--

Implement a function that returns the sum of the $n$-th row.

--hints--

numberOfNames should be function.

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

numberOfNames(5) should equal 7.

assert.equal(numberOfNames(5), 7);

numberOfNames(12) should equal 77.

assert.equal(numberOfNames(12), 77);

numberOfNames(18) should equal 385.

assert.equal(numberOfNames(18), 385);

numberOfNames(23) should equal 1255.

assert.equal(numberOfNames(23), 1255);

numberOfNames(42) should equal 53174.

assert.equal(numberOfNames(42), 53174);

numberOfNames(123) should equal 2552338241.

assert.equal(numberOfNames(123), 2552338241);

--seed--

--seed-contents--

function numberOfNames(num) {

  return true;
}

--solutions--

function numberOfNames(num) {
  const cache = [
    [1]
  ];
  for (let l = cache.length; l < num + 1; l++) {
    let Aa;
    let Mi;
    const r = [0];
    for (let x = 1; x < l + 1; x++) {
      r.push(r[r.length - 1] + (Aa = cache[l - x < 0 ? cache.length - (l - x) : l - x])[(Mi = Math.min(x, l - x)) < 0 ? Aa.length - Mi : Mi]);
    }
    cache.push(r);
  }
  return cache[num][cache[num].length - 1];
}