--- title: 9 billion names of God the integer id: 5949b579404977fbaefcd736 challengeType: 5 forumTopicId: 302219 --- ## 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": 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.
## Tests
```yml tests: - text: numberOfNames should be function. testString: assert(typeof numberOfNames === 'function'); - text: numberOfNames(5) should equal 7. testString: assert.equal(numberOfNames(5), 7); - text: numberOfNames(12) should equal 77. testString: assert.equal(numberOfNames(12), 77); - text: numberOfNames(18) should equal 385. testString: assert.equal(numberOfNames(18), 385); - text: numberOfNames(23) should equal 1255. testString: assert.equal(numberOfNames(23), 1255); - text: numberOfNames(42) should equal 53174. testString: assert.equal(numberOfNames(42), 53174); - text: numberOfNames(123) should equal 2552338241. testString: assert.equal(numberOfNames(123), 2552338241); ```
## Challenge Seed
```js function numberOfNames(num) { return true; } ```
## Solution
```js 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]; } ```