# JavaScript Challenge - Stairmaster

This community-built FAQ covers the “Stairmaster” code challenge in JavaScript. You can find that challenge here, or pick any challenge you like from our list.

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I took this challenge to the next level by listing all the ways in an array instead of just the number of ways. When we have all the permutations, we can just get the number of ways by returning the length of the array. Furthermore, to pass this test, we have to define the case when `n === 0`.

I wrote a function called` theseAddToSum` that takes in an array of `steps`, such as `[1, 2, 3]`, and a `sum`, here is `4`— using recursion. I sketched it out on paper.

And here is my `theseAddToSum` function
``````function theseAddToSum(steps = [], sum) {

if (sum < 0) return []; // No solution
if (sum === 0) return [[]]; // A solution

let results = [];

if (steps.length < 1) return 'error';

for (let i = 0; i < steps.length; i++) {
let cur = steps[i];
let remaining = sum - cur;

for (let solution of c) {
solution.push(cur);
results.push(solution);
}
}

return results;
};
``````

This is the solution that helped me pass this challenge:

function theseAddToSum (steps = [], sum) { // Base cases if (sum < 0) return []; // No Solution if (sum === 0) return [[]]; // A Solution let results = []; if (steps.length < 1) return 'error'; for (let i = 0; i < steps.length; i++) { let cur = steps[i]; let remaining = sum - cur; let c = theseAddToSum(steps, remaining); for (let solution of c) { solution.push(cur); results.push(solution); } } return results; } function stairmaster(n) { // Base cases if (n === 0) return 0; let ways = theseAddToSum([1, 2, 3], n); console.log(ways); return ways.length; } console.log(stairmaster(4)); // Leave this so we can test your code: module.exports = stairmaster;
3 Likes
``````function stairmaster(n) {
let count = 0;

function step(stairs) {
for (let i = 1; i <= 3 && i <= n; i++) {
if (stairs + i > n) {
break;
} else if (stairs + i === n) {
count++;
break;
} else {
step(stairs+i);
}
}
}
step(0);
return count;
}
``````

I did the same thing as the other posts, I used recursion.

recursive version
``````function stairmaster(n) {
const maxSteps = 3;
if (n <= 0) { return 0; }
else if (n <= maxSteps) { // base case
return Math.pow(2, n - 1);
}
else { // recursive case
let total = 0;
for (let j = 1; j <= maxSteps; j++) {
total += stairmaster(n - j);
}
}
}
``````

But I guess storing the values already computed would be useful if the number of stairs is big.

version using memoziation
``````const mastered = [0, 1, 2, 4];

function stairmaster(n) {
const maxSteps = 3;
if (n <= 0) { return 0; }
else if (n < mastered.length) { // memoization used
return mastered[n];
}
else { // recursive case
let total = 0;
for (let j = 1; j <= maxSteps; j++) {
total += stairmaster(n - j);
}
mastered[n] = total;
}
return mastered[n];
}
``````

Very fast calculation with combinations, pemutations, factorial and caching.

``````function cachedFactorial() {
let cache = new Map();
return function fact(n) {
if (cache.has(n)) return cache.get(n);
let res = n ? n * fact(n - 1) : 1;
cache.set(n, res);
return res;
};
}
const factorial = cachedFactorial();

function stairmaster(n) {
if (n === 0) return 0;
let result = 0;
for (let i = 0; i <= n; i++) {
for (let j = 0; j <= Math.floor(n / 2); j++) {
for (let k = 0; k <= Math.floor(n / 3); k++) {
if (i + j * 2 + k * 3 === n) {
const permutations = factorial(i + j + k) / (factorial(i) * factorial(j) * factorial(k));
result += permutations;
}
}
}
}
return result;
}
``````
1 Like
``````function stairmaster(n) {
if (!(Number.isInteger(n) && n > 0))
return '[] The argument passed must be a positive integer greater than zero'

if (n === 1) return '[1]'

const stepsToUp = [1, 2, 3]

let array = []
let arr = []
let sum = 0

arr.push(x)
sum = arr.reduce((a, b) => a + b)

if (sum === n) {
array.push([...arr])
}

if (sum < n) {
let currentArr = [...arr]
arr = [...currentArr]
if (x > 1) {
while(x--) {
}
}
}
arr = []
}

stepsToUp.forEach(e => {
})

const copy = [...array]

copy.forEach(e => {
for (let i = 0; i < e.length; i++) {
for (let j = 0; j < e.length - 1; j++) {
let next = e[j+1]
if (next !== e[j]) {
e[j+1] = e[j]
e[j] = next
}
array.push([...e])
}
}
})

let set = {}
array.forEach(e => set[e] = e)
const values = Object.values(set).sort((a, b) => b.length - a.length)
let stringReturn = ''
values.forEach((e) => stringReturn += `[\${e}] `)

return stringReturn.trim()
// If you want it returned as an array of arrays, just return the "values" or return an Object.values(freq).
// If you just want how many combinations/permutations there are, just return "values.length" or Object.values(freq).length
}
``````
``````function stairmaster(n) {
function climb(num) {
if(num < 0) return 0;
if(num === 0) return 1;
return climb(num-1) + climb(num-2) + climb(num-3);
}
if(n===0) return 0;
return climb(n)
}

console.log(stairmaster(4));
``````