JavaScript Challenge - Stairmaster

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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; 
 
    let c = theseAddToSum(steps, remaining);
    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);
    }
    return total;
  }
}

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