Find Number of Ways to Reach the K-th Stair - Combinatorics [JS]

Description 

Solution: Combinatorics

We can use at most 1 more decrement operation than jumps.
Go through each amount of jumps until we reach a stair where we don't have enough decrement operations to go back to k.
For each amount of jumps, we reach a stair nextStair, and nextStair - k decrement operations to go back to k.

The decrement operations can be performed in any order as long as they are not consecutive.
There are a total of jumps + 1 positions where a decrement operation can be performed.
Use the n-choose-r formula to calculate the number of combinations of putting r decrement operations in jumps + 1 different positions.

e.g. k = 6, and we use 2 jumps
1 -> 2 -> 4 -> 8
decrement operations = 8 - 6 = 2
positions: 1 _ 2 _ 4 _ 8 _ (4 available positions for 2 decrement operations)
4 choose 2 = 6

Time Complexity: O(log2(k))
Space Complexity: O(1)

var waysToReachStair = function(k) {
  let jump = 0, stair = 1, ways = k <= 1 ? 1 : 0;
  while (true) {
    let nextStair = stair + 2 ** jump;
    jump++;
    if (nextStair >= k) {
      let decrements = nextStair - k;
      let positions = jump + 1;
      if (decrements > positions) break; // only going to grow larger from here, so we can never go back to k after this
      ways += nCr(positions, decrements);
    }
    stair = nextStair;
  }
  return ways;
};

function nCr(n, r) {
  return factorial(n) / (factorial(n - r) * factorial(r));
}

function factorial(n) {
  let ans = 1;
  while (n > 1) {
    ans *= n;
    n--;
  }
  return ans;
}