Python Challenge - Stairmaster

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This was a tricky one!

Essentially, you want to add the number of ways for stairmaster( n -1) , stairmaster(n-2), and stairmaster(n-3), to get the value of stairmaster(n). Since it’s really easy to calculate stairmaster() for n = 1, 2, or 3, we can just include that at the beginning of the function to give us a starting point for bigger numbers.

Then we just run a for loop for 4 to n and on each iteration:

  1. We set d = sum of the three previous values of a, b, and c
  2. We update the values of a, b, c so that they are the values needed for the next value of i.

We are then counting up to stairmaster (n) eventually. Once i reaches n, d is the sum of the 3 previous values of d, which is the number of permutations of 1, 2, 3 that equal n number of stairs, so we terminate the loop and return d as our answer.

def stairmaster(n): a = 1 b = 2 c = 4 d = 0 # fourth variable (placehold for when n > 3) if (n == 0 or n == 1 or n == 2): return n if (n == 3): return c for i in range(4,n+1): # starting from 4 because we already know the solution for, 1, 2, or 3 stairs d = c + b + a # already counted for 3 stairs a = b b = c c = d return d print(stairmaster(10))

Solved this one using recursion. Codebyte option did not work so here is just the raw code:

def stairmaster(n):
  if n == 0 or  n < 0:
    return 0
  elif n == 1:
    return 1
  elif n == 2:
    return 2
  elif n == 3:
    return 4
  else:
    return  stairmaster(n - 1) + stairmaster(n - 2) + stairmaster(n - 3)

print(stairmaster(3))

Or this:

def stairmasterDP(n):

    memo = {0:1,1:1,2:2,3:4}

    if n in memo:

        return memo[n]

    else:

        for i in range(4,n+1):

            memo[i] = memo[i-1] + memo[i-2] + memo[i-3]

    return memo[n]

print(stairmasterDP(10))

This challenge is like trying to find Fibonacci numbers, but with base cases: 1, 2, 4.

I used loops instead of recursion, and I stored the relevant previous values in a list.

def stairmaster(n, k = 3):
  if (n <= 0):
    return 0;
  elif (n <= k):
    return 2 ** (n - 1)
  recursive = [2 ** i for i in range(k)]
  for i in range(k + 1, n + 1):
    current = sum(recursive)
    recursive.pop(0)
    recursive.append(current)
  return current

A list probably wasn’t the best data structure to use for this. Maybe a deque would be better.