Replacing recursion with while loop (stair climbin

Question

I'm practicing replacing recursion with while loops, and I'm stuck on the following problem.

How many ways can you go up a staircase of length n if you can only take the stairs 1 or 2 at a time?

The recursive solution is pretty simple:

def stairs(n):
  if n <= 1:
    return 1
  else:
    return stairs(n-2) + stairs(n-1)

I feel like the structure for the iterative program should go something like this:

def stairs_iterative(n):
  ways = 0
  while n > 1:
    # do something
    ways +=1
  return ways

But I don't know what I need to put in the #do something part. Can someone help me? Pseudocode is fine!

Answer 1

This amounts to the top-down (recursive) approach vs. the bottom-up (iterative) approach for dynamic programming.

Since you know that for input n you need all values of stairs(p) for 0 <= p <= n. You can iteratively compute stairs(p) starting at p = 0 until you reach p = n, as follows:

def stairs(n):
    table = [1, 1]  # p = 0 and p = 1
    for i in range(2, n + 1):
        table.append(table[i - 2] + table[i - 1])
    return table[n]

Answer 2

A different approach than @univerio is to use a list as stack:

def stairs_it(n):
    res = 0
    stack = [n]
    while len(stack) > 0:
        curr = stack[0]
        stack.remove(curr)
        if curr == 0:
            res += 0
        elif curr == 1:
            res += 1
        elif curr == 2:
            res += 2
        else:
            stack.append(curr-1)
            stack.append(curr-2)
    return res