Recursion on staircase

Question

I\'m trying to understand the solution provided in a book to the following question:

\"A child is running up a staircase with n steps and can hop either 1 step, 2 st

Answer 1

You need to think about it has a tree with 3 possible options on each node. If the size of the staircase is 4 we will have something like this:

(4)--1-->(3)--..(Choose a step and keep branching)...
  |__2-->(2)--..(Until you get size of zero)............
  |__3-->(1)--1-->(0) # <--Like this <--

At the end if you count all the leafs with size of zero you will get all the possible ways.

So you can think it like this, what if you take a step and then consider update the size of the stair like this size-step, where your steps can be (1,2,3)

Doing that you can code something like this:

choices = (1, 2, 3)
counter = 0

def test(size):
    global counter

    if size == 0:
        counter += 1

    for choice in choices:
        if size - choice >= 0:
            test(size - choice)

    return counter