Counting ways to climb up the stairs

Question

"You are climbing a staircase. It takes n steps to reach the top.
Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?"

I saw this question on leetcode and I was trying to work it out using recursive way. I am not sure if this is the proper approach but I still gave it a try. Can someone please help me and tell me if this is even a right approach?

WAYS = 0   # global variable?

def climb_stairs(n):
    if n == 0:
        WAYS += 1
        return
    else:
        climb_stairs(n-1)
        if n % 2 == 0:
            climb_stairs(n-2)
    return WAYS

Asked By: mybraincellssayhi

||

Source

Answer 1

IIUC, you simply need to use:

def climb_stairs(n): # n > 0
    if n <= 2:
        return n # if 2 steps, 2 ways (1 or 2), else only 1
    else:
        # if we chose 1, then there are n-1 steps left
        # if we chose 2, then there are n-2 steps left
        return climb_stairs(n-1)+climb_stairs(n-2)

climb_stairs(2)
# 2 (1+1 or 2)

climb_stairs(5)
# 8 (1+1+1+1+1, 2+1+1+1, 1+2+1+1, 1+1+2+1, 1+1+1+2, 2+2+1, 2+1+2, 1+2+2)

complexity

I assumed the question was before all a coding training. If you want to improve efficiency you can use a cache, wuch as functools.lru_cache (or functools.cache for python ≥ 3.9):

from functools import lru_cache

@lru_cache(maxsize=None)
def climb_stairs(n):
    if n <= 2:
        return n # if 2 steps, 2 ways (1 or 2), else only 1
    else:
        return climb_stairs(n-1)+climb_stairs(n-2)
    
climb_stairs(40)
# 165580141

Answered By: mozway

Answer 2

This is far from optimal solution – the complexity will be O(2^n). To achieve an optimal programatic solution you should use DP / memoization:

def climb_stairs_updated(n):
    climb_stairs_updated.counter += 1
    if n <= 2:
        memoization[n] = n
        return n
    if n in memoization:
        return memoization[n]
    else:
        result = climb_stairs_updated(n-1) + climb_stairs_updated(n-2)
        memoization[n] = result
        return result
    
memoization = {}
climb_stairs_updated.counter = 0
climb_stairs_updated(5)

The counter in the function counts the call for evaluation purposes – the solution is in O(n).

Answered By: matwasilewski

Answer 3

If you approach from top to bottom it will be easier. Imagine you have 5 stair cases:

If you are at the last staircase, it takes 1 step to reach top. that is default. If you are at the (n-1)th staircase in this case you are at 4th staircase, it would take 1 step to reach the top.

now you ask how many different ways you can reach the top from 3rd staircase. If take 1 step up, you get 4th and you already know from the 4th step it takes 1 step to reach the top. If you take 2 steps up, you reach the top and from top it takes 1 step to reach by default. So, you can see that from 3rd step, it takes 1+1 to reach the top. based on this logic:

class Solution:
    def climbStairs(self, n: int) -> int:
        one,two=1,1
        # since we initialized one,two, we have to compute only n-1 values. remember n-1 is not inclusive
        for i in range(n-1):
            one,two=two,one+two
        return two

Answered By: Yilmaz

Counting ways to climb up the stairs

Question:

Answers:

complexity