'''

Problem:

You are climbing a stair case. It takes n steps to reach to the top.

Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Solution:

Let k be the number of 2-steps taken

Then there are (n-2*k) 1-steps and k 2-steps to reach the top.

In total there are n/2 different k's for a given n, because then there are 0 1-steps and n/2 2-steps.

Using the binomial coeficcient the permutations of 1-steps and 2-steps for a given k can be calculated as

(n-(2*k)+k) choose k = (n-k) choose k

The total number of ways is thus

sum (n-k) choose k where k=0..n/2

The solution of this sum equals the (n+1)th fibonacci number as seen here: https://www.wolframalpha.com/input/?i=sum+%28n-k%29+choose+k+for+k%3D0..n%2F2

That way the problem becomes rather easy.

'''

class Solution:

def climbStairs(self, n: int) -> int:

return self.fib(n + 1)

fibStore = {}

def fib(self, n: int) -> int:

if n in self.fibStore:

return self.fibStore[n]

if n <= 2:

return 1

value = self.fib(n - 1) + self.fib(n - 2)

self.fibStore[n] = value

return value

print(Solution().climbStairs(30))