"""
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?

Example 1:

Input: 2
Output: 2
Explanation: There are two ways to climb to the top.
1. 1 step + 1 step
2. 2 steps
Example 2:

Input: 3
Output: 3
Explanation: There are three ways to climb to the top.
1. 1 step + 1 step + 1 step
2. 1 step + 2 steps
3. 2 steps + 1 step
 

Constraints:

1 <= n <= 45
"""

"""
In the above approach we have used dpdp array where dp[i]=dp[i-1]+dp[i-2]dp[i]=dp[i−1]+dp[i−2]. It can be easily analysed that dp[i]dp[i] is nothing but i^{th}i 
th
  fibonacci number.

Fib(n)=Fib(n-1)+Fib(n-2)
Fib(n)=Fib(n−1)+Fib(n−2)

Now we just have to find n^{th}n 
th
  number of the fibonacci series having 11 and 22 their first and second term respectively, i.e. Fib(1)=1Fib(1)=1 and Fib(2)=2Fib(2)=2.
"""


class Solution:
    def climbStairs(self, n: int) -> int:
        if n == 1:
            return 1
        
        first = 1
        second = 2
        
        for i in range(3, n+1):
            third = first + second
            first = second
            second = third
            
            
        print(second)
        return second
        
