## Description

https://leetcode.com/problems/climbing-stairs/

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?

**Example 1:**

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

**Example 2:**

Input:n = 3Output:3Explanation: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`

## Explanation

The total distinct ways to climb to i th stair is actually the sum of the distinct ways of i – 2 and i – 1 stairs.

## Java Solution

```
class Solution {
public int climbStairs(int n) {
if (n <= 1) {
return 1;
}
int[] ways = new int[n + 1];
ways[0] = 1;
ways[1] = 1;
for (int i = 2; i <= n; i++) {
ways[i] = ways[i - 2] + ways[i - 1];
}
return ways[n];
}
}
```

## Python Solution

```
class Solution:
def climbStairs(self, n: int) -> int:
f = []
f.append(1)
f.append(2)
for i in range(2, n):
f.append(f[i - 1] + f[i - 2])
return f[n - 1]
```

- Time complexity: O(n). Single loop up to n.
- Space complexity: O(n). dp array of size n is used.

I found that solution is very popular and helpful : https://www.youtube.com/watch?v=zJ6fNciu_DM