# Climbing Stairs Leetcode Solution

In this post, we are going to solve the Climbing Stairs Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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

Example 2:

```Input: n = 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`

Now, lets see the leetcode solution of Climbing Stairs Leetcode Solution.

### Climbing Stairs Leetcode Solution in Python

```class Solution:
def climbStairs(self, n: int) -> int:
# dp[i] := # Of distinct ways to climb to i-th stair
dp = [1, 1] +  * (n - 1)

for i in range(2, n + 1):
dp[i] = dp[i - 1] + dp[i - 2]

return dp[n]
```

### Climbing Stairs Leetcode Solutionin CPP

```class Solution {
public:
int climbStairs(int n) {
// dp[i] := # of distinct ways to climb to i-th stair
vector<int> dp(n + 1);
dp = 1;
dp = 1;

for (int i = 2; i <= n; ++i)
dp[i] = dp[i - 1] + dp[i - 2];

return dp[n];
}
};
```

### Climbing Stairs Leetcode Solution in Java

```class Solution {
public int climbStairs(int n) {
// dp[i] := # of distinct ways to climb to i-th stair
int[] dp = new int[n + 1];
dp = 1;
dp = 1;

for (int i = 2; i <= n; ++i)
dp[i] = dp[i - 1] + dp[i - 2];

return dp[n];
}
}
```

Note: This problem Climbing Stairs is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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