# 70. Climbing Stairs

### Description

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?

### Constraints

n >= 1

### Approach

Fibonacci

### Links

* [GeeksforGeeks](https://www.geeksforgeeks.org/count-ways-reach-nth-stair)
* [Leetcode](https://leetcode.com/problems/climbing-stairs)
* ProgramCreek

### Examples

{% tabs %}
{% tab title="Example 1" %}
**Input:** 2

**Output:** 2

**Explanation:** There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps
   {% endtab %}

{% tab title="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
   {% endtab %}
   {% endtabs %}

### Solutions

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : O(2^n).Size of recursion tree will be 2^n.
 *
 */
 
public class Solution {
    public int climbStairs(int n) {
        return climb_Stairs(0, n);
    }
    public int climb_Stairs(int i, int n) {
        if (i > n) {
            return 0;
        }
        if (i == n) {
            return 1;
        }
        return climb_Stairs(i + 1, n) + climb_Stairs(i + 2, n);
    }
}
```

{% endtab %}

{% tab title="Solution 2" %}

```java
/**
 * Time complexity : O(n). Single loop upto n.
 * Space complexity : O(n). dp array of size n is used.
 */

public class Solution {
    public int climbStairs(int n) {
        if (n == 1) {
            return 1;
        }
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}
```

{% endtab %}

{% tab title="Solution 3" %}

```java
/**
 * Time complexity : O(n). Single loop upto n is required to calculate nth fibonacci number.
 * Space complexity : O(1). Constant space is used.
 */
 
class Solution {
    public int climbStairs(int n) {
        int first = 1;
        int second = 1;
        while(--n > 0) {
            int t = first;
            first = second;
            second += t;
        }
        return second;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://code-snippets.hbamithkumara.com/leetcode/problems/1-100/climbing-stairs.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.
