# Climbing Stairs

You are climbing a staircase. 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?

**Note:**&#x47;iven \_n \_will be a positive integer.

**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
```

## Analysis

**DP - Bottom Up Approach**

Break this problem into subproblems, with optimal substructure, thus using Dynamic Programming (Bottom Up approach):

> **State:**&#x20;
>
> `dp[i]`- number of ways to reach the `ith` step
>
> **State Transfer Function:**
>
> `dp[i] = dp[i - 1] + dp[i - 2]`- taking a step from (i - 1), or taking a step of 2 from (i - 2)
>
> **Initial State:**
>
> `dp[0] = 0;`
>
> `dp[1] = 1;`
>
> `dp[2] = 2;`
>
> **Answer:**
>
> `dp[n]`
>
> **Complexity Analysis**
>
> * Time complexity: O(n). Single loop up to `n`.
> * Space complexity : O(n). `dp` array of size `n`is used.

**DP - Top-Down Approach - Recursion with memorization**

We could define `memo[]` to store the number of ways to `ith` step, it helps pruning recursion. And the recursive function could be defined as `climb_Stairs(int i, int n, int memo[])` that returns the number of ways from `ith` step to `nth` step.

> **Complexity Analysis**&#x20;
>
> * Time complexity : O(n). Single loop upto n.
> * Space complexity : O(n). dpdp array of size n is used.

**DP - Fibonacci Number - Optimize Space Complexity**

`Fib(n)=Fib(n−1)+Fib(n−2)`

That the nth number only has to do with its previous two numbers, thus we don't have to maintain a whole array of results, just the last 2 results are enough. Thus optimized the space for O(1).

**Complexity Analysis**

* Time complexity : `O(n)`. Single loop upto n is required to calculate `n th` fibonacci number.
* Space complexity : `O(1)`. Constant space is used.

## Solution

Dynamic Programming - Bottom Up - O(n) space, O(n) time (3ms 32.02% AC)

```java
class Solution {
    public int climbStairs(int n) {
        if (n == 0) return 0;
        if (n == 1) return 1;
        if (n == 2) return 2;
        int[] steps = new int[n + 1];
        steps[0] = 0;
        steps[1] = 1;
        steps[2] = 2;
        for (int i = 3; i < n + 1; i++) {
            steps[i] = steps[i - 1] + steps[i - 2];
        }
        return steps[n];
    }
}
```

Dynamic Programming - Recursion with memorization - Top Down - O(n) Space, O(n) Time (3ms 32.02%)

```java
public class Solution {
    public int climbStairs(int n) {
        int memo[] = new int[n + 1];
        return climb_Stairs(0, n, memo);
    }
    public int climb_Stairs(int i, int n, int memo[]) {
        if (i > n) {
            return 0;
        }
        if (i == n) {
            return 1;
        }
        if (memo[i] > 0) {
            return memo[i];
        }
        memo[i] = climb_Stairs(i + 1, n, memo) + climb_Stairs(i + 2, n, memo);
        return memo[i];
    }
}
```

DP - Fibonacci Numbers - Space Optimized (2ms 92.09%)

```java
public class Solution {
    public int climbStairs(int n) {
        if (n == 1) {
            return 1;
        }
        int first = 1;
        int second = 2;
        for (int i = 3; i <= n; i++) {
            int third = first + second;
            first = second;
            second = third;
        }
        return second;
    }
}
```

## Reference

[https://leetcode.com/problems/climbing-stairs/solution/](https://leetcode.com/problems/climbing-stairs/solution/)
