# 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:Given _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:

`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;`

`dp = 1;`

`dp = 2;`

`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

• 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)

``````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;
steps = 1;
steps = 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%)

``````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%)

``````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/