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] = 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 nis 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] = 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%)

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/