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

• Leetcode

• ProgramCreek

Examples

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

Solutions

Solution 1

/**
 * 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);
    }
}

Solution 2

/**
 * 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];
    }
}

Solution 3

/**
 * 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;
    }
}

Follow up