Climbing Stairs

Problem Statement:

You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?

Example 1:

Input: n = 2 Output: 2 Explanation: There are two ways to climb to the top.

1. 1 step + 1 step
2. 2 steps

Example 2:

Input: n = 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

Constraints:

• 1 <= n <= 45

Solution:

1. Recursion and memoization:

class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n + 1];
        //dp[i] represents the total ways to reach the i'th step.
        Arrays.fill(dp, -1);
        return climbStairsUtil(n, dp);
    }
 
    public int climbStairsUtil(int n, int[] dp) {
        if(n == 1 || n == 2 || n == 0){
            return n;
        }
        if(dp[n] != -1){
            return dp[n];
        }
 
        return dp[n] = climbStairsUtil(n-1, dp) + climbStairsUtil(n-2, dp);
    }
}

2. Tabulation:

class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 0; //There're no ways to reach to 0th step from 0th step because we're already there.
        dp[1] = 1;
 
        for(int i = 2; i <= n; i++){
            if(i == 2){
                dp[i] = 1 + dp[i - 1];
            }
            if(i > 2){
                dp[i] = dp[i - 1] + dp[i - 2];
            }
        }
        return dp[n];
    }
}

3. Space optimized:

class Solution {
    public int climbStairs(int n) {
        // Special cases for small values of n
        if (n <= 1) {
            return 1;
        }
 
        // Variables to store the number of ways to reach the previous step and the step before the previous step
        int twoStepsBefore = 0;
        int oneStepBefore = 1;
 
        // Iterate from the second step to the nth step
        for (int step = 2; step <= n; step++) {
            int currentWays = 0;
            if (step == 2) {
                // Special case for the second step
                currentWays = 1 + oneStepBefore;
            } else {
                // General case for steps greater than 2
                currentWays = oneStepBefore + twoStepsBefore;
            }
            // Update the steps for the next iteration
            twoStepsBefore = oneStepBefore;
            oneStepBefore = currentWays;
        }
        return oneStepBefore;
    }
}