Climbing Stairs Problem

Leetcode Lesson: Climbing Stairs

1. 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

2. Conceptual Understanding

This problem is about finding the number of different combinations to reach a goal (the top of the stairs) given a set of fixed moves (1 or 2 steps at a time).

Think of it like this: You're standing at the bottom of a staircase, and for each step, you have two choices: take 1 step or take 2 steps. We need to count all the possible sequences of these choices that lead to the top.

This problem introduces the concept of dynamic programming, where we build the solution to a larger problem from solutions to smaller subproblems.

3. Approach Brainstorming

Let's consider a few approaches:

  1. Recursive: Solve the problem by breaking it down into smaller subproblems.
  2. Dynamic Programming: Build up the solution iteratively, storing intermediate results.
  3. Mathematical: Recognize the pattern and use a formula (this is actually the Fibonacci sequence).

Each approach has trade-offs in terms of time and space complexity, as well as ease of understanding.

4. Optimal Solution Walkthrough

We'll use the Dynamic Programming approach:

  1. Create an array dp where dp[i] represents the number of ways to climb to the i-th step.
  2. Initialize the base cases: dp[1] = 1 (one way to climb 1 step) and dp[2] = 2 (two ways to climb 2 steps).
  3. For steps 3 to n, the number of ways to reach step i is the sum of ways to reach the previous step (and then take 1 step) and the ways to reach two steps back (and then take 2 steps). So, dp[i] = dp[i-1] + dp[i-2]
  4. Return dp[n] as the final answer.

This approach is optimal because it solves each subproblem only once and uses the results to build up to the final solution.

5. Python Implementation

def climbStairs(n):
    if n <= 2:
        return n
    
    dp = [0] * (n + 1)
    dp[1] = 1
    dp[2] = 2
    
    for i in range(3, n + 1):
        dp[i] = dp[i-1] + dp[i-2]
    
    return dp[n]
  • We use a list dp to store the number of ways for each step.
  • We initialize the base cases for 1 and 2 steps.
  • We then iterate from 3 to n, filling in the dp array.
  • The final answer is in dp[n].

6. Time and Space Complexity Analysis

Time Complexity: O(n)

  • We iterate through the steps once, performing constant-time operations at each step.

Space Complexity: O(n)

  • We use an array of size n+1 to store intermediate results.

Note: We can optimize the space complexity to O(1) by only keeping track of the last two values instead of the entire array.

7. Edge Cases and Testing

Consider these test cases:

  • n = 1 (base case)
  • n = 2 (base case)
  • n = 3 (first non-base case)
  • n = 5 (a medium-sized case)
  • n = 45 (the maximum input according to the constraints)

8. Common Pitfalls and Mistakes

  • Forgetting to handle the base cases (n = 1 and n = 2) separately.
  • Confusing the problem with counting the number of steps, rather than the number of ways to take steps.
  • Trying to generate all possible combinations explicitly, which is inefficient for larger n.

9. Optimization Opportunities

We can optimize the space complexity:

def climbStairs(n):
    if n <= 2:
        return n
    
    prev, curr = 1, 2
    for _ in range(3, n + 1):
        prev, curr = curr, prev + curr
    
    return curr

This solution uses only O(1) extra space.

  • Fibonacci Sequence (this problem follows the same pattern)
  • "House Robber" problem (another dynamic programming problem)
  • Other dynamic programming problems like "Coin Change" or "Minimum Cost Climbing Stairs"

11. Reflection Questions

  1. How does this problem relate to the Fibonacci sequence?
  2. Can you think of a real-world scenario that follows a similar pattern?
  3. How would the solution change if you could take 1, 2, or 3 steps at a time?
  4. Can you solve this problem recursively? How would the time complexity compare?

12. Additional Resources