Efficient Solutions to LeetCode’s Climbing Stairs Problem

The Climbing Stairs problem is a classic algorithmic challenge that can be approached in several ways. In this tutorial, we will explore three powerful techniques: recursion, memoization, and bottom-up dynamic programming. By the end of this guide, you will have a solid understanding of these concepts and how to apply them to solve the problem efficiently.

Prerequisites

Before diving into the solutions, it’s helpful to have a basic understanding of the following concepts:

• Recursion: A method where the solution to a problem depends on solutions to smaller instances of the same problem.
• Dynamic Programming: An optimization technique that solves problems by breaking them down into simpler subproblems and storing the results to avoid redundant calculations.
• Fibonacci Sequence: Understanding this sequence will help you grasp the Climbing Stairs problem, as it is closely related.

Understanding the Climbing Stairs Problem

The problem can be stated as follows: You are climbing a staircase with n steps, and you can either take 1 step or 2 steps at a time. Your goal is to find out how many distinct ways you can climb to the top.

For example, if n = 3, you can climb the stairs in the following ways:

• 1 step + 1 step + 1 step
• 1 step + 2 steps
• 2 steps + 1 step

Thus, there are 3 distinct ways to climb 3 steps.

Solution 1: Recursion

The simplest way to solve this problem is through recursion. The idea is to break down the problem into smaller subproblems. The number of ways to reach the nth step can be expressed as:

ways(n) = ways(n-1) + ways(n-2)

This means that to reach the nth step, you can either come from the (n-1)th step or the (n-2)th step.

Here’s how you can implement this in Python:

def climb_stairs(n):
    if n <= 1:
        return 1
    return climb_stairs(n - 1) + climb_stairs(n - 2)

While this approach is straightforward, it can be inefficient for larger values of n due to repeated calculations.

Solution 2: Memoization

To improve the efficiency of our recursive solution, we can use memoization. This technique involves storing the results of expensive function calls and reusing them when the same inputs occur again.

Here’s how you can implement memoization in Python:

def climb_stairs_memo(n, memo={}):
    if n in memo:
        return memo[n]
    if n <= 1:
        return 1
    memo[n] = climb_stairs_memo(n - 1, memo) + climb_stairs_memo(n - 2, memo)
    return memo[n]

With memoization, we significantly reduce the number of calculations, making this approach much faster.

Solution 3: Bottom-Up Dynamic Programming

The final approach we will discuss is bottom-up dynamic programming. Instead of using recursion, we build our solution iteratively from the ground up.

We can create an array to store the number of ways to reach each step:

def climb_stairs_dp(n):
    if n <= 1:
        return 1
    dp = [0] * (n + 1)
    dp[0], dp[1] = 1, 1
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]
    return dp[n]

This approach has a time complexity of O(n) and a space complexity of O(n), making it efficient for larger values of n.

Conclusion

In this tutorial, we explored three different methods to solve the Climbing Stairs problem: recursion, memoization, and bottom-up dynamic programming. Each method has its own advantages and trade-offs. Understanding these techniques will not only help you tackle this specific problem but also enhance your problem-solving skills in general.

For further reading and practice, check out the original problem on LeetCode: Continue reading on appkodersolution ». Happy coding!