Let's dive into a classic problem that perfectly illustrates the power of dynamic programming: the LeetCode Stair Climbing problem. If you're new to dynamic programming or looking to sharpen your skills, this is an excellent place to start. We'll break down the problem, explore different approaches, and provide a clear, step-by-step solution. So, grab your coding gear, and let's get climbing!

Understanding the Stair Climbing Problem

Okay, so what's the big deal about climbing stairs? Here's the problem:

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?

For example:

• If n = 1, there is only one way to climb (1 step).
• If n = 2, there are two ways to climb (1 step + 1 step, or 2 steps).
• If n = 3, there are three ways to climb (1+1+1, 1+2, 2+1).

Constraints:

• 1 <= n <= 45

That's it! Simple, right? But don't let the simplicity fool you. This problem is a gateway to understanding dynamic programming.

Why Dynamic Programming?

You might be thinking, "Can't I just use recursion?" Yes, you could, but you'll quickly run into performance issues as n increases. A naive recursive solution would recalculate the same values over and over again, leading to exponential time complexity. That's where dynamic programming comes to the rescue.

Dynamic programming is an optimization technique that involves breaking down a problem into smaller overlapping subproblems, solving each subproblem only once, and storing the results to avoid redundant computations. In essence, it's like learning from your past mistakes (or, in this case, calculations) to work smarter, not harder.

For this problem, the key insight is that the number of ways to reach step n is the sum of the number of ways to reach step n-1 (by taking one step) and the number of ways to reach step n-2 (by taking two steps). This overlapping subproblem structure makes it perfect for dynamic programming.

Approaches to Solving the Problem

There are two primary approaches to solving dynamic programming problems:

1. Top-Down (Memoization): This approach starts with the original problem and recursively breaks it down into subproblems. We store the results of each subproblem in a memo (usually a dictionary or array) to avoid recalculating them.
2. Bottom-Up (Tabulation): This approach starts with the smallest subproblems and iteratively builds up to the solution of the original problem. We store the results of each subproblem in a table (usually an array).

Let's explore both approaches.

1. Top-Down (Memoization) Approach

In the top-down approach, we'll define a recursive function that calculates the number of ways to reach step n. Before making any recursive calls, we'll check if the result for n is already stored in our memo. If it is, we simply return the stored value. Otherwise, we calculate the result recursively, store it in the memo, and then return it.

Here's the Python code:

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

def climbStairs(n):
    memo = {}
    return climbStairs_memo(n, memo)

Explanation:

• climbStairs_memo(n, memo): This is our recursive helper function.
  • It takes n (the number of steps) and memo (the memoization dictionary) as input.
  • It first checks if the result for n is already in memo. If so, it returns the stored value.
  • If n is 1 or 2, it returns n directly (base cases).
  • Otherwise, it recursively calculates the result as climbStairs_memo(n - 1, memo) + climbStairs_memo(n - 2, memo), stores it in memo, and returns it.
• climbStairs(n): This is the main function.
  • It initializes an empty dictionary memo.
  • It calls climbStairs_memo(n, memo) to start the recursive calculation.

2. Bottom-Up (Tabulation) Approach

In the bottom-up approach, we'll create an array dp of size n + 1. dp[i] will store the number of ways to reach step i. We'll initialize dp[1] to 1 and dp[2] to 2 (base cases). Then, we'll iterate from i = 3 to n, calculating dp[i] as dp[i - 1] + dp[i - 2]. Finally, we'll return dp[n].

Here's the Python code:

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]

Explanation:

• climbStairs(n): This is the main function.
  • It first handles the base cases: if n is 1 or 2, it returns n directly.
  • It creates an array dp of size n + 1 filled with 0s.
  • It initializes dp[1] to 1 and dp[2] to 2.
  • It iterates from i = 3 to n, calculating dp[i] as dp[i - 1] + dp[i - 2].
  • Finally, it returns dp[n].

Comparing the Approaches

Both the top-down and bottom-up approaches have their pros and cons:

• Top-Down (Memoization):
  • Pros:
    • More intuitive and easier to understand for some people, as it directly follows the recursive definition of the problem.
    • Only calculates the values that are actually needed.
  • Cons:
    • Can have higher overhead due to recursive function calls.
    • May lead to stack overflow errors for very large values of n.
• Bottom-Up (Tabulation):
  • Pros:
    • More efficient, as it avoids recursive function calls.
    • No risk of stack overflow errors.
  • Cons:
    • May be less intuitive for some people.
    • Calculates all values up to n, even if some of them are not needed.

In general, the bottom-up approach is often preferred for dynamic programming problems because it's more efficient and avoids stack overflow errors. However, the top-down approach can be a good choice if you find it more intuitive or if you only need to calculate a small subset of the possible values.

Optimizing for Space Complexity

Notice that in the bottom-up approach, we only need the values of dp[i - 1] and dp[i - 2] to calculate dp[i]. This means we don't need to store the entire dp array. We can optimize the space complexity by using only two variables to store the previous two values.

Here's the optimized Python code:

def climbStairs(n):
    if n <= 2:
        return n
    one_step_before = 2
    two_steps_before = 1
    for i in range(3, n + 1):
        current = one_step_before + two_steps_before
        two_steps_before = one_step_before
        one_step_before = current
    return one_step_before

Explanation:

• one_step_before: Stores the number of ways to reach the step one step before the current step.
• two_steps_before: Stores the number of ways to reach the step two steps before the current step.
• In each iteration, we calculate the number of ways to reach the current step (current) as the sum of one_step_before and two_steps_before.
• Then, we update two_steps_before to one_step_before and one_step_before to current to prepare for the next iteration.

This optimized solution has a time complexity of O(n) and a space complexity of O(1), which is a significant improvement over the original bottom-up approach.

Conclusion

The LeetCode Stair Climbing problem is a fantastic introduction to dynamic programming. We've explored two different approaches (top-down and bottom-up) and even optimized the space complexity. Remember, the key to mastering dynamic programming is to identify overlapping subproblems and store their solutions to avoid redundant calculations.

So, keep practicing, keep climbing, and you'll be conquering those coding challenges in no time! Now that you understand this problem, try tackling other similar dynamic programming problems on LeetCode. Happy coding, guys!