[Dynamic Programming] Count Ways to Climb Stairs - Easy

Dynamic programming is a strong method we use to solve tough problems. It helps us break down these problems into easier parts. For the “Count Ways to Climb Stairs” problem, we want to find out how many different ways we can reach the top of a staircase with a certain number of steps. We can take either one or two steps at a time. By using dynamic programming, we can find the number of ways to climb the stairs quickly and avoid doing the same work over again.

In this article, we will look at different ways to solve the climbing stairs problem with dynamic programming. We will give examples in Java, Python, and C++. We will explain the problem in detail. We will also check out a better way to use space and a recursive solution with memoization for each programming language. Here is what we will cover:

  • [Dynamic Programming] Counting Ways to Climb Stairs with Dynamic Programming
  • Understanding the Problem Statement for Climbing Stairs
  • Dynamic Programming Approach to Count Ways in Java
  • Dynamic Programming Approach to Count Ways in Python
  • Dynamic Programming Approach to Count Ways in C++
  • Optimizing Space Complexity in Climbing Stairs Problem
  • Recursive Solution with Memoization in Java
  • Recursive Solution with Memoization in Python
  • Recursive Solution with Memoization in C++
  • Frequently Asked Questions

If you want to learn more about dynamic programming techniques, you can check other articles like Dynamic Programming Fibonacci Number and Dynamic Programming Min Cost Climbing Stairs.

Understanding the Problem Statement for Climbing Stairs

The Climbing Stairs problem is a well-known challenge in dynamic programming. It helps us find how many different ways we can climb a staircase with n steps. We can take either 1 step or 2 steps at a time. Our goal is to count how many different combinations of these steps will get us to the top.

Problem Statement

We have an integer n. This number shows how many steps are in the staircase. We need to find out how many distinct ways we can reach the top.

Example

  • If n = 2, we have 2 ways to climb the stairs: (1, 1) or (2).
  • If n = 3, we have 3 ways to climb the stairs: (1, 1, 1), (1, 2), (2, 1).

Recurrence Relation

We can define the number of ways to reach the nth step using a recursive formula: - ways(n) = ways(n-1) + ways(n-2) This means to get to the nth step, we can: - Come from the (n-1)th step by taking a single step, or - Come from the (n-2)th step by taking a double step.

Base Cases

  • ways(0) = 1: There is one way to stay on the ground (by doing nothing).
  • ways(1) = 1: There is one way to reach the first step (by taking one step).

We can solve this problem well using dynamic programming methods. We will talk about this in the next sections. We will include Java, Python, and C++ examples.

For more related information, visit Dynamic Programming - Climbing Stairs.

Dynamic Programming Approach to Count Ways in Java

To solve the “Count Ways to Climb Stairs” problem with dynamic programming in Java, we use an array. This array will store how many ways we can reach each step. The basic idea is that the number of ways to reach the nth step is the total of the ways to reach the (n-1)th step and the (n-2)th step.

Java Code Implementation

public class ClimbStairs {
    public static int countWays(int n) {
        if (n <= 1) return 1;

        int[] dp = new int[n + 1];
        dp[0] = 1; // 1 way to stay at the ground
        dp[1] = 1; // 1 way to reach the first step

        for (int i = 2; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }

        return dp[n];
    }

    public static void main(String[] args) {
        int n = 5; // Example input
        System.out.println("Number of ways to climb " + n + " stairs: " + countWays(n));
    }
}

Explanation of the Code

  • Base Cases:
    • If n is 0 or 1, there is only 1 way to climb.
  • Dynamic Programming Array:
    • We create an array dp to store the number of ways to reach each step from 0 to n.
  • Loop:
    • We start from the 2nd step to find the number of ways for each step. We add the values from the two steps before it.
  • Output:
    • The result for the nth step is in dp[n].

This method has a time complexity of O(n) and a space complexity of O(n). To make it better, we can reduce the space complexity to O(1). We just need two variables to keep the last two results instead of using an array.

For more insight into dynamic programming techniques, refer to Dynamic Programming - Fibonacci Number.

Dynamic Programming Approach to Count Ways in Python

In Python, we can count how many ways we can climb stairs using a dynamic programming method. This method uses overlapping subproblems to find the result quickly.

Dynamic Programming Implementation

We can make a function that takes the number of stairs n as input. It will return how many distinct ways we can reach the top. We represent the states with the number of ways to reach each step. We calculate this step by step.

def countWays(n):
    if n <= 1:
        return 1

    # Initialize an array for dynamic programming
    dp = [0] * (n + 1)
    dp[0], dp[1] = 1, 1  # Base cases

    # Fill the dp array
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2]  # Ways to reach current step

    return dp[n]

# Example usage:
n = 5
print(f"Ways to climb {n} stairs: {countWays(n)}")

Time and Space Complexity

  • Time Complexity: O(n) - We find each step once.
  • Space Complexity: O(n) - We use an array of size n.

Optimizing Space Complexity

To make space use better, we can remember only the last two steps. The current step only needs the last two steps to calculate.

def countWaysOptimized(n):
    if n <= 1:
        return 1

    first, second = 1, 1

    for i in range(2, n + 1):
        current = first + second
        first, second = second, current  # Update for the next step

    return second

# Example usage:
n = 5
print(f"Ways to climb {n} stairs (Optimized): {countWaysOptimized(n)}")

This optimized solution keeps O(1) space complexity. It still has O(n) time complexity. If you want to read more about related dynamic programming problems, you can check the Dynamic Programming - Climbing Stairs article.

Dynamic Programming Approach to Count Ways in C++

In C++, we can use dynamic programming to count distinct ways to climb stairs. We can do this in a simple way using an iterative method. The problem says we can take either 1 or 2 steps at a time. We need to find the total ways to reach the top of n stairs.

Code Implementation

#include <iostream>
#include <vector>

int countWays(int n) {
    if (n <= 1) return 1;

    std::vector<int> dp(n + 1);
    dp[0] = 1; // 1 way to stay at the ground
    dp[1] = 1; // 1 way to climb to the first step

    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2]; // Sum of ways from the last two steps
    }

    return dp[n];
}

int main() {
    int n = 5; // Example: number of stairs
    std::cout << "Number of ways to climb " << n << " stairs is: " << countWays(n) << std::endl;
    return 0;
}

Explanation of the Code

  • Initialization: We create an array dp. Here, dp[i] stores the number of ways to reach the i-th step.
  • Base Cases:
    • We set dp[0] to 1 because there is one way to stay at the ground (do nothing).
    • We also set dp[1] to 1 since there is only one way to reach the first step.
  • Dynamic Programming Relation: For each step i, the number of ways to reach that step is the sum of the ways to reach the two previous steps: dp[i] = dp[i - 1] + dp[i - 2].
  • Output: The function gives back the value of dp[n], which shows the number of distinct ways to climb n stairs.

Time and Space Complexity

  • Time Complexity: O(n), where n is the number of stairs.
  • Space Complexity: O(n) for the dp array.

This C++ code is good for counting the ways to climb stairs. We can make it even better by reducing space. For example, we can use just two variables instead of the whole array. If we want to learn more about dynamic programming, we can read about the Dynamic Programming Fibonacci Number and Dynamic Programming Min Cost Climbing Stairs articles for more insights.

Optimizing Space Complexity in Climbing Stairs Problem

In the Climbing Stairs problem, we can make space better. We only need to remember the last two values we calculated. We do not need to save all previous values. This change helps us lower space from O(n) to O(1).

Implementation in Java

public class ClimbingStairs {
    public int climbStairs(int n) {
        if (n <= 2) return n;
        int first = 1, second = 2;
        for (int i = 3; i <= n; i++) {
            int current = first + second;
            first = second;
            second = current;
        }
        return second;
    }
}

Implementation in Python

class ClimbingStairs:
    def climb_stairs(self, n: int) -> int:
        if n <= 2:
            return n
        first, second = 1, 2
        for i in range(3, n + 1):
            current = first + second
            first = second
            second = current
        return second

Implementation in C++

class ClimbingStairs {
public:
    int climbStairs(int n) {
        if (n <= 2) return n;
        int first = 1, second = 2;
        for (int i = 3; i <= n; i++) {
            int current = first + second;
            first = second;
            second = current;
        }
        return second;
    }
};

With this space saving method, we keep the same time of O(n). But we use much less space. This makes our solution work better for bigger inputs. If you want to learn more about similar topics in dynamic programming, check out Dynamic Programming: Fibonacci Number and Dynamic Programming: Min Cost Climbing Stairs.

Recursive Solution with Memoization in Java

We can improve the recursive way to count how many ways to climb stairs by using memoization. This helps us save results we have already found. It makes the process faster, especially for bigger numbers.

Code Implementation

Here is a simple way to show the recursive solution with memoization in Java:

import java.util.HashMap;

public class ClimbStairs {
    private HashMap<Integer, Integer> memo = new HashMap<>();

    public int countWays(int n) {
        if (n <= 1) return 1; // Base case: 1 way to climb 0 or 1 stair
        if (memo.containsKey(n)) return memo.get(n); // Return cached result

        // Recursive step: count ways for n-1 and n-2
        int ways = countWays(n - 1) + countWays(n - 2);
        memo.put(n, ways); // Cache the result
        return ways;
    }

    public static void main(String[] args) {
        ClimbStairs cs = new ClimbStairs();
        int n = 5; // Example: Climbing 5 stairs
        System.out.println("Number of ways to climb " + n + " stairs: " + cs.countWays(n));
    }
}

Explanation of the Code

  • Base Case: If n is 0 or 1, there is only one way to climb the stairs.
  • Memoization: We use a HashMap to save the results we got before.
  • Recursion: The method countWays(n) finds out how many ways to climb n stairs by adding the ways to climb n-1 and n-2 stairs.
  • Caching: Before doing the math, we check if we already have the result for n. If we do, we just return that result to save time.

This way to write the code helps us deal with bigger numbers and still keeps the logic clear and easy to follow. If you want to learn more about dynamic programming methods, you can check out the article on Dynamic Programming: Fibonacci with Memoization.

Recursive Solution with Memoization in Python

In the Climbing Stairs problem, we can use a recursive method with memoization. This helps us count the ways to reach the top of the stairs quickly. We break down the problem into smaller parts. Then we save their results to avoid doing the same work again.

Problem Definition

We have n stairs. We can climb either 1 or 2 stairs at a time. Our goal is to find out how many different ways we can reach the top.

Recursive Formula

Let f(n) be the number of ways to climb to the nth stair. The formula is: - f(n) = f(n-1) + f(n-2), and we have base cases: - f(0) = 1 (There is 1 way to stay on the ground) - f(1) = 1 (There is 1 way to reach the first stair)

Implementation in Python

We can use a dictionary to keep the computed values (memoization).

def climbStairs(n):
    memo = {}

    def helper(n):
        if n in memo:
            return memo[n]
        if n == 0:
            return 1
        if n == 1:
            return 1
        memo[n] = helper(n - 1) + helper(n - 2)
        return memo[n]

    return helper(n)

# Example usage
n = 5
print(f"Ways to climb {n} stairs: {climbStairs(n)}")

Explanation of the Code

  • memo: This is a dictionary that saves the number of ways to climb to each stair level.
  • helper(n): This is the recursive function that finds the number of ways to reach the nth stair.
  • We check base cases first to return known values.
  • We store the result for each stair in memo to make future calls faster.

This recursive method with memoization makes things much faster than using a simple recursive method. It avoids recalculating, so it works better for larger values of n.

For more learning on dynamic programming methods, check this dynamic programming climbing stairs article.

Recursive Solution with Memoization in C++

We can use the recursive solution with memoization to solve the “Count Ways to Climb Stairs” problem in C++. This method helps us break the problem into simpler parts. We store the results of these parts so we do not calculate them again.

Problem Statement

We have n stairs. We can climb either 1 or 2 stairs at once. Our goal is to find the total number of different ways to reach the top.

Recursive Approach

We can define the recursive relation like this: - If n == 0, there is 1 way to stay on the ground. - If n == 1, there is 1 way to reach the first step. - If n == 2, there are 2 ways to reach the second step (1+1 or 2).

For n > 2, we have:

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

C++ Implementation

Here is the C++ code that uses the recursive solution with memoization:

#include <iostream>
#include <vector>

class ClimbStairs {
public:
    int countWays(int n) {
        std::vector<int> memo(n + 1, -1);
        return countWaysUtil(n, memo);
    }

private:
    int countWaysUtil(int n, std::vector<int>& memo) {
        // Base cases
        if (n == 0) return 1;
        if (n == 1) return 1;
        
        // Check if already computed
        if (memo[n] != -1) return memo[n];

        // Recursive case with memoization
        memo[n] = countWaysUtil(n - 1, memo) + countWaysUtil(n - 2, memo);
        return memo[n];
    }
};

int main() {
    ClimbStairs cs;
    int n = 5; // Example: 5 stairs
    std::cout << "Number of ways to climb " << n << " stairs: " << cs.countWays(n) << std::endl;
    return 0;
}

Explanation of Code

  • Memoization Array: We use a vector called memo. It stores results of subproblems. It starts with -1 to show the value is not computed yet.
  • Base Cases: We handle cases for 0 and 1 stairs directly.
  • Recursive Function: The function countWaysUtil checks if we already computed the result for n. If not, it calculates it using the recursive relation and saves it in the memo array.

This method makes the time taken much less than the simple recursive method. So it works better for larger values of n.

For more insights into dynamic programming solutions, you can check out Dynamic Programming - Fibonacci with Memoization.

Frequently Asked Questions

1. What is the climbing stairs problem in dynamic programming?

The climbing stairs problem is a well-known challenge in dynamic programming. We need to find the number of different ways to reach the top of a staircase that has n steps. You can go up either 1 step or 2 steps at a time. This problem is similar to the Fibonacci sequence because of how it works. For more information on related problems, look at our article on dynamic programming Fibonacci numbers.

2. How can I implement the dynamic programming approach in Python for climbing stairs?

To solve the climbing stairs problem with dynamic programming in Python, we need to make a list. This list will hold the number of ways to reach each step. We start with the first two steps. Then we fill the list using the rule: ways[i] = ways[i-1] + ways[i-2]. For a full code example, check out our Python section in the article on dynamic programming climbing stairs.

3. What is the time complexity of the dynamic programming solution for climbing stairs?

The time complexity of the dynamic programming solution to count ways to climb stairs is O(n). Here, n means the number of steps. This is because we calculate the number of ways for each step just one time and save the results. This makes our calculation fast. For more complex versions of this problem, look at our article on min cost climbing stairs.

4. Can I use recursion to solve the climbing stairs problem?

Yes, we can use a recursive method to solve the climbing stairs problem. But this way can lead to slow performance. This happens because of overlapping subproblems. To make it better, we can use memoization to save results we already calculated. For an example of this method in Java, go to our section on recursive solutions with memoization in the climbing stairs article.

5. How can I optimize space complexity when solving the climbing stairs problem?

To make space complexity better in the climbing stairs problem, we do not need to keep all the values for each step. We only need to remember the last two computed values. This changes the space complexity from O(n) to O(1). This makes our solution more efficient. For more ideas on other dynamic programming problems and how to save space, check our article on maximum subarray using Kadane’s algorithm.