[Dynamic Programming] Climbing Stairs - Easy

The Climbing Stairs problem is a well-known challenge in dynamic programming. It asks us to find out how many different ways we can reach the top of a staircase with a certain number of steps. We can climb one step or two steps at a time. Our goal is to figure out how many unique ways we can get to the top. We can solve this problem using different methods. These include recursive methods, memoization, and iterative dynamic programming.

In this article, we will look closely at the Climbing Stairs problem. We will discuss good solutions and different ways to solve it. We will talk about the recursive approach, dynamic programming with memoization, the bottom-up dynamic programming method, and a space-saving solution. We will also show how to implement the Climbing Stairs problem in Java, Python, and C++. At the end, we will answer some common questions about this problem.

  • Dynamic Programming Efficient Solutions for Climbing Stairs - Easy
  • Understanding the Climbing Stairs Problem
  • Recursive Approach to Climbing Stairs
  • Dynamic Programming Approach Using Memoization
  • Bottom-Up Dynamic Programming Approach
  • Space Optimized Dynamic Programming Solution
  • Climbing Stairs Problem in Java
  • Climbing Stairs Problem in Python
  • Climbing Stairs Problem in C++
  • Frequently Asked Questions

If you are interested in more topics about dynamic programming, you can check our articles on the Dynamic Programming Fibonacci Number and Dynamic Programming Fibonacci with Memoization. These articles will give you more insights into dynamic programming and how it works.

Understanding the Climbing Stairs Problem

The Climbing Stairs problem is a well-known challenge in dynamic programming. We can state the problem like this:

We have a staircase with n steps. We can take either 1 step or 2 steps at a time. Our goal is to find out how many different ways we can climb to the top.

Problem Breakdown:

  • Input: An integer n, which shows how many steps are in the staircase.
  • Output: An integer that tells us the total number of ways to reach the top.

Example:

  • If n = 2, the ways to climb are:
    1. 1 step + 1 step
    2. 2 steps So, the output is 2.
  • If n = 3, the ways are:
    1. 1 step + 1 step + 1 step
    2. 1 step + 2 steps
    3. 2 steps + 1 step The output is 3.

Relation to Fibonacci:

The number of ways to climb the stairs is similar to the Fibonacci sequence: - Ways to climb n stairs = Ways to climb n-1 stairs + Ways to climb n-2 stairs. - This is because the last step we take can be from step n-1 (1 step) or from n-2 (2 steps).

For more information on Fibonacci and how it is used in dynamic programming, check these links: Dynamic Programming Fibonacci Number - Easy and Dynamic Programming Fibonacci with Memoization - Easy.

Recursive Approach to Climbing Stairs

We can solve the Climbing Stairs problem using a simple recursive way. The problem is about finding how many ways we can reach the top of a staircase with n steps. We can take either 1 step or 2 steps at a time.

Recursive Formula

We can express the recursive relation like this:

  • If n == 0, we have 1 way to stay on the ground. We just do nothing.
  • If n == 1, we have 1 way to climb. We take one step.
  • If n >= 2, the number of ways to reach the n-th step is the sum of ways to reach the (n-1)-th step and the (n-2)-th step:

[ f(n) = f(n-1) + f(n-2) ]

Recursive Code Example

Here is a simple code in Python:

def climbStairs(n: int) -> int:
    if n == 0:
        return 1
    if n == 1:
        return 1
    return climbStairs(n - 1) + climbStairs(n - 2)

# Example usage
print(climbStairs(5))  # Output: 8

Performance Consideration

The recursive way has a time complexity of ( O(2^n) ). This happens because of overlapping subproblems.

This way is not good for big values of n. It recalculates values many times.

For a better solution, we can look into the Dynamic Programming Approach Using Memoization. This way uses storage to avoid doing the same calculations again. It really makes performance much better.

Dynamic Programming Approach Using Memoization

We can solve the Climbing Stairs problem well using a method called memoization. This method helps us improve the simple recursive solution. It does this by saving results we have already worked out. This way, we don’t have to do the same calculations again.

Problem Definition

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

Recursive Function with Memoization

With memoization, we save the results of smaller problems in an array. This lets us get the values back quickly when we need them.

Code Implementation

Here is how we can use the memoization approach in different programming languages:

Java:

public class ClimbingStairs {
    private int[] memo;

    public int climbStairs(int n) {
        memo = new int[n + 1];
        return climb(n);
    }

    private int climb(int n) {
        if (n <= 1) return 1;
        if (memo[n] != 0) return memo[n];
        memo[n] = climb(n - 1) + climb(n - 2);
        return memo[n];
    }
}

Python:

class ClimbingStairs:
    def __init__(self):
        self.memo = {}

    def climb_stairs(self, n: int) -> int:
        if n <= 1:
            return 1
        if n in self.memo:
            return self.memo[n]
        self.memo[n] = self.climb_stairs(n - 1) + self.climb_stairs(n - 2)
        return self.memo[n]

C++:

class ClimbingStairs {
public:
    int climbStairs(int n) {
        vector<int> memo(n + 1, 0);
        return climb(n, memo);
    }

private:
    int climb(int n, vector<int>& memo) {
        if (n <= 1) return 1;
        if (memo[n] != 0) return memo[n];
        memo[n] = climb(n - 1, memo) + climb(n - 2, memo);
        return memo[n];
    }
};

Time and Space Complexity

  • Time Complexity: O(n) - We calculate each step from 1 to n one time.
  • Space Complexity: O(n) - We need extra space for the memoization array.

This way helps us cut down the number of function calls compared to the simple recursive method. It makes it faster for bigger values of n.

If you want to learn more about dynamic programming methods, you can check out articles on the Fibonacci Number and Fibonacci with Memoization.

Bottom-Up Dynamic Programming Approach

The Bottom-Up Dynamic Programming approach helps us solve the Climbing Stairs problem. We build the solution step by step. We start from the basic cases. This way, we do not have to use recursion. Instead, we keep an array to store how many ways we can reach each step.

Implementation Steps:

  1. Initialization:
    • We create an array dp. In this array, dp[i] shows the number of ways to reach the i-th step.
    • We set our base cases. We have dp[0] = 1 which means there is 1 way to stay on the ground. And dp[1] = 1 means there is 1 way to go to the first step.
  2. Iterate:
    • We go through each step from 2 to n. For each step, we calculate how many ways we can reach it. We use this formula:
      • dp[i] = dp[i-1] + dp[i-2] This formula works because to get to the i-th step, we can come from the (i-1)-th step or the (i-2)-th step.
  3. Result:
    • The value of dp[n] tells us the total number of ways to climb to the n-th step.

Code Example (Python):

def climbStairs(n: int) -> int:
    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]

Code Example (Java):

public class ClimbingStairs {
    public int climbStairs(int n) {
        if (n <= 1) return 1;

        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;

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

        return dp[n];
    }
}

Code Example (C++):

class Solution {
public:
    int climbStairs(int n) {
        if (n <= 1) return 1;

        vector<int> dp(n + 1);
        dp[0] = 1;
        dp[1] = 1;

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

        return dp[n];
    }
};

This approach runs in O(n) time and needs O(n) space because of the dp array. It works well for bigger values of n compared to the simple recursive way. If we want a better way to save space, we can look at the Space Optimized Dynamic Programming Solution.

For more about dynamic programming, we can read the Fibonacci Number and Fibonacci with Memoization articles.

Space Optimized Dynamic Programming Solution

In the Climbing Stairs problem, we can make our dynamic programming solution use less space. We do this by reducing the number of variables we need. Instead of using a full array to keep track of the number of ways to get to each step, we only need to remember the last two values. Each step only depends on the last two steps.

Approach

  1. Use Two Variables: We use two variables to store the number of ways to reach the last two steps.
  2. Go Through Steps: We update these variables step by step until we reach the target step.

Code Implementation

Here is how we can write the space-optimized solution in different programming languages:

Java

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

Python

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

C++

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

Complexity Analysis

  • Time Complexity: O(n). This is because we go through the steps one time.
  • Space Complexity: O(1). We only use two variables for storage.

This space-optimized dynamic programming solution is fast and uses the properties of the Fibonacci sequence in the Climbing Stairs problem. For more about related dynamic programming methods, we can check Dynamic Programming - Fibonacci Number and Dynamic Programming - Fibonacci with Memoization.

Climbing Stairs Problem in Java

The Climbing Stairs problem is a well-known example of dynamic programming. It is about finding how many different ways we can climb to the top of a staircase with n steps. We can take either 1 step or 2 steps at a time.

Java Implementation

Here are some ways to solve the Climbing Stairs problem in Java.

Recursive Approach

The recursive method is simple but not good for big inputs. It has an exponential time complexity.

public class ClimbingStairs {
    public int climbStairs(int n) {
        if (n <= 1) {
            return 1;
        }
        return climbStairs(n - 1) + climbStairs(n - 2);
    }
}

Dynamic Programming with Memoization

With memoization, we can save results we already calculated. This helps to avoid doing the same work again.

public class ClimbingStairs {
    private int[] memo;

    public int climbStairs(int n) {
        memo = new int[n + 1];
        return climbStairsMemo(n);
    }

    private int climbStairsMemo(int n) {
        if (n <= 1) {
            return 1;
        }
        if (memo[n] != 0) {
            return memo[n];
        }
        memo[n] = climbStairsMemo(n - 1) + climbStairsMemo(n - 2);
        return memo[n];
    }
}

Bottom-Up Dynamic Programming Approach

This method uses a loop to build the solution from the simplest cases.

public class ClimbingStairs {
    public int climbStairs(int n) {
        if (n <= 1) return 1;
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;

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

Space Optimized Dynamic Programming Solution

This method saves space by only keeping track of the last two steps.

public class ClimbingStairs {
    public int climbStairs(int n) {
        if (n <= 1) return 1;
        int first = 1, second = 1, current = 0;

        for (int i = 2; i <= n; i++) {
            current = first + second;
            first = second;
            second = current;
        }
        return current;
    }
}

These methods show how we can solve the Climbing Stairs problem in Java. We use different ways, like recursive, memoization, bottom-up, and space-optimized techniques. For more about dynamic programming, you can read the articles on Dynamic Programming Fibonacci Number - Easy and Dynamic Programming Fibonacci with Memoization - Easy.

Climbing Stairs Problem in Python

The Climbing Stairs problem is a well-known challenge in dynamic programming. The problem says we can climb to the top of a staircase with n steps. We can take either 1 step or 2 steps at a time. Our goal is to find the number of different ways to reach the top.

Recursive Approach

A simple way to solve the Climbing Stairs problem is by using recursion. This method is easy to understand but not very efficient. It has a high time complexity because it does many repeated calculations.

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

Dynamic Programming Approach Using Memoization

To make the recursive method better, we can use memoization. This technique saves the results of costly function calls. We can use these results when the same inputs come again.

def climb_stairs_memoization(n, memo={}):
    if n <= 1:
        return 1
    if n not in memo:
        memo[n] = climb_stairs_memoization(n - 1, memo) + climb_stairs_memoization(n - 2, memo)
    return memo[n]

Bottom-Up Dynamic Programming Approach

The bottom-up method builds the solution step by step. It has a linear time complexity and uses O(n) space.

def climb_stairs_bottom_up(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]

Space Optimized Dynamic Programming Solution

To save space more, we can reduce the space complexity to O(1). We only need to keep track of the last two steps.

def climb_stairs_space_optimized(n):
    if n <= 1:
        return 1
    first, second = 1, 1
    for i in range(2, n + 1):
        first, second = second, first + second
    return second

Example Usage

We can call any of the above functions with a specific number of stairs:

n = 5
print(climb_stairs_recursive(n))          # Output: 8
print(climb_stairs_memoization(n))        # Output: 8
print(climb_stairs_bottom_up(n))          # Output: 8
print(climb_stairs_space_optimized(n))    # Output: 8

For more reading on similar problems, check Dynamic Programming: Fibonacci Number - Easy and Dynamic Programming: Fibonacci with Memoization - Easy.

Climbing Stairs Problem in C++

We can solve the Climbing Stairs problem in C++ in smart ways. The goal is to find out how many different ways we can climb to the top of a staircase with n steps. We can take either 1 or 2 steps at a time.

Recursive Approach

A simple way to solve this is with a recursive method. But this method has a big time complexity.

int climbStairs(int n) {
    if (n <= 1) return 1;
    return climbStairs(n - 1) + climbStairs(n - 2);
}

Dynamic Programming Approach Using Memoization

This method improves the recursive solution by saving the results we already calculated. This way we avoid doing the same work again.

#include <vector>

int climbStairsMemo(int n, std::vector<int>& memo) {
    if (n <= 1) return 1;
    if (memo[n] != -1) return memo[n];
    memo[n] = climbStairsMemo(n - 1, memo) + climbStairsMemo(n - 2, memo);
    return memo[n];
}

int climbStairs(int n) {
    std::vector<int> memo(n + 1, -1);
    return climbStairsMemo(n, memo);
}

Bottom-Up Dynamic Programming Approach

This method builds the solution from the bottom up. It uses less space than the recursive method with memoization.

int climbStairs(int n) {
    if (n <= 1) return 1;
    std::vector<int> dp(n + 1);
    dp[0] = 1;
    dp[1] = 1;
    for (int i = 2; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    return dp[n];
}

Space Optimized Dynamic Programming Solution

We can make it even better by using just two variables to keep the last two results. This way we save more space.

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

These ways help us solve the Climbing Stairs problem in C++ very well. They use dynamic programming for better performance. If you want to learn more about dynamic programming, you can check out the Dynamic Programming Fibonacci Number and Fibonacci with Memoization articles.

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. It asks how many different ways we can climb to the top of a staircase with n steps. We can take either 1 step or 2 steps at a time. We can solve this problem in many ways. Some methods include recursion, memoization, and bottom-up dynamic programming.

2. How does the recursive approach work for the Climbing Stairs problem?

In the recursive approach, we define the solution using smaller problems. We can say that the number of ways to climb n stairs equals the ways to climb n-1 stairs plus the ways to climb n-2 stairs. This method is simple but not very efficient. It can take a lot of time because of repeated calculations.

3. What is memoization, and how does it optimize the Climbing Stairs solution?

Memoization is a way to make our solution faster. It stores results of expensive function calls. When the same inputs come again, it uses the saved results. For the Climbing Stairs problem, using memoization changes the time needed from a lot to just a little. We save values we calculated before, so we don’t do the same work again.

4. Can you explain the bottom-up dynamic programming approach for Climbing Stairs?

The bottom-up dynamic programming approach builds the solution step by step. We start from the base cases and calculate the number of ways to climb each step up to n. We use results we got before. This method keeps results in an array. It has a time complexity of O(n) and space complexity of O(n). We can also make it use less space.

5. How can I implement the Climbing Stairs problem in different programming languages?

We can implement the Climbing Stairs problem in many programming languages like Java, Python, and C++. Each language has its own way, but the main idea stays the same. We can find example codes and more details for Java here. For Python and C++, you can check their parts in this article.

By answering these common questions, we can understand the Climbing Stairs problem better. This helps improve our programming skills and knowledge of algorithms.