[Dynamic Programming] Staircase Problem with 1,2,3 Steps - Easy

The Staircase Problem with steps of 1, 2, or 3

The Staircase Problem with steps of 1, 2, or 3 is a well-known problem in dynamic programming. It asks how many different ways we can reach the top of a staircase that has n steps. To find the answer, we see that the number of ways to get to the nth step is the total of the ways to get to the (n-1)th, (n-2)th, and (n-3)th steps. This gives us a formula we can use to solve the problem with dynamic programming.

In this article, we will look closely at the Staircase Problem with 1, 2, and 3 steps. We will explain what the problem means. We will also show a dynamic programming method and give recursive solutions in Java, Python, and C++. Plus, we will answer some common questions to help you understand better. Here is what we will talk about:

  • Dynamic Programming Solution to Staircase Problem with 1 2 3 Steps
  • Understanding the Staircase Problem with 1 2 3 Steps
  • Dynamic Programming Approach for Staircase Problem
  • Recursive Solution for Staircase Problem in Java
  • Dynamic Programming Solution for Staircase Problem in Java
  • Recursive Solution for Staircase Problem in Python
  • Dynamic Programming Solution for Staircase Problem in Python
  • Recursive Solution for Staircase Problem in C++
  • Dynamic Programming Solution for Staircase Problem in C++
  • Frequently Asked Questions

If you want to learn more about dynamic programming, we have some articles for you: Dynamic Programming Fibonacci Number, Dynamic Programming Fibonacci with Memoization, Dynamic Programming Climbing Stairs, and Dynamic Programming Min Cost Climbing Stairs.

Understanding the Staircase Problem with 1 2 3 Steps

The Staircase Problem is about finding how many different ways we can climb a staircase with n steps. We can take 1, 2, or 3 steps at a time. We can use dynamic programming to solve this problem. The number of ways to get to step n depends on the number of ways to get to the last three steps.

Problem Definition

  • Input: An integer n, which is the total number of steps.
  • Output: The number of different ways to reach the top of the stairs.

Recurrence Relation

We can define the number of ways to reach step n like this: [ ways(n) = ways(n-1) + ways(n-2) + ways(n-3) ] We have some base cases: - ( ways(0) = 1 ) (There is 1 way to stay on the ground) - ( ways(1) = 1 ) (There is 1 way to reach the first step) - ( ways(2) = 2 ) (There are 2 ways: (1+1) or (2)) - ( ways(3) = 4 ) (There are 4 ways: (1+1+1), (1+2), (2+1), (3))

This relation is important for both recursive and dynamic programming solutions.

Example

If we have n = 4, the different ways to reach the fourth step are: - From step 3 (1 step) - From step 2 (2 steps) - From step 1 (3 steps)

So, we combine the ways to get to the last three steps: [ ways(4) = ways(3) + ways(2) + ways(1) = 4 + 2 + 1 = 7 ]

The combinations for n = 4 are: - (1, 1, 1, 1) - (1, 1, 2) - (1, 2, 1) - (2, 1, 1) - (2, 2) - (1, 3) - (3, 1)

This problem is a well-known example of dynamic programming. We can implement it easily in different programming languages.

For more reading, we can check out related articles about Fibonacci numbers and climbing stairs: Fibonacci Number - Easy and Climbing Stairs - Easy.

Dynamic Programming Approach for Staircase Problem

We can solve the Staircase Problem with 1, 2, and 3 steps using dynamic programming. Our goal is to find how many different ways we can reach the top of a staircase with n steps. We can take 1, 2, or 3 steps at a time.

Problem Definition

Let ways(n) be the number of ways to get to the nth step. We can write the recurrence relation like this:

ways(n) = ways(n-1) + ways(n-2) + ways(n-3)

This means to reach the nth step, we can come from the (n-1)th, (n-2)th, or (n-3)th steps.

Base Cases

  • ways(0) = 1: There is one way to stay on the ground. We do nothing.
  • ways(1) = 1: There is only one way to step to the first step.
  • ways(2) = 2: There are two ways (1+1 or 2).
  • ways(3) = 4: There are four ways (1+1+1, 1+2, 2+1, 3).

Dynamic Programming Implementation

We can show this approach in different programming languages. Below, we have the implementations in Java, Python, and C++.

Java Implementation

public class Staircase {
    public static int countWays(int n) {
        if (n == 0) return 1;
        if (n == 1) return 1;
        if (n == 2) return 2;
        
        int[] dp = new int[n + 1];
        dp[0] = 1; dp[1] = 1; dp[2] = 2;
        
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
        }
        
        return dp[n];
    }
}

Python Implementation

def count_ways(n):
    if n == 0:
        return 1
    if n == 1:
        return 1
    if n == 2:
        return 2
    
    dp = [0] * (n + 1)
    dp[0], dp[1], dp[2] = 1, 1, 2
    
    for i in range(3, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]
    
    return dp[n]

C++ Implementation

#include <iostream>
#include <vector>
using namespace std;

int countWays(int n) {
    if (n == 0) return 1;
    if (n == 1) return 1;
    if (n == 2) return 2;

    vector<int> dp(n + 1);
    dp[0] = 1; dp[1] = 1; dp[2] = 2;

    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
    }
    
    return dp[n];
}

Complexity Analysis

  • Time Complexity: O(n), because we find solution for each step from 0 to n.
  • Space Complexity: O(n) for keeping the number of ways to reach each step.

Using dynamic programming, we can solve the Staircase Problem with 1, 2, and 3 steps fast. This method is much better than a simple recursive approach. For more information about dynamic programming, we can check articles on Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.

Recursive Solution for Staircase Problem in Java

The Staircase Problem is about finding how many different ways we can climb to the top of a staircase with n steps. We can take 1, 2, or 3 steps at a time. The recursive way helps us to break this problem into smaller parts.

Recursive Function

We can define the recursive function like this:

public class StaircaseProblem {
    public static int countWays(int n) {
        // Base cases
        if (n < 0) return 0;
        if (n == 0) return 1;
        
        // Recursive cases
        return countWays(n - 1) + countWays(n - 2) + countWays(n - 3);
    }

    public static void main(String[] args) {
        int n = 4; // Example number of steps
        System.out.println("Number of ways to climb " + n + " steps: " + countWays(n));
    }
}

Explanation

  • Base Cases:
    • If there are no steps (n == 0), we have one way to stay on the ground which is to do nothing.
    • If n is negative, we have no valid ways to climb.
  • Recursive Cases:
    • The function calls itself with n-1, n-2, and n-3. This shows we can take 1, 2, or 3 steps from where we are.

Complexity

  • Time Complexity: O(3^n) because each time we call the function, it makes three new calls.
  • Space Complexity: O(n) because of the call stack.

This recursive method is easy to understand but it can be slow for big values of n. If we want a better solution, we can think about using dynamic programming or memoization. For more on dynamic programming, we can look at the Dynamic Programming Climbing Stairs article.

Dynamic Programming Solution for Staircase Problem in Java

The Staircase Problem is about finding how many different ways we can climb to the top of a staircase with n steps. We can take 1, 2, or 3 steps at a time. With a dynamic programming method, we can quickly find the number of ways to reach the top.

Dynamic Programming Approach

  1. State Definition: We define dp[i] as the number of ways to reach the i-th step.

  2. Recurrence Relation: We can write the relationship like this: [ dp[i] = dp[i-1] + dp[i-2] + dp[i-3] ] Here:

    • dp[i-1] shows the ways to reach the i-th step from the (i-1)-th step.
    • dp[i-2] shows the ways to reach the i-th step from the (i-2)-th step.
    • dp[i-3] shows the ways to reach the i-th step from the (i-3)-th step.

  3. Base Cases:

    • dp[0] = 1 (1 way to stay on the ground)
    • dp[1] = 1 (1 way to reach the first step)
    • dp[2] = 2 (2 ways to reach the second step: (1,1) or (2))
    • dp[3] = 4 (4 ways to reach the third step: (1,1,1), (1,2), (2,1), (3))

Java Code Implementation

public class StaircaseProblem {
    public static int countWays(int n) {
        if (n == 0) return 1;
        if (n == 1) return 1;
        if (n == 2) return 2;
        if (n == 3) return 4;

        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        dp[2] = 2;
        dp[3] = 4;

        for (int i = 4; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
        }
        return dp[n];
    }

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

Explanation of Code

  • The countWays method counts how many different ways we can reach step n.
  • We use an array dp to keep the number of ways to reach each step.
  • The for loop fills the dp array based on the relation we defined.
  • Finally, it returns dp[n], which is the total number of ways to climb to the n-th step.

This dynamic programming solution is efficient. It has a time complexity of (O(n)) and a space complexity of (O(n)). For more details on similar problems, check articles on Climbing Stairs and Fibonacci Numbers.

Recursive Solution for Staircase Problem in Python

The Staircase Problem is about finding how many ways we can climb to the top of a staircase with n steps. We can take 1, 2, or 3 steps at a time. We can use a recursive way to solve this problem.

Recursive Function

The main idea of the recursive solution is simple. For any step n, the number of ways to get to that step is the sum of the ways to get to the three steps before it:

  • ways(n) = ways(n-1) + ways(n-2) + ways(n-3)

We have some base cases: - ways(0) = 1 (one way to stay on the ground) - ways(1) = 1 (one way to reach the first step) - ways(2) = 2 (two ways to reach the second step: (1+1) or (2)) - ways(3) = 4 (four ways to reach the third step: (1+1+1), (1+2), (2+1), (3))

Python Code Implementation

Here is a simple Python code for the recursive solution:

def countWays(n):
    if n < 0:
        return 0
    if n == 0:
        return 1
    if n == 1:
        return 1
    if n == 2:
        return 2
    if n == 3:
        return 4
    
    return countWays(n-1) + countWays(n-2) + countWays(n-3)

# Example usage
steps = 5
print(f"Number of ways to climb {steps} steps: {countWays(steps)}")

This function counts the number of ways to climb n steps using a simple recursive method. But this method has exponential time complexity because it does repeated calculations for the same step counts.

For bigger values of n, we should think about using memoization or dynamic programming to make the solution better. The recursive way helps us understand the problem clearly. But for better performance, we can check the Dynamic Programming Approach for Staircase Problem.

For more information on similar problems, we can look at the Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.

Dynamic Programming Solution for Staircase Problem in Python

The Staircase Problem is about finding how many ways we can reach the top of a staircase with n steps. We can take 1, 2, or 3 steps at a time. We can use dynamic programming to solve this problem. This method helps us store results so we do not do the same calculations over and over.

Dynamic Programming Approach

  1. State Definition: We say dp[i] is the number of ways to reach the i-th step.
  2. Recurrence Relation:
    • We can define the relation like this: [ dp[i] = dp[i-1] + dp[i-2] + dp[i-3] ]
    • Base Cases:
      • dp[0] = 1 (1 way to stay at the ground)
      • dp[1] = 1 (1 way to reach the first step)
      • dp[2] = 2 (2 ways: (1,1) or (2))
      • dp[3] = 4 (4 ways: (1,1,1), (1,2), (2,1), (3))

Python Code Implementation

def countWays(n):
    if n == 0:
        return 1
    elif n == 1:
        return 1
    elif n == 2:
        return 2
    elif n == 3:
        return 4

    dp = [0] * (n + 1)
    dp[0], dp[1], dp[2], dp[3] = 1, 1, 2, 4

    for i in range(4, n + 1):
        dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3]

    return dp[n]

# Example usage
n = 5
print("Number of ways to reach the top:", countWays(n))

Explanation of the Code

  • The function countWays calculates the number of ways to reach the top of a staircase with n steps.
  • It starts by creating an array dp to keep the values we computed.
  • The loop goes from 4 to n, and it updates the dp array using the relation we defined.
  • At the end, it gives us the value of dp[n], which is the number of ways to climb the staircase.

This dynamic programming solution is fast. It has a time complexity of (O(n)) and space complexity of (O(n)).

For more information on similar problems, we can check the Dynamic Programming: Climbing Stairs article.

Recursive Solution for Staircase Problem in C++

We can solve the Staircase Problem using a recursive method. This method helps us find how many different ways we can reach the top of the stairs by taking 1, 2, or 3 steps at a time.

We have some base cases. These are when the number of steps is zero, one, or two. In these cases, we can return the answer directly. For other numbers, we will call the function again to add the ways from the last three steps.

Here is a simple C++ code for the recursive solution:

#include <iostream>
using namespace std;

int countWays(int n) {
    // Base cases
    if (n < 0) return 0; // No way to step on negative stairs
    if (n == 0) return 1; // One way to stay at the ground (do nothing)
    if (n == 1) return 1; // One way to take one step
    if (n == 2) return 2; // Two ways: (1+1) or (2)

    // Recursive case: sum the ways from the last three steps
    return countWays(n - 1) + countWays(n - 2) + countWays(n - 3);
}

int main() {
    int n = 5; // Example: number of steps
    cout << "Number of ways to reach the top: " << countWays(n) << endl;
    return 0;
}

Explanation of the Code:

  • countWays Function: This function finds the number of ways to reach the nth step.
  • Base Cases:
    • If n < 0, we return 0 (no steps).
    • If n == 0, we return 1 (one way to stay on the ground).
    • If n == 1, we return 1 (only one way to step).
    • If n == 2, we return 2 (we can take two steps of one or one step of two).
  • Recursive Call: If n > 2, we add the results of the last three steps.

This recursive method is not very fast. It has an exponential time of O(3^n). This happens because of overlapping problems. We can make it better using dynamic programming. In the next section, we will discuss a more efficient way to solve this problem.

Dynamic Programming Solution for Staircase Problem in C++

To solve the Staircase Problem with 1, 2, or 3 steps using dynamic programming in C++, we can say it like this: Given n steps, the number of ways to reach the top is the total of the ways to reach the top from the last three steps. We can write this as:

ways(n) = ways(n-1) + ways(n-2) + ways(n-3)

We also need some base cases: - ways(0) = 1: There is one way to stay on the ground (do nothing). - ways(1) = 1: One way to reach the first step. - ways(2) = 2: Two ways to reach the second step (1+1 or 2). - ways(3) = 4: Four ways to reach the third step (1+1+1, 1+2, 2+1, 3).

Here is the C++ code for the dynamic programming solution:

#include <iostream>
#include <vector>

using namespace std;

int countWays(int n) {
    if (n == 0) return 1;
    if (n == 1) return 1;
    if (n == 2) return 2;
    
    vector<int> dp(n + 1);
    dp[0] = 1;
    dp[1] = 1;
    dp[2] = 2;

    for (int i = 3; i <= n; i++) {
        dp[i] = dp[i - 1] + dp[i - 2] + dp[i - 3];
    }
    
    return dp[n];
}

int main() {
    int n;
    cout << "Enter the number of steps: ";
    cin >> n;
    cout << "Number of ways to climb to the top: " << countWays(n) << endl;
    return 0;
}

In this code: - We make a dynamic programming array dp where dp[i] saves the number of ways to reach the i-th step. - The loop goes from step 3 to n, filling in the number of ways based on the last three steps. - Finally, the function gives back the number of ways to reach the n-th step.

This dynamic programming method gives us a good solution with a time complexity of O(n) and a space complexity of O(n). If we want to read more about similar dynamic programming problems, we can check out Dynamic Programming: Climbing Stairs.

Frequently Asked Questions

What is the Staircase Problem with 1, 2, and 3 Steps?

The Staircase Problem with 1, 2, and 3 steps is about finding how many ways we can reach the top of a staircase. We can take either 1, 2, or 3 steps at a time. We can solve this problem well using dynamic programming. This method helps us keep track of results we already calculated. For more details, we can check this dynamic programming example.

How do I implement a recursive solution for the Staircase Problem in Java?

To create a recursive solution for the Staircase Problem in Java, we can make a function that counts the total ways to climb the stairs. It does this by adding the ways to get to the last three steps. This method is simple and helps us understand the problem. But it may not work well for big numbers because it does the same calculations many times. For more detailed ways, we can look into dynamic programming too.

What are the advantages of using dynamic programming for the Staircase Problem?

Using dynamic programming for the Staircase Problem helps us save time compared to just using recursion. We store results in a table. This way, we only calculate each step once. This leads to a faster O(n) time complexity. This speed is very important when we work with larger numbers, so many people like to use this method for the Staircase Problem.

Can the Staircase Problem be solved using Python?

Yes, we can solve the Staircase Problem easily with Python. We can use both recursive and dynamic programming methods in Python. Python is simple and easy to read. In dynamic programming, we create a list to keep track of the number of ways to reach each step. This helps us compute things quickly and perform better. For examples, we can check our dynamic programming solutions in Python.

Is there a relation between the Staircase Problem and the Fibonacci sequence?

Yes, there is a link between the Staircase Problem and the Fibonacci sequence. The ways to climb stairs with 1, 2, or 3 steps can come from a changed Fibonacci sequence. In this case, each number is the sum of the last three numbers. This shows how dynamic programming can help solve many problems, including those with the Fibonacci sequence. For more about Fibonacci in dynamic programming, we can look at this Fibonacci with memoization.