[Dynamic Programming] Count Ways to Climb Stairs with Step Size Constraints - Medium

To count how many ways we can climb stairs with certain step sizes, we can use dynamic programming. This problem is about finding how many different paths we can take to reach the top of a staircase with limits on the sizes of the steps we can take. By breaking down the problem into smaller parts and keeping track of results, we can find out the total ways to reach the top.

In this article, we will look at how to count the ways to climb stairs with step size limits using dynamic programming. We will explain the problem, show different dynamic programming methods in Java, Python, and C++, and talk about how to save space. We will also check out recursive methods with memoization, iterative methods, and why testing and checking our solutions is important. Here are the topics we will discuss:

  • Dynamic Programming Count Ways to Climb Stairs with Step Size Constraints Solution Overview
  • Understanding the Problem Statement for Counting Ways to Climb Stairs
  • Dynamic Programming Approach to Count Ways to Climb Stairs in Java
  • Dynamic Programming Approach to Count Ways to Climb Stairs in Python
  • Dynamic Programming Approach to Count Ways to Climb Stairs in C++
  • Optimizing Space Complexity in Dynamic Programming for Stair Climbing
  • Recursive Approach with Memoization for Counting Stair Climbing Ways
  • Iterative Approach to Count Ways to Climb Stairs with Step Size Constraints
  • Testing and Validating the Stair Climbing Solutions
  • Frequently Asked Questions

For more reading on dynamic programming ideas, you can check articles like Dynamic Programming: Fibonacci Number and Dynamic Programming: Climbing Stairs.

Understanding the Problem Statement for Counting Ways to Climb Stairs

We want to count how many ways a person can climb stairs. The person can take steps of different sizes. For example, if there are n steps, they might be able to take 1, 2, or up to k steps at once.

Problem Definition:

  • Input:
    • An integer n. This is the total number of steps.
    • A list of integers. These are the allowed step sizes, like [1, 2, 3].
  • Output:
    • An integer that shows the total number of ways to reach the top of the stairs.

Example:

  • If n = 4 and the allowed step sizes are [1, 2], the ways to reach the top are:
    • (1, 1, 1, 1)
    • (1, 1, 2)
    • (1, 2, 1)
    • (2, 1, 1)
    • (2, 2)

So, the output will be 5.

Constraints:

  • The values for n should be non-negative integers.
  • The step size list must not be empty and should have positive integers.

We need to understand this problem to make a good dynamic programming solution. This helps us count the ways to climb stairs with given step sizes. A clear plan can make our calculations easier. It also helps us deal with overlaps in subproblems. This is very important in dynamic programming.

If we want to learn more about this topic, we can check out the Dynamic Programming: Count Ways to Climb Stairs with Step Size Constraints for more information.

Dynamic Programming Approach to Count Ways to Climb Stairs in Java

We can count the ways to climb stairs with step size limits using dynamic programming in Java. We will use an array to keep track of how many ways we can reach each step. We build the solution from the base cases.

Problem Definition

We have n stairs and can take steps from a set {1, 2, ..., k}. We want to find the total number of different ways to reach the top.

Solution Approach

  1. Base Case:
    • There is 1 way to stay on the ground (0 stairs).
    • If there are no stairs (n = 0), the answer is 1.
  2. Recurrence Relation:
    • To reach the i-th step, we can come from the (i-1)-th, (i-2)-th, …, or (i-k)-th steps. So, the relation becomes: [ dp[i] = dp[i-1] + dp[i-2] + … + dp[i-k] ]
    • We will go from 1 to n, calculating dp[i] using values we already found.

Java Implementation

public class ClimbStairs {
    public static int countWays(int n, int k) {
        // Base case
        if (n == 0) return 1;
        if (n < 0) return 0;

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

        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= k; j++) {
                if (i - j >= 0) {
                    dp[i] += dp[i - j];
                }
            }
        }
        return dp[n];
    }

    public static void main(String[] args) {
        int n = 5; // Number of stairs
        int k = 2; // Maximum step size
        System.out.println("Total ways to climb " + n + " stairs with max step " + k + ": " + countWays(n, k));
    }
}

Explanation of the Code

The countWays function starts a dynamic programming array dp. Here, dp[i] shows how many ways we can reach the i-th stair.

The outer loop goes through the stairs. The inner loop adds ways from previous stairs based on the maximum step size k.

Finally, the result is in dp[n], which we print in the main method.

This dynamic programming way finds the number of ways to climb stairs by building up solutions to smaller problems. This saves time and resources. For more information, you can check related articles like Count Ways to Climb Stairs - Easy.

Dynamic Programming Approach to Count Ways to Climb Stairs in Python

In Python, we solve the problem of counting ways to climb stairs using dynamic programming. We define a function that uses a dynamic programming array to keep track of the number of ways to reach each step.

Problem Definition

We have n stairs and we can take steps of size 1 or 2. Our goal is to find the total number of different ways to reach the top of the stairs.

Dynamic Programming Solution

  1. Initialization: We create an array dp. Here, dp[i] shows the number of ways to reach the i-th step. We set dp[0] = 1 because there is one way to stay on the ground. We also set dp[1] = 1 since there is one way to reach the first step.

  2. Recurrence Relation: For each step i from 2 to n, we calculate: [ dp[i] = dp[i-1] + dp[i-2] ] This means the number of ways to reach step i is the sum of the ways to reach the two steps before it.

  3. Result: The value at dp[n] gives us the total number of ways to climb n stairs.

Python Code Implementation

def count_ways(n):
    if n == 0:
        return 1
    if n == 1:
        return 1
    
    dp = [0] * (n + 1)
    dp[0] = 1  # One way to stay on the ground
    dp[1] = 1  # One way to reach the first step

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

    return dp[n]

# Example Usage
n = 5
print(f"Number of ways to climb {n} stairs: {count_ways(n)}")

Complexity Analysis

  • Time Complexity: O(n). We go through the stairs one time.
  • Space Complexity: O(n). We need space for the dp array.

Space Optimization

We can make it better by using only the last two values instead of the whole array:

def count_ways_optimized(n):
    if n == 0:
        return 1
    if n == 1:
        return 1

    prev1, prev2 = 1, 1

    for i in range(2, n + 1):
        current = prev1 + prev2
        prev2 = prev1
        prev1 = current

    return prev1

# Example Usage
n = 5
print(f"Number of ways to climb {n} stairs with optimized space: {count_ways_optimized(n)}")

This version uses O(1) space but keeps O(n) time complexity. If you want to learn more about dynamic programming, you can check out similar problems like Dynamic Programming Fibonacci Number.

Dynamic Programming Approach to Count Ways to Climb Stairs in C++

To solve the problem of counting ways to climb stairs with step size limits, we can use dynamic programming in C++. We will use a bottom-up approach. This method helps us find the total number of ways to reach the top of the stairs based on allowed step sizes.

Problem Statement

We have n stairs and a list of allowed step sizes. Our goal is to count how many different ways we can reach the top.

Dynamic Programming Solution

We will create a dynamic programming array called dp. In this array, dp[i] shows the number of ways to reach the i-th stair. To find the number of ways to reach stair i, we add the ways to reach previous stairs based on the allowed steps.

Here is a C++ code example:

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

int countWays(int n, vector<int>& steps) {
    vector<int> dp(n + 1, 0);
    dp[0] = 1; // base case: 1 way to stay at the ground

    for (int i = 1; i <= n; ++i) {
        for (int step : steps) {
            if (i - step >= 0) {
                dp[i] += dp[i - step];
            }
        }
    }
    
    return dp[n];
}

int main() {
    int n = 5; // number of stairs
    vector<int> steps = {1, 2}; // allowed steps
    cout << "Total ways to climb " << n << " stairs: " << countWays(n, steps) << endl;
    return 0;
}

Explanation of the Code

  1. We start by making a vector dp with size n + 1 and set all values to zero. We set dp[0] to 1. This means there is one way to stay on the ground.
  2. We go through each stair from 1 to n.
  3. For each stair, we check each allowed step size. If the previous stair is not negative, we add the ways from the previous stair to the current stair.
  4. In the end, dp[n] gives us the total number of ways to reach the top.

This dynamic programming way helps us count the number of ways to climb stairs while thinking of step size limits. The time complexity is O(n * m), where m is the number of allowed steps.

For more about dynamic programming, you can read Dynamic Programming Count Ways to Climb Stairs with Constraints.

Optimizing Space Complexity in Dynamic Programming for Stair Climbing

When we solve the problem of counting ways to climb stairs with step size limits using dynamic programming, we need to optimize space complexity. This helps us perform better, especially with big inputs. The usual dynamic programming method keeps an array to store results for smaller problems. But we can use less space by looking at the problem structure.

Key Observations

  • The ways to reach the n-th step only depend on the last few steps we calculated. This means we only need to look at the last k steps, where k is the biggest step size allowed.
  • So, we do not need to save results for all steps before, just the last k results.

Space Optimization Approach

  1. Use a Fixed-Size Array: Instead of making a big DP array of size n, we can use a smaller array of size k + 1. This will store only the results we need.
  2. Sliding Window Technique: When we find the number of ways to reach each step, we update the array in a circular way. This means we overwrite older values that we do not need anymore.

Implementation Example in Java

public class StairClimbing {
    public int countWays(int n, int[] steps) {
        int k = steps.length;
        int[] dp = new int[k + 1];
        dp[0] = 1; // base case
        
        for (int i = 1; i <= n; i++) {
            int current = 0;
            for (int j = 0; j < k; j++) {
                if (i - steps[j] >= 0) {
                    current += dp[(i - steps[j]) % (k + 1)];
                }
            }
            dp[i % (k + 1)] = current;
        }
        
        return dp[n % (k + 1)];
    }
}

Implementation Example in Python

def count_ways(n, steps):
    k = len(steps)
    dp = [0] * (k + 1)
    dp[0] = 1  # base case
    
    for i in range(1, n + 1):
        current = 0
        for step in steps:
            if i - step >= 0:
                current += dp[(i - step) % (k + 1)]
        dp[i % (k + 1)] = current
    
    return dp[n % (k + 1)]

Implementation Example in C++

#include <vector>
using namespace std;

class Solution {
public:
    int countWays(int n, vector<int>& steps) {
        int k = steps.size();
        vector<int> dp(k + 1, 0);
        dp[0] = 1; // base case
        
        for (int i = 1; i <= n; i++) {
            int current = 0;
            for (int j = 0; j < k; j++) {
                if (i - steps[j] >= 0) {
                    current += dp[(i - steps[j]) % (k + 1)];
                }
            }
            dp[i % (k + 1)] = current;
        }
        
        return dp[n % (k + 1)];
    }
};

Summary of Benefits

  • Reduced Space Usage: We change the space usage from O(n) to O(k), where k is the maximum step size.
  • Efficiency: This method can make our program much faster for larger inputs. It is good for competitive programming and real-world cases.

This way to optimize fits well with dynamic programming principles for counting ways to climb stairs easily. For more info on dynamic programming methods, we can check articles like Dynamic Programming Fibonacci Number and Dynamic Programming Count Ways to Climb Stairs.

Recursive Approach with Memoization for Counting Stair Climbing Ways

We can use the recursive approach with memoization to count how many ways we can climb stairs. This method helps us break the problem down into smaller parts. We also store results we have already calculated. This way, we do not do the same work twice.

Problem Definition

We have n stairs and a list of allowed step sizes. Our job is to find the total number of different ways to reach the top.

Recursive Function

We can define the recursive function like this:

  • Base Case:
    • If n == 0, we return 1 (one way to stay on the ground).
    • If n < 0, we return 0 (no way to climb).
  • Recursive Case:
    • For each step size in the allowed steps, we calculate the number of ways to reach the top from n - step.

Implementation

Now let’s see how we can implement this recursive approach with memoization in Python, Java, and C++.

Python

def countWays(n, allowed_steps):
    memo = {}

    def recurse(steps_left):
        if steps_left in memo:
            return memo[steps_left]
        if steps_left == 0:
            return 1
        if steps_left < 0:
            return 0
        
        total_ways = 0
        for step in allowed_steps:
            total_ways += recurse(steps_left - step)
        
        memo[steps_left] = total_ways
        return total_ways

    return recurse(n)

# Example usage
n = 5
allowed_steps = [1, 2]
print(countWays(n, allowed_steps))  # Output: number of ways to climb 5 stairs

Java

import java.util.HashMap;

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

    public static int countWays(int n, int[] allowedSteps) {
        if (memo.containsKey(n)) return memo.get(n);
        if (n == 0) return 1;
        if (n < 0) return 0;

        int totalWays = 0;
        for (int step : allowedSteps) {
            totalWays += countWays(n - step, allowedSteps);
        }
        memo.put(n, totalWays);
        return totalWays;
    }

    public static void main(String[] args) {
        int n = 5;
        int[] allowedSteps = {1, 2};
        System.out.println(countWays(n, allowedSteps));  // Output: number of ways to climb 5 stairs
    }
}

C++

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

class StairClimbing {
public:
    unordered_map<int, int> memo;

    int countWays(int n, vector<int>& allowedSteps) {
        if (memo.find(n) != memo.end()) return memo[n];
        if (n == 0) return 1;
        if (n < 0) return 0;

        int totalWays = 0;
        for (int step : allowedSteps) {
            totalWays += countWays(n - step, allowedSteps);
        }
        memo[n] = totalWays;
        return totalWays;
    }
};

int main() {
    StairClimbing sc;
    int n = 5;
    vector<int> allowedSteps = {1, 2};
    cout << sc.countWays(n, allowedSteps);  // Output: number of ways to climb 5 stairs
}

Complexity

  • Time Complexity: O(n * m), where n is the number of stairs and m is the number of allowed step sizes.
  • Space Complexity: O(n) because of the memoization storage.

This approach helps us save time. We avoid doing the same calculations again. We use recursion to look at all ways to climb stairs with the steps we can take.

Iterative Approach to Count Ways to Climb Stairs with Step Size Constraints

We can count the number of ways to climb stairs with step size limits using an iterative method. This way is more efficient than using recursion. We will use a dynamic programming table to keep track of results. Then we build the solution step by step.

Problem Definition

We have n stairs and some allowed step sizes. Our goal is to find out how many different ways we can climb to the top of the stairs using these steps.

Iterative Solution

The main idea is to use an array dp. Here, dp[i] shows how many ways we can reach the i-th stair. We set the first values like this: - dp[0] = 1: There is one way to stay on the ground (do nothing). - For each stair i, we calculate the ways to reach it by adding the ways to reach each of the last k stairs (where k is the biggest step size).

Java Implementation

public class ClimbStairs {
    public static int countWays(int n, int[] steps) {
        int[] dp = new int[n + 1];
        dp[0] = 1; // Base case

        for (int i = 1; i <= n; i++) {
            for (int step : steps) {
                if (i - step >= 0) {
                    dp[i] += dp[i - step];
                }
            }
        }
        return dp[n];
    }

    public static void main(String[] args) {
        int n = 5; // Total number of stairs
        int[] steps = {1, 2}; // Allowed steps
        System.out.println("Number of ways to climb stairs: " + countWays(n, steps));
    }
}

Python Implementation

def count_ways(n, steps):
    dp = [0] * (n + 1)
    dp[0] = 1  # Base case
    
    for i in range(1, n + 1):
        for step in steps:
            if i - step >= 0:
                dp[i] += dp[i - step]
    return dp[n]

n = 5  # Total number of stairs
steps = [1, 2]  # Allowed steps
print("Number of ways to climb stairs:", count_ways(n, steps))

C++ Implementation

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

int countWays(int n, vector<int>& steps) {
    vector<int> dp(n + 1, 0);
    dp[0] = 1; // Base case

    for (int i = 1; i <= n; i++) {
        for (int step : steps) {
            if (i - step >= 0) {
                dp[i] += dp[i - step];
            }
        }
    }
    return dp[n];
}

int main() {
    int n = 5; // Total number of stairs
    vector<int> steps = {1, 2}; // Allowed steps
    cout << "Number of ways to climb stairs: " << countWays(n, steps) << endl;
    return 0;
}

Explanation of the Code

  • The outer loop goes through each stair from 1 to n.
  • The inner loop checks each allowed step size.
  • If we can use the step size (that is i - step >= 0), we add the number of ways to get to i - step to dp[i].
  • Finally, dp[n] gives us the total number of ways to reach the top of the stairs.

This method is efficient. It has a time complexity of O(n * k), where k is the number of allowed step sizes. The space complexity is O(n). For more about dynamic programming, we can check out count ways to climb stairs.

Testing and Validating the Stair Climbing Solutions

We need to check if our dynamic programming solutions for counting ways to climb stairs are correct. This means we should do a lot of testing and validation. We will use unit tests with different inputs, edge cases, and performance checks.

Test Cases

  1. Basic Test Cases:
    • Input: n = 3, steps = [1, 2]
      • Expected Output: 3
      • Explanation: We can reach the 3rd step in these ways: (1,1,1), (1,2), (2,1).
  2. Edge Cases:
    • Input: n = 0
      • Expected Output: 1
      • Explanation: There is one way to stay on the ground (do nothing).
    • Input: n = 1, steps = [1]
      • Expected Output: 1
      • Explanation: There is only one way to get to the first step.
  3. Performance Testing:
    • Input: n = 50, steps = [1, 2]
      • We should check how long it takes to run and see if it meets our performance goals.

Sample Code for Testing in Python

Here is how we can write the unit tests in Python using the unittest framework:

import unittest

def countWays(n, steps):
    dp = [0] * (n + 1)
    dp[0] = 1

    for i in range(1, n + 1):
        for step in steps:
            if i - step >= 0:
                dp[i] += dp[i - step]

    return dp[n]

class TestCountWays(unittest.TestCase):
    
    def test_basic_cases(self):
        self.assertEqual(countWays(3, [1, 2]), 3)
        self.assertEqual(countWays(4, [1, 2]), 5)
    
    def test_edge_cases(self):
        self.assertEqual(countWays(0, [1, 2]), 1)
        self.assertEqual(countWays(1, [1]), 1)
    
    def test_performance(self):
        import time
        start_time = time.time()
        countWays(50, [1, 2])
        end_time = time.time()
        self.assertLess(end_time - start_time, 1)  # Should take less than 1 second

if __name__ == '__main__':
    unittest.main()

Java and C++ Testing

For Java and C++, we will do similar things. Here are short examples for unit testing:

Java:

import org.junit.Test;
import static org.junit.Assert.assertEquals;

public class CountWaysTest {
    
    public int countWays(int n, int[] steps) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        
        for (int i = 1; i <= n; i++) {
            for (int step : steps) {
                if (i - step >= 0) {
                    dp[i] += dp[i - step];
                }
            }
        }
        
        return dp[n];
    }

    @Test
    public void testBasicCases() {
        assertEquals(3, countWays(3, new int[]{1, 2}));
    }
    
    // More tests can be added here...
}

C++:

#include <cassert>
#include <vector>

int countWays(int n, const std::vector<int>& steps) {
    std::vector<int> dp(n + 1, 0);
    dp[0] = 1;

    for (int i = 1; i <= n; ++i) {
        for (int step : steps) {
            if (i - step >= 0) {
                dp[i] += dp[i - step];
            }
        }
    }
    return dp[n];
}

int main() {
    assert(countWays(3, {1, 2}) == 3);
    // More checks can be added here...
    return 0;
}

Validation Techniques

  • Assertions: We can use assertions to check if our output matches what we expect.
  • Boundary Testing: We should test with the smallest and largest inputs.
  • Random Testing: We can create random test cases to find any hidden issues.

By using these testing methods, we can make sure our dynamic programming solution for counting ways to climb stairs is strong and works well in many situations.

Frequently Asked Questions

1. What is the dynamic programming approach to count ways to climb stairs with step size constraints?

The dynamic programming approach counts the ways to climb stairs with step size limits. We create an array to keep track of how many ways we can reach each step. We fill this array step by step. We add the ways to get to the last steps based on the allowed step sizes. This way, we find the total combinations without doing the same work over and over. It works well for bigger inputs.

2. How can I optimize space complexity in the dynamic programming solution for climbing stairs?

To make space usage better in the dynamic programming solution for counting ways to climb stairs, we can make the array smaller. Instead of keeping an array for all steps, we can only store the last few values we need. This cuts the space from O(n) to O(1). This is helpful when the step size limits are small.

3. Can I implement the stair climbing problem using recursion with memoization?

Yes, we can use recursion with memoization for the stair climbing problem. This method calculates the number of ways to reach a step. At the same time, we save the results we already found in a hash map or an array. This helps us not to calculate the same thing again. It makes our solution faster, especially for bigger inputs.

4. What are the differences between iterative and recursive approaches in counting ways to climb stairs?

The iterative approach for counting ways to climb stairs builds the answer step by step. We use a loop to fill an array with the number of ways to each step. On the other hand, the recursive approach checks all possible ways to climb the stairs by breaking the problem into smaller parts. Recursion is often easier to follow, but the iterative method is usually better in space and time use.

5. What programming languages can I use to implement the dynamic programming solution for stair climbing?

We can use many programming languages to implement the dynamic programming solution for counting ways to climb stairs. Some of these languages are Java, Python, and C++. Each language has its own style and features. But the main idea of the algorithm stays the same. For example, we can find detailed examples in Java, Python, and C++.