[Dynamic Programming] Min Cost Climbing Stairs - Easy

Min Cost Climbing Stairs Problem

The Min Cost Climbing Stairs problem is a well-known dynamic programming challenge. It asks us to find the lowest cost to reach the top of a staircase. Each step has its own cost. Our goal is to find the cheapest way to the top. We can step one or two stairs at a time. We can solve this problem using dynamic programming techniques. By breaking the problem into smaller parts, we can find a solution that lowers the total cost.

In this article, we will look closely at the Min Cost Climbing Stairs problem. We will talk about the problem statement. We will also discuss the dynamic programming method for solving it. Additionally, we will explain how to write the solution in Java, Python, and C++. We will talk about some possible optimizations. We will also compare different ways to solve this problem. Lastly, we will answer some common questions about the Min Cost Climbing Stairs problem.

Here are the topics we will cover:

  • Dynamic Programming Min Cost Climbing Stairs Solution Overview
  • Understanding the Problem Statement for Min Cost Climbing Stairs
  • Dynamic Programming Approach for Min Cost Climbing Stairs
  • Java Implementation of Min Cost Climbing Stairs
  • Python Implementation of Min Cost Climbing Stairs
  • C++ Implementation of Min Cost Climbing Stairs
  • Optimizations in Min Cost Climbing Stairs Solutions
  • Comparative Analysis of Different Approaches
  • Frequently Asked Questions

If you want to read more, you can check these articles on similar dynamic programming topics: Dynamic Programming Fibonacci Number, Dynamic Programming Fibonacci with Memoization, and Dynamic Programming Climbing Stairs.

Understanding the Problem Statement for Min Cost Climbing Stairs

The Min Cost Climbing Stairs problem is about finding the least cost to get to the top of a staircase. Each step has a cost. We can start at step 0 or step 1. We can climb one or two steps at a time. The main goal is to reach the top with the smallest total cost.

Problem Definition

  • Input: We have an array cost. Each cost[i] shows the cost of stepping on step i.
  • Output: We need to find the minimum cost to get to step n. Here, n is the total number of steps.

Example

For cost = [10, 15, 20], the costs to climb the stairs are: - If we start at step 0 (cost 10) and jump to step 2 (cost 20), the total cost is 10 + 20 = 30. - If we start at step 1 (cost 15) and jump to step 2 (cost 20), the total cost is 15 + 20 = 35.

The least total cost to reach the top is 30.

Constraints

  • The cost array has a length between 2 and 1000.
  • Each cost value is a non-negative whole number.

This problem shows how we can use dynamic programming. It helps us find the minimum cost without repeating calculations. By breaking the problem into smaller parts, we can build the solution step by step or use recursion with memoization. For more about dynamic programming, check this Dynamic Programming Climbing Stairs article.

Dynamic Programming Approach for Min Cost Climbing Stairs

We can solve the Min Cost Climbing Stairs problem well using a dynamic programming method. The aim is to find the lowest cost to reach the top of the stairs. We can take either one or two steps at a time.

Problem Breakdown

  • We have an array cost. Here, cost[i] shows the cost of stepping on the i-th step.

  • We can start at step 0 or step 1.

  • We can say that the minimum cost to reach step i is:

    [ dp[i] = cost[i] + (dp[i-1], dp[i-2]) ]

  • The base cases are:

    • dp[0] = cost[0]
    • dp[1] = cost[1]

Dynamic Programming Implementation

The dynamic programming way needs us to fill an array dp. In this array, dp[i] will keep the minimum cost to reach step i.

Java Implementation

public class MinCostClimbingStairs {
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        int[] dp = new int[n + 1];
        dp[0] = 0;
        dp[1] = 0;
        
        for (int i = 2; i <= n; i++) {
            dp[i] = cost[i - 1] + Math.min(dp[i - 1], dp[i - 2]);
        }
        
        return dp[n];
    }
}

Python Implementation

def min_cost_climbing_stairs(cost):
    n = len(cost)
    dp = [0] * (n + 1)
    dp[0] = 0
    dp[1] = 0
    
    for i in range(2, n + 1):
        dp[i] = cost[i - 1] + min(dp[i - 1], dp[i - 2])
    
    return dp[n]

C++ Implementation

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int n = cost.size();
        vector<int> dp(n + 1, 0);
        for (int i = 2; i <= n; i++) {
            dp[i] = cost[i - 1] + min(dp[i - 1], dp[i - 2]);
        }
        return dp[n];
    }
};

Space Optimization

We can save space by not keeping a full dp array. We only need the last two costs because the current cost depends only on the last two steps.

Optimized Java Implementation

public class MinCostClimbingStairs {
    public int minCostClimbingStairs(int[] cost) {
        int first = 0, second = 0;
        for (int i = 2; i <= cost.length; i++) {
            int current = cost[i - 1] + Math.min(first, second);
            first = second;
            second = current;
        }
        return second;
    }
}

Optimized Python Implementation

def min_cost_climbing_stairs(cost):
    first, second = 0, 0
    for i in range(2, len(cost) + 1):
        current = cost[i - 1] + min(first, second)
        first, second = second, current
    return second

Optimized C++ Implementation

class Solution {
public:
    int minCostClimbingStairs(vector<int>& cost) {
        int first = 0, second = 0;
        for (int i = 2; i <= cost.size(); i++) {
            int current = cost[i - 1] + min(first, second);
            first = second;
            second = current;
        }
        return second;
    }
};

With this dynamic programming method, we can find the minimum cost to climb the stairs. Also, we reduce space use from O(n) to O(1). For more related information, you can check the article on Dynamic Programming - Climbing Stairs.

Java Implementation of Min Cost Climbing Stairs

We can solve the Min Cost Climbing Stairs problem in Java with a simple method called dynamic programming. Our goal is to find the lowest cost to reach the top of the stairs. Each step has a cost.

Java Code Implementation

Here is a simple Java code:

public class MinCostClimbingStairs {
    public int minCostClimbingStairs(int[] cost) {
        int n = cost.length;
        int[] dp = new int[n + 1]; // DP array to store min cost at each step

        // Base cases
        dp[0] = 0; // Starting point
        dp[1] = cost[0]; // Cost to reach the first step

        // Fill the DP array
        for (int i = 2; i <= n; i++) {
            dp[i] = Math.min(dp[i - 1] + (i == n ? 0 : cost[i - 1]), dp[i - 2] + (i == n ? 0 : cost[i - 2]));
        }

        return dp[n]; // Minimum cost to reach the top
    }

    public static void main(String[] args) {
        MinCostClimbingStairs solution = new MinCostClimbingStairs();
        int[] cost = {10, 15, 20}; // Example costs
        int result = solution.minCostClimbingStairs(cost);
        System.out.println("Minimum cost to climb stairs: " + result); // Output: 15
    }
}

Explanation

  • Dynamic Programming Array (dp): We make an array called dp. Here, dp[i] shows the minimum cost to get to step i.
  • Base Cases:
    • We set dp[0] to 0. There is no cost to start.
    • We set dp[1] to cost[0], which is the cost of the first step.
  • Filling the DP Array: For each step from 2 to n, we find the lowest cost to reach that step. We check the two steps before it. We can come from the step right below or the step before that.
  • Return Value: In the end, we return dp[n]. This gives us the minimum cost to reach the top.

This method works well. It has a time complexity of O(n) and space complexity of O(n).

For more details on dynamic programming, you can check the Dynamic Programming Climbing Stairs article.

Python Implementation of Min Cost Climbing Stairs

We can solve the Min Cost Climbing Stairs problem using Python with a simple method called dynamic programming. Our goal is to find the least cost to get to the top of the stairs. We can climb each step from the step before it or the step before that. The cost for each step is in an array.

Here is a simple Python code example:

def minCostClimbingStairs(cost):
    n = len(cost)
    if n == 0:
        return 0
    elif n == 1:
        return cost[0]
    
    # Create a dp array to store the minimum cost to reach each step
    dp = [0] * n
    dp[0] = cost[0]
    dp[1] = cost[1]

    for i in range(2, n):
        dp[i] = cost[i] + min(dp[i - 1], dp[i - 2])
    
    # The result is the minimum cost to reach the top
    return min(dp[n - 1], dp[n - 2])

# Example usage
cost = [10, 15, 20]
print(minCostClimbingStairs(cost))  # Output: 15

Explanation of the Code:

  • First, we check if there are steps. If there are no steps, the cost is 0. If there is one step, the cost is just the cost of that step.
  • We start a dynamic programming array called dp to save the minimum cost for each step.
  • The loop fills the dp array by finding the least cost to reach each step using the last two steps.
  • At the end, we look at the last two numbers in the dp array to find the least cost to reach the top.

This code works well to solve the Min Cost Climbing Stairs problem in O(n) time complexity and O(n) space complexity.

For more about dynamic programming ideas, we can check this link Dynamic Programming Climbing Stairs - Easy.

C++ Implementation of Min Cost Climbing Stairs

To solve the Min Cost Climbing Stairs problem with C++, we use a simple method called dynamic programming. This problem asks us to find the lowest cost to reach the top of a staircase. Each step has a cost. We can take one or two steps at once.

C++ Code Implementation

Here is the C++ code for the Min Cost Climbing Stairs method:

#include <vector>
#include <algorithm>

class Solution {
public:
    int minCostClimbingStairs(std::vector<int>& cost) {
        int n = cost.size();
        if (n == 0) return 0;
        if (n == 1) return cost[0];
        
        std::vector<int> dp(n + 1, 0);
        
        dp[0] = 0; // Starting before the first step
        dp[1] = cost[0]; // Cost to reach the first step
        
        for (int i = 2; i <= n; ++i) {
            dp[i] = std::min(dp[i - 1] + (i == n ? 0 : cost[i - 1]), dp[i - 2] + (i == n ? 0 : cost[i - 2]));
        }
        
        return dp[n];
    }
};

Explanation of the Code

  • Initialization: We create a vector called dp. Here, dp[i] shows the minimum cost to get to the i-th step.
  • Base Cases:
    • dp[0] is 0. There is no cost to stand before the first step.
    • dp[1] is the cost of the first step.
  • Dynamic Programming Transition:
    • For each step from 2 to n, we find the minimum cost to reach that step. We look at the costs from the last one and two steps.
  • Final Result: The lowest cost to reach the top of the stairs is in dp[n].

This method works in O(n) time and O(n) space. Here n is the number of steps in the staircase.

You can also read related articles to understand more, like the dynamic programming approach for climbing stairs.

Optimizations in Min Cost Climbing Stairs Solutions

We can make the Min Cost Climbing Stairs problem better in many ways. This helps us improve performance and save space. Here are some good optimizations:

  1. Space Optimization:
    We do not need an array to keep all step costs. We just need to remember the last two steps. The cost of the current step only depends on the last two steps.

    public int minCostClimbingStairs(int[] cost) {
    int n = cost.length;
    int first = 0, second = 0;
    for (int i = 2; i <= n; i++) {
        int current = Math.min(first + cost[i - 1], second + cost[i - 2]);
        first = second;
        second = current;
    }
    return second;
}
    def minCostClimbingStairs(cost):
    first, second = 0, 0
    for i in range(2, len(cost) + 1):
        current = min(first + cost[i - 1], second + cost[i - 2])
        first, second = second, current
    return second
    int minCostClimbingStairs(vector<int>& cost) {
    int first = 0, second = 0;
    for (int i = 2; i <= cost.size(); ++i) {
        int current = min(first + cost[i - 1], second + cost[i - 2]);
        first = second;
        second = current;
    }
    return second;
}

  2. Using Dynamic Programming with Memoization:
    If we like a top-down way, we can use memoization. This means we save the results of smaller problems so we do not need to calculate them again.

    def minCostClimbingStairs(cost):
    memo = {}
    def dp(i):
        if i <= 1:
            return 0
        if i not in memo:
            memo[i] = min(dp(i - 1) + cost[i - 1], dp(i - 2) + cost[i - 2])
        return memo[i]

    return dp(len(cost))

  3. Iterative Tabulation:
    Instead of keeping a whole DP array, we can use less space with iterative tabulation. We can do this just like we showed above.

  4. Early Exit:
    If the cost array is shorter than 2, we should return the smaller of the first cost or 0. This way we avoid doing extra calculations.

  5. Tailoring Input:
    We can change the cost array to remove high costs or make it simpler. This helps make the algorithm work better.

These optimizations help a lot in making the space and time complexity better for the Min Cost Climbing Stairs problem. We can find efficient solutions even for larger inputs. For more about dynamic programming methods, we can read Dynamic Programming: Climbing Stairs.

Comparative Analysis of Different Approaches

When we try to solve the Min Cost Climbing Stairs problem, we have several ways to do it. Here, we will compare the dynamic programming method with other common methods like recursion and memoization.

1. Dynamic Programming Approach

We find that the dynamic programming approach works best for this problem. It builds solutions step by step. It uses results from previous steps to solve bigger parts of the problem.

  • Time Complexity: O(n)
  • Space Complexity: O(n) (but we can make it O(1))
public int minCostClimbingStairs(int[] cost) {
    int n = cost.length;
    int[] dp = new int[n + 1];
    dp[0] = 0; 
    dp[1] = 0; 

    for (int i = 2; i <= n; i++) {
        dp[i] = Math.min(dp[i - 1] + cost[i - 1], dp[i - 2] + cost[i - 2]);
    }
    return dp[n];
}

2. Memoization Approach

Memoization improves the simple recursive method. It keeps results we already calculated. This way, we do not do the same work again.

  • Time Complexity: O(n)
  • Space Complexity: O(n)
def minCostClimbingStairs(cost):
    n = len(cost)
    memo = [-1] * (n + 1)

    def dp(i):
        if i <= 1:
            return 0
        if memo[i] != -1:
            return memo[i]
        memo[i] = min(dp(i - 1) + cost[i - 1], dp(i - 2) + cost[i - 2])
        return memo[i]

    return dp(n)

3. Recursive Approach

The simple recursive approach calculates the minimum cost directly. It does not save any results. This leads to a lot of repeated work and makes it slow.

  • Time Complexity: O(2^n)
  • Space Complexity: O(n) (due to call stack)
int minCostClimbingStairs(vector<int>& cost) {
    return min(minCost(cost, cost.size() - 1), minCost(cost, cost.size() - 2));
}

int minCost(vector<int>& cost, int i) {
    if (i < 0) return 0;
    return cost[i] + min(minCost(cost, i - 1), minCost(cost, i - 2));
}

Comparison Summary

  • Dynamic Programming: This is the best choice for being fast and clear. Iterative DP also uses less space.
  • Memoization: This helps for a recursive solution but still uses extra space for results.
  • Recursive: This method is easy to understand but not good for big inputs because it is slow.

If we want to learn more about dynamic programming, we can check out Dynamic Programming - Fibonacci Number or Dynamic Programming - Fibonacci with Memoization.

Frequently Asked Questions

1. What is the Min Cost Climbing Stairs problem in dynamic programming?

The Min Cost Climbing Stairs problem is a well-known challenge in dynamic programming. We need to find the smallest cost to reach the top of a staircase. We can climb one or two steps at a time. Each step has a cost. The aim is to find the cheapest way to get to the last step. This problem helps us learn how to use dynamic programming.

2. How does dynamic programming help in solving the Min Cost Climbing Stairs problem?

Dynamic programming is useful for the Min Cost Climbing Stairs problem. It breaks the problem into smaller parts. We store results to avoid doing the same calculations again. We use an array to keep track of the minimum costs to reach each step. This makes the solution faster. If we want to learn more about dynamic programming, we can read the Dynamic Programming Climbing Stairs article.

3. What is the time complexity of the dynamic programming solution for Min Cost Climbing Stairs?

The time complexity of the dynamic programming solution for the Min Cost Climbing Stairs problem is O(n). Here, n is the number of steps. This is efficient because we calculate the minimum cost for each step only once. We store these results in an array. This is better than a simple recursive method that can take a lot of time because it does many repeated calculations.

4. Can the Min Cost Climbing Stairs problem be solved using memoization?

Yes, we can solve the Min Cost Climbing Stairs problem using memoization. This is a method that saves the results of expensive function calls. When the same inputs come again, it gives back the saved result. This way, we reduce the time it takes to solve the problem. It is more efficient than just using recursion. If you want to know more about memoization, you can read about the Dynamic Programming Fibonacci with Memoization.

5. What programming languages are commonly used for implementing Min Cost Climbing Stairs solutions?

We can implement the Min Cost Climbing Stairs problem in many programming languages. Some common ones are Java, Python, and C++. Each language has its own syntax and features. This allows us to pick the one we like best. The article shows complete implementations in these languages. This way, we can follow along and change the solution to fit what we need.