[Dynamic Programming] Minimum Cost to Buy Candies (DP Tiling) - Medium

Dynamic programming (DP) is a strong method we use to solve problems. It helps us break big problems into smaller, easier parts. When we want to find the lowest cost to buy candies, the DP tiling method helps us figure out the cheapest way to buy candies. We can do this while following certain rules and setups. By looking at the costs for different ways to arrange candies, we can find the best solution that lowers our total spending.

In this article, we will look at the problem of finding the minimum cost to buy candies. We will explain the dynamic programming idea for calculating costs. We will also show how to implement it in Java, Python, and C++. We will talk about how to make the DP solution faster and how to test it. We will answer common questions about this topic. The following headers show what we will discuss:

  • [Dynamic Programming] Minimum Cost to Buy Candies Using DP Tiling Approach
  • Understanding the Problem Statement for Minimum Cost to Buy Candies
  • Dynamic Programming Concept for Minimum Cost Calculation
  • Java Solution for Minimum Cost to Buy Candies
  • Python Implementation of Minimum Cost to Buy Candies
  • C++ Code for Minimum Cost to Buy Candies
  • Optimizing the DP Solution for Performance
  • Testing and Validating the Minimum Cost Solution
  • Frequently Asked Questions

If you want to learn more about related dynamic programming ideas, you can read articles about the Fibonacci number, Climbing Stairs, and the Minimum Cost Climbing Stairs.

Understanding the Problem Statement for Minimum Cost to Buy Candies

We can think about the problem of finding the least cost to buy candies as a dynamic programming task. Our goal is to spend the least amount of money while following some rules. We have a list of candy costs and we can buy them in different ways. We need to find how much money we really need to buy a certain number of candies.

Problem Details

  • Input: An array that shows the costs of candies and a number that tells how many candies we want to buy.
  • Output: The least cost needed to buy the number of candies we want.

Constraints

  1. We can buy any number of candies from the list.
  2. There can be different ways to buy the same amount of candies that may cost differently.
  3. We should use dynamic programming methods to find the minimum cost and avoid doing the same work again.

Example

Let’s look at this example:

  • Candy costs: [1, 3, 5]
  • Target candies: 5

Here are some ways to reach the target of 5 candies:

  • Buy 5 candies at cost 1: Total cost = 5
  • Buy 1 candy at cost 1 and 2 candies at cost 3: Total cost = 7
  • Buy 1 candy at cost 1, 1 candy at cost 3, and 1 candy at cost 5: Total cost = 9
  • Buy 5 candies at cost 3: Total cost = 15

So, the least cost to buy 5 candies is 5.

We can solve this problem well using dynamic programming (DP) methods. The DP Tiling approach works well here. It breaks the problem into smaller parts and builds the solution step by step. The next sections will explain the dynamic programming idea and show how to implement it in Java, Python, and C++.

For more reading on similar dynamic programming problems, we can check these articles: - Dynamic Programming: Minimum Path Sum in a Grid - Dynamic Programming: Coin Change - Dynamic Programming: Minimum Cost Climbing Stairs

Dynamic Programming Concept for Minimum Cost Calculation

Dynamic programming (DP) is a strong technique we use to solve problems by breaking them into easier parts. We store solutions to avoid doing the same work again. For finding the minimum cost to buy candies, we can use the DP method to make our calculations faster.

Problem Breakdown

To find the minimum cost to buy candies, we can look at the problem with states and transitions:

  1. State Definition: Let dp[i] show the minimum cost to buy the first i candies.
  2. Transition Formula: We can calculate the cost for the i-th candy based on earlier states. For example:
    • If we buy one candy at a time, the formula is: [ dp[i] = min(dp[i-1] + cost[i], dp[i-2] + cost[i-1] + cost[i]) ]
    • Here, cost[i] is the cost of the i-th candy. We take the minimum cost from the ways to buy it (either alone or with the previous one).

Initialization

  • Base Cases:
    • dp[0] = 0: The cost is zero for no candies.
    • dp[1] = cost[0]: This is the cost for the first candy.

Implementation Example

Here is a simple way to show the dynamic programming solution to find the minimum cost to buy candies in Java, Python, and C++.

Java Implementation:

public class MinimumCostCandies {
    public int minCost(int[] cost) {
        int n = cost.length;
        if (n == 0) return 0;
        if (n == 1) return cost[0];

        int[] dp = new int[n];
        dp[0] = cost[0];
        dp[1] = Math.min(cost[0] + cost[1], cost[1]);

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

Python Implementation:

def min_cost(cost):
    n = len(cost)
    if n == 0:
        return 0
    if n == 1:
        return cost[0]

    dp = [0] * n
    dp[0] = cost[0]
    dp[1] = min(cost[0] + cost[1], cost[1])

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

C++ Implementation:

class Solution {
public:
    int minCost(vector<int>& cost) {
        int n = cost.size();
        if (n == 0) return 0;
        if (n == 1) return cost[0];

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

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

Complexity Analysis

  • Time Complexity: O(n), where n is the number of candies. We go through the cost array one time.
  • Space Complexity: O(n) for the DP array. We can make it O(1) by keeping only the last two states.

Using this dynamic programming method helps us find the minimum cost quickly. We do not repeat calculations. For more on dynamic programming, check articles on dynamic programming concepts.

Java Solution for Minimum Cost to Buy Candies

We can solve the Minimum Cost to Buy Candies problem using Dynamic Programming in Java. We need to create a method that finds the minimum cost based on the number of candies and their costs. The DP array helps us track the minimum costs step by step.

Here is a simple Java solution:

public class MinimumCostCandies {
    public static int minCost(int[] cost) {
        int n = cost.length;
        if (n == 0) return 0;
        
        // DP array to store minimum costs
        int[] dp = new int[n + 1];
        dp[0] = 0; // Base case, no cost for zero candies
        
        // Fill the dp array
        for (int i = 1; i <= n; i++) {
            dp[i] = Integer.MAX_VALUE;
            // Calculate minimum cost for buying i candies
            for (int j = 0; j < i; j++) {
                dp[i] = Math.min(dp[i], dp[j] + cost[i - 1]);
            }
        }
        return dp[n];
    }

    public static void main(String[] args) {
        int[] cost = {1, 2, 3, 4, 5}; // Example costs
        System.out.println("Minimum Cost to Buy Candies: " + minCost(cost));
    }
}

Explanation of the Code

  • The minCost method takes an array of integers called cost. Each element shows the cost of each candy.
  • We create an array dp to keep the minimum cost to buy i candies.
  • The outer loop goes through each candy count. The inner loop finds the minimum cost by looking at all previous purchases.
  • At the end, dp[n] has the minimum cost to buy all candies.

This Java solution works well and uses dynamic programming. It makes sure we calculate the minimum cost only once for each state. This makes it good for larger inputs. For more on similar dynamic programming problems, we can look at the Dynamic Programming - Coin Change article.

Python Implementation of Minimum Cost to Buy Candies

To solve the problem of finding the minimum cost to buy candies in Python, we can use a bottom-up way. We will make a DP array. Each part of this array shows the minimum cost to buy candies up to that point.

Problem Statement

We have a list of costs. Each index in this list shows the cost of one candy. Our job is to find the minimum cost to buy all the candies while following some rules for buying.

Implementation

def min_cost_to_buy_candies(costs):
    n = len(costs)
    if n == 0:
        return 0
    if n == 1:
        return costs[0]

    # Create a DP array
    dp = [0] * n
    dp[0] = costs[0]
    dp[1] = costs[1] + costs[0]

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

    return dp[-1]

# Example Usage
costs = [1, 2, 3, 4, 5]
print(f'Minimum cost to buy candies: {min_cost_to_buy_candies(costs)}')

Explanation

  • We start with the dp array. Here dp[i] shows the minimum cost to buy candies up to index i.
  • First, we set up base cases for the first and second candy purchases.
  • For each candy after that, we find the minimum cost. We add the cost of the current candy to the minimum cost of either buying the last candy or skipping it.
  • In the end, the last part of the dp array gives us the minimum cost to buy all the candies.

This way of solving the problem uses dynamic programming. It helps us use the best parts of the problem and avoids doing the same work again.

For more about dynamic programming, you can check articles on Dynamic Programming: Minimum Cost Climbing Stairs and Dynamic Programming: Minimum Falling Path Sum.

C++ Code for Minimum Cost to Buy Candies

We want to find the minimum cost to buy candies. We will use a dynamic programming approach in C++. We can make a function that uses a DP array. This array will help us keep track of the minimum costs at each step. We will calculate the cost based on different options we have for buying candies.

Here is a simple C++ code example:

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

int minCostToBuyCandies(vector<int>& cost) {
    int n = cost.size();
    if (n == 0) return 0;

    // DP array to store minimum cost up to each index
    vector<int> dp(n);
    
    // Base case
    dp[0] = cost[0];
    
    // Fill the DP array
    for (int i = 1; i < n; i++) {
        dp[i] = cost[i];
        // Check previous costs to minimize current cost
        for (int j = 0; j < i; j++) {
            dp[i] = min(dp[i], dp[j] + cost[i]);
        }
    }
    
    // The minimum cost to buy all candies will be in dp[n-1]
    return dp[n - 1];
}

int main() {
    vector<int> cost = {1, 2, 3, 4, 5}; // Example costs
    int result = minCostToBuyCandies(cost);
    cout << "Minimum cost to buy candies: " << result << endl;
    return 0;
}

Explanation of the Code:

We have the function minCostToBuyCandies. This function takes a vector of integers called cost. This vector shows the cost of each candy.

We start by making a dynamic programming array called dp. This array will store the minimum costs. Then we check each candy one by one. We update the minimum cost by looking at the current cost and the costs of the previous candies added to the current candy’s cost.

At the end, we return the minimum cost to buy all the candies. This cost is found at dp[n - 1].

This C++ code helps us find the minimum cost using dynamic programming. It works well and is efficient. If you want to learn more, you can read about the Dynamic Programming - Coin Change Problem.

Optimizing the DP Solution for Performance

We want to make the dynamic programming solution for the Minimum Cost to Buy Candies problem better. We can do this by using less space and not repeating calculations. Here are some simple ways to improve the performance:

  1. Space Optimization:
    • Instead of keeping a full DP table, we can use a rolling array technique. This way, we only need to remember the last two states. This is enough to find the current state.
  2. Avoiding Redundant Calculations:
    • We can use memoization to save results that we already calculated. This will help us do fewer calculations.

Implementation

Here is a better Java implementation:

public class MinimumCostCandies {
    public int minCost(int[] cost) {
        if (cost.length == 0) return 0;
        int n = cost.length;
        int[] dp = new int[2];
        
        dp[0] = cost[0]; // Cost for the first candy
        for (int i = 1; i < n; i++) {
            dp[i % 2] = cost[i] + Math.min(dp[(i - 1) % 2], (i > 1 ? dp[(i - 2) % 2] : 0));
        }
        return dp[(n - 1) % 2];
    }
}

Python Implementation

Here is the better Python code using similar ideas:

class MinimumCostCandies:
    def minCost(self, cost):
        n = len(cost)
        if n == 0:
            return 0
        dp = [0] * 2
        dp[0] = cost[0]
        
        for i in range(1, n):
            dp[i % 2] = cost[i] + min(dp[(i - 1) % 2], dp[(i - 2) % 2] if i > 1 else 0)
        return dp[(n - 1) % 2]

C++ Implementation

Here is a better version in C++:

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

Complexity Analysis

  • Time Complexity: O(n), where n is the number of candies. We check each candy’s cost one time.
  • Space Complexity: O(1), because we use only a small amount of space for the DP array.

By using these simple techniques, we can make the dynamic programming solution for the Minimum Cost to Buy Candies work better in time and space. If you want to learn more about dynamic programming, you can check these links: Dynamic Programming: Fibonacci Number and Dynamic Programming: Coin Change.

Testing and Validating the Minimum Cost Solution

We need to make sure our solution for finding the minimum cost to buy candies using dynamic programming is correct. For this, we have to test and validate it well. This means we should create test cases that check many different situations. These include edge cases, common cases, and big inputs.

Test Cases

  1. Basic Test Cases:
    • Input: costs = [1, 2, 3], Expected Output: 3
    • Input: costs = [5, 10, 20], Expected Output: 5
  2. Edge Cases:
    • Input: costs = [], Expected Output: 0 (No candies)
    • Input: costs = [1], Expected Output: 1 (Single candy)
    • Input: costs = [0, 0, 0], Expected Output: 0 (All free)
  3. Large Input Test Case:
    • Input: costs = [i for i in range(1, 1001)], Expected Output: 1000 (Sum of the first 1000 integers)

Validation Methodology

  • Unit Testing: We can use unit tests with a testing framework like JUnit for Java or unittest for Python. This helps us automate testing.

  • Assertions: We should use assertions to check if the output from the function is the same as the expected output.

Example Code for Testing in Python

def minCost(costs):
    if not costs:
        return 0
    n = len(costs)
    dp = [0] * n
    dp[0] = costs[0]
    for i in range(1, n):
        dp[i] = costs[i] + min(dp[i - 1], dp[i - 2] if i > 1 else 0)
    return dp[-1]

# Test Cases
assert minCost([1, 2, 3]) == 3
assert minCost([5, 10, 20]) == 5
assert minCost([]) == 0
assert minCost([1]) == 1
assert minCost([0, 0, 0]) == 0
assert minCost([i for i in range(1, 1001)]) == 1000

Performance Testing

  • Time Complexity: We should check the time it takes to run with big inputs. This helps us see if our solution works well enough.

  • Memory Usage: We need to watch how much memory our algorithm uses. This ensures it runs within limits.

Debugging

If a test does not pass, we should debug to find the problem. We can use print statements or logging to follow the values step-by-step, especially looking at important variables while building the dynamic programming table.

Conclusion

By following these steps for testing and validation, we can make sure our minimum cost solution is correct and works well. Good testing is very important for dynamic programming solutions. It helps us handle different inputs and edge cases properly.

Frequently Asked Questions

1. What is the Minimum Cost to Buy Candies problem in dynamic programming?

The Minimum Cost to Buy Candies problem is about finding the cheapest way to buy a certain number of candies when we have different prices for different amounts. We can solve this problem well using dynamic programming techniques. One way is the DP tiling approach. This method breaks the problem into smaller parts. This helps us calculate the cost better.

2. How does the DP Tiling approach work for solving the Minimum Cost to Buy Candies?

The DP Tiling approach for the Minimum Cost to Buy Candies makes a table. Each part of the table shows the least cost to buy a specific number of candies. We look at different combinations and use costs we calculated before. This way, we find the smallest amount of money we need to buy the candies we want.

3. Can you provide a Python implementation for the Minimum Cost to Buy Candies problem?

Sure! Here is a simple Python code for the Minimum Cost to Buy Candies problem using dynamic programming:

def minCost(candies):
    n = len(candies)
    dp = [float('inf')] * (n + 1)
    dp[0] = 0
    
    for i in range(1, n + 1):
        for j in range(i):
            dp[i] = min(dp[i], dp[j] + candies[i - j - 1])
    
    return dp[n]

This code sets up a DP array. It then calculates the minimum cost for each number of candies by looking at results from before.

4. What are some common mistakes to avoid when implementing dynamic programming solutions?

When we make dynamic programming solutions for the Minimum Cost to Buy Candies, there are some mistakes we should not make. We should define the state transitions well. We must also start the DP table correctly. It is important to look at all combinations to find the best solution. If we are clear about the problem and test with different cases, we can avoid these mistakes.

5. How can I optimize the DP solution for performance in the Minimum Cost to Buy Candies problem?

To make the DP solution better for the Minimum Cost to Buy Candies problem, we can save space. We only keep the necessary results instead of the whole DP table. Using memoization can stop us from doing the same calculations again. This makes the whole solution faster, especially with larger data sets.

If you want to learn more about dynamic programming strategies, you can read articles like Dynamic Programming: Minimum Cost Climbing Stairs and Dynamic Programming: Coin Change Problem.