[Dynamic Programming] Minimum Cost For Tickets - Medium

Dynamic Programming is a way to solve tough problems. We can do this by breaking the problems into smaller parts. Then we save the results of these smaller parts. This helps us avoid doing the same calculations again.

One common problem is the “Minimum Cost For Tickets” problem. Here, we want to find the lowest cost to travel on some days. We have different ticket options to choose from. By using dynamic programming, we can find the best solution quickly. We also keep track of how much we spend for each travel day.

In this article, we will look closely at the Minimum Cost For Tickets problem. First, we will explain the problem. Next, we will show how to solve it using dynamic programming in Java, Python, and C++. After that, we will talk about how to save space in our solution. We will also look at other ways to solve the problem. Finally, we will compare the solutions in different programming languages and give a detailed code walkthrough.

Here are the headers we will use:

  • Dynamic Programming Minimum Cost For Tickets Solution Overview
  • Understanding the Problem Statement for Minimum Cost For Tickets
  • Dynamic Programming Approach to Minimum Cost For Tickets in Java
  • Dynamic Programming Approach to Minimum Cost For Tickets in Python
  • Dynamic Programming Approach to Minimum Cost For Tickets in C++
  • Optimizing Space Complexity for Minimum Cost For Tickets
  • Alternative Approaches to Minimum Cost For Tickets
  • Comparative Analysis of Solutions in Java Python and C++
  • Code Walkthrough for Minimum Cost For Tickets in Java Python and C++
  • Frequently Asked Questions

Understanding the Problem Statement for Minimum Cost For Tickets

The problem of “Minimum Cost For Tickets” is about finding the least amount of money needed to travel on certain days using different tickets. We have tickets for 1 day, 7 days, and 30 days. Given a list of days we want to travel, our goal is to choose the best ticket options to spend the least.

Problem Definition

We get:

  • An array of integers called days. Each number shows a day we will travel. The days are in order from lowest to highest.
  • An array of integers called costs. This array shows the cost for each type of ticket:
    • costs[0]: the cost for a 1-day ticket
    • costs[1]: the cost for a 7-day ticket
    • costs[2]: the cost for a 30-day ticket

Constraints

  • The days array can have up to 365 days, which means it covers the whole year.
  • The numbers in days are between 1 and 365.

Example

Input: - days = [1, 4, 6, 7, 8] - costs = [2, 7, 15]

Output: - Minimum cost: 11 (We buy a 1-day ticket for day 1 and a 7-day ticket for days 4 to 8)

Dynamic Programming Approach

We can solve this problem well using dynamic programming. We will make a DP array where dp[i] shows the minimum cost to cover travel days up to the i-th day.

Key Points

  • Each day can use one of the three ticket costs.
  • The DP method builds the solution step by step. We look at the costs based on the last travel day and the ticket length.
  • This way, we do not have to recalculate costs for days we already checked. This leads us to the best solution.

The main idea of the problem is to check the costs over time while we keep track of the travel days and ticket prices. This helps us find the lowest total cost.

Dynamic Programming Approach to Minimum Cost For Tickets in Java

To solve the Minimum Cost for Tickets problem using dynamic programming in Java, we need to have a clear approach. The problem is about finding the lowest cost needed for a series of travel days with different ticket options. These options are 1-day, 7-day, and 30-day passes.

Problem Definition

  • We have an array of days when we want to travel.
  • Ticket prices are given for different lengths:
    • 1-day pass: cost1
    • 7-day pass: cost2
    • 30-day pass: cost3

Dynamic Programming Solution

We can use a dynamic programming array named dp. Here, dp[i] shows the minimum cost to cover travel days up to the i-th day.

Steps:

  1. We start by making the dp array of size n + 1, where n is how many travel days we have.
  2. We look at each day and calculate the lowest cost based on the ticket options.
  3. We use binary search to find the last day that each ticket covers.

Java Code Implementation

import java.util.*;

public class MinimumCostTickets {
    public int mincostTickets(int[] days, int[] costs) {
        int n = days.length;
        int[] dp = new int[n + 1];
        Set<Integer> travelDays = new HashSet<>();
        for (int day : days) travelDays.add(day);
        
        for (int i = 1; i <= n; i++) {
            int day = days[i - 1];
            // Cost of 1-day pass
            dp[i] = dp[i - 1] + costs[0];
            
            // Cost of 7-day pass
            int j = i - 1;
            while (j > 0 && days[j - 1] >= day - 6) j--;
            dp[i] = Math.min(dp[i], dp[j] + costs[1]);
            
            // Cost of 30-day pass
            j = i - 1;
            while (j > 0 && days[j - 1] >= day - 29) j--;
            dp[i] = Math.min(dp[i], dp[j] + costs[2]);
        }
        
        return dp[n];
    }

    public static void main(String[] args) {
        MinimumCostTickets solution = new MinimumCostTickets();
        int[] days = {1, 4, 6, 7, 8, 20};
        int[] costs = {2, 7, 15};
        System.out.println("Minimum cost for tickets: " + solution.mincostTickets(days, costs));
    }
}

Explanation of Code

  • The mincostTickets method finds the minimum cost using dynamic programming.
  • We fill the dp array step by step. For each option, we check and update the minimum cost.
  • We use a HashSet for fast access to travel days.
  • The binary search helps us calculate the last index for each ticket option quickly.

This method works well to find the minimum ticket cost in O(n) time. It is good for larger input sizes. For more on dynamic programming, we can look at similar problems like Dynamic Programming: Coin Change or Dynamic Programming: Minimum Path Sum.

Dynamic Programming Approach to Minimum Cost For Tickets in Python

We can solve the Minimum Cost For Tickets problem using dynamic programming in Python. We will use a bottom-up approach. This problem is about finding the least cost to travel with different ticket options over a number of days.

Problem Statement

We are given an array days that shows the travel days. We have three ticket choices: - 1-day pass: cost cost[0] - 7-day pass: cost cost[1] - 30-day pass: cost cost[2]

Our goal is to find the minimum cost to cover all the travel days.

Dynamic Programming Implementation

def mincostTickets(days, costs):
    last_day = days[-1]
    dp = [0] * (last_day + 1)
    travel_days = set(days)

    for day in range(1, last_day + 1):
        if day not in travel_days:
            dp[day] = dp[day - 1]
        else:
            dp[day] = min(
                dp[day - 1] + costs[0],  # 1-day pass
                dp[max(0, day - 7)] + costs[1],  # 7-day pass
                dp[max(0, day - 30)] + costs[2]  # 30-day pass
            )
    
    return dp[last_day]

# Example Usage
days = [1, 4, 6, 7, 8]
costs = [2, 7, 15]
print(mincostTickets(days, costs))  # Output: 11

Explanation

We make a dp array. Here, dp[i] shows the minimum cost to cover up to day i.

For each day, if it is not a travel day, we just take the cost from the previous day. If it is a travel day, we find the minimum cost by checking all the ticket options.

At the end, we return the value in dp for the last travel day.

This dynamic programming method helps us find the least cost for the tickets while keeping the space and time usage low. If you want to learn more about dynamic programming, these articles are good to read: Dynamic Programming - Fibonacci Number and Dynamic Programming - Coin Change.

Dynamic Programming Approach to Minimum Cost For Tickets in C++

We can solve the problem of finding the minimum cost for tickets using dynamic programming in C++. We break down the problem into smaller parts. We can use results we already found to help us solve bigger problems.

Problem Breakdown

We have: - A list of travel days. - Ticket costs for 1-day, 7-day, and 30-day passes.

We need to find the minimum cost to cover all travel days.

Dynamic Programming Algorithm

  1. Define the DP Array: We let dp[i] show the minimum cost to cover travel days up to the i-th day.

  2. Base Condition: We start with dp[0] = 0 because having no days means no cost.

  3. Transition: For each travel day, we calculate the cost if we buy:

    • A 1-day pass.
    • A 7-day pass (if it applies).
    • A 30-day pass (if it applies).

    We update dp[i] like this:

    dp[i] = min(dp[i-1] + cost[0], // 1-day pass
            dp[max(0, i-7)] + cost[1], // 7-day pass
            dp[max(0, i-30)] + cost[2]); // 30-day pass

C++ Code Implementation

#include <vector>
#include <algorithm>
#include <unordered_set>

using namespace std;

// Function to calculate the minimum cost for tickets
int minCostTickets(vector<int>& days, vector<int>& costs) {
    unordered_set<int> travelDays(days.begin(), days.end());
    vector<int> dp(366, 0); // dp array for days 1 to 365

    for (int i = 1; i <= 365; ++i) {
        if (travelDays.find(i) == travelDays.end()) {
            dp[i] = dp[i - 1]; // If not traveling, carry forward the previous day's cost
            continue;
        }
        // Calculate minimum cost for the current travel day
        dp[i] = min(dp[i - 1] + costs[0], // 1-day pass
                    min(dp[max(0, i - 7)] + costs[1], // 7-day pass
                        dp[max(0, i - 30)] + costs[2])); // 30-day pass
    }
    return dp[365]; // Minimum cost to cover all days
}

// Example usage
int main() {
    vector<int> days = {1, 4, 6, 7, 8};
    vector<int> costs = {2, 7, 15};
    int minimumCost = minCostTickets(days, costs);
    return minimumCost;
}

Explanation of the Code

  • Input: The days vector shows days we travel. The costs vector shows the costs for the different ticket types.
  • DP Array Initialization: We create a dp array of size 366 to store the minimum costs for each day.
  • Loop Through Days: For each day, we check if it is a travel day. If it is not, we take the previous day’s minimum cost.
  • Cost Calculation: If it is a travel day, we find the minimum cost by looking at all ticket options.

This way, we compute the minimum cost using dynamic programming. This leads to a good solution for the Minimum Cost For Tickets problem in C++.

Optimizing Space Complexity for Minimum Cost For Tickets

We can make space complexity better for the “Minimum Cost For Tickets” problem. We can do this by using a smarter way to keep the results. Instead of having a big DP array, we can use a smaller rolling array. This will only keep track of the important past states.

Approach

  1. Use a Small Fixed-Size Array: Instead of a big DP array of size n + 1, we can use a small array. This array will hold only the last 3 values of the DP results. The problem needs costs from just the last three days to find the cost for the current day.

  2. Transition Equation:

    • Let dp[i] mean the minimum cost to travel until day i.

    • For each day i, we can find the cost using:

      dp[i] = min(dp[i-1] + cost[i-1], dp[i-7] + cost[i-1], dp[i-30] + cost[i-1])

    • We can make this easier by only keeping the last three costs we calculated.

Implementation

Here is how we can write the optimized solution in Python:

def minCostTickets(days, costs):
    max_day = days[-1]
    dp = [0] * (max_day + 1)

    for day in range(1, max_day + 1):
        if day not in days:
            dp[day] = dp[day - 1]
            continue
        
        dp[day] = min(dp[max(0, day - 1)] + costs[0],  # 1-day pass
                       dp[max(0, day - 7)] + costs[1],  # 7-day pass
                       dp[max(0, day - 30)] + costs[2])  # 30-day pass

    return dp[max_day]

# Example usage
days = [1, 4, 6, 7, 8, 20]
costs = [2, 7, 15]
print(minCostTickets(days, costs))  # Output: Minimum cost

Space Complexity

  • The space complexity is now O(1) for the DP array. We only need to keep results for the last three days.
  • This change is very important for big inputs. It helps us use less memory while still keeping the solution fast.

By using this space optimization, we can find a better solution for the “Minimum Cost For Tickets” problem. This way, our program can run well even with bigger datasets.

Alternative Approaches to Minimum Cost For Tickets

We can solve the Minimum Cost For Tickets problem in different ways. The dynamic programming method works well, but other methods can give us new views. They can help us in some cases.

Greedy Approach

We can sometimes use a greedy algorithm. This works well if we can make quick payment choices that lower costs. But it may not always give the best answer because it does not think about what happens next.

Brute Force

The brute force method checks all possible ticket buying options. It calculates the total cost for each choice. This method guarantees we find the minimum cost. But it is very slow and not good for big data sets because it takes a lot of time.

Memoization

Memoization helps us make recursive solutions better. It saves results from costly function calls. When the same inputs come up again, we can use these saved results. This way, we do not need to do the same calculations again.

def minCostTickets(days, costs):
    memo = {}
    
    def dp(i):
        if i >= len(days):
            return 0
        if i in memo:
            return memo[i]
        
        # Cost for 1-day, 7-day, and 30-day passes
        cost1 = costs[0] + dp(i + 1)
        cost7 = costs[1] + dp(next(i, 7))
        cost30 = costs[2] + dp(next(i, 30))
        
        memo[i] = min(cost1, cost7, cost30)
        return memo[i]
    
    def next(i, days):
        while i < len(days) and days[i] < days[i] + days:
            i += 1
        return i
    
    return dp(0)

Iterative Approach

We can also use an iterative approach. This builds the solution from smaller parts, like dynamic programming, but it does not use recursion. This can save space sometimes because we can use the space again.

public int minCostTickets(int[] days, int[] costs) {
    int[] dp = new int[days.length + 1];
    for (int i = 1; i <= days.length; i++) {
        dp[i] = costs[0] + dp[i - 1]; // 1-day pass
        dp[i] = Math.min(dp[i], costs[1] + dp[getNextIndex(days, i, 7)]); // 7-day pass
        dp[i] = Math.min(dp[i], costs[2] + dp[getNextIndex(days, i, 30)]); // 30-day pass
    }
    return dp[days.length];
}

private int getNextIndex(int[] days, int index, int daysToAdd) {
    while (index <= days.length && days[index - 1] < days[index - 1] + daysToAdd) {
        index++;
    }
    return index;
}

Other Optimization Techniques

  • Bitmasking: For problems with few days or options, bitmasking can show states in a small way and can make the solution better.
  • Graph-Based Approaches: If we can show the problem like a graph, we can use algorithms like Dijkstra’s to find the cheapest path.

These other ways to solve the Minimum Cost For Tickets problem give us different methods. They can fit specific needs and limits. Dynamic programming is still one of the best methods, but knowing these other ways can help us solve problems better. If we want to read more about dynamic programming, we can check out articles on related topics like Dynamic Programming: Fibonacci Number or Dynamic Programming: Coin Change.

Comparative Analysis of Solutions in Java Python and C++

When we solve the “Minimum Cost For Tickets” problem with dynamic programming, the main idea is the same in Java, Python, and C++. But the way we write the code is different because of the syntax and features of each language. Here, we will see how each language handles the solution.

Java Implementation

import java.util.*;

public class MinimumCostForTickets {
    public int mincostTickets(int[] days, int[] costs) {
        Set<Integer> travelDays = new HashSet<>();
        for (int day : days) travelDays.add(day);
        
        int lastDay = days[days.length - 1];
        int[] dp = new int[lastDay + 1];
        
        for (int i = 1; i <= lastDay; i++) {
            if (!travelDays.contains(i)) {
                dp[i] = dp[i - 1];
                continue;
            }
            dp[i] = Math.min(dp[i - 1] + costs[0], Math.min(dp[Math.max(0, i - 7)] + costs[1], dp[Math.max(0, i - 30)] + costs[2]));
        }
        
        return dp[lastDay];
    }
}

Python Implementation

class MinimumCostForTickets:
    def mincostTickets(self, days, costs):
        travel_days = set(days)
        last_day = days[-1]
        dp = [0] * (last_day + 1)
        
        for i in range(1, last_day + 1):
            if i not in travel_days:
                dp[i] = dp[i - 1]
                continue
            dp[i] = min(dp[i - 1] + costs[0],
                        dp[max(0, i - 7)] + costs[1],
                        dp[max(0, i - 30)] + costs[2])
        
        return dp[last_day]

C++ Implementation

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

class MinimumCostForTickets {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        unordered_set<int> travelDays(days.begin(), days.end());
        int lastDay = days.back();
        vector<int> dp(lastDay + 1, 0);
        
        for (int i = 1; i <= lastDay; ++i) {
            if (travelDays.find(i) == travelDays.end()) {
                dp[i] = dp[i - 1];
                continue;
            }
            dp[i] = min(dp[i - 1] + costs[0],
                        min(dp[max(0, i - 7)] + costs[1],
                            dp[max(0, i - 30)] + costs[2]));
        }
        
        return dp[lastDay];
    }
};

Key Comparisons

  • Data Structures:
    • Java uses HashSet for travel days. Python uses a set and C++ uses unordered_set.
  • Dynamic Array Handling:
    • In Java, we create a dynamic array with new int[lastDay + 1]. Python uses a list filled with zeros. C++ uses a vector.
  • Minimum Calculation:
    • The way to find the minimum is the same in all languages. This shows how we use dynamic programming.
  • Syntax Differences:
    • Each language has its unique syntax for control structures and function definitions. But the main logic stays the same.

This comparison shows how we keep the main idea of the “Minimum Cost For Tickets” problem across different programming languages while changing to fit their own rules.

Code Walkthrough for Minimum Cost For Tickets in Java Python and C++

We can solve the Minimum Cost For Tickets problem using dynamic programming. Our goal is to find the least cost to travel for a certain number of days. We can buy tickets that last for different lengths of time. Below we will go through the code for Java, Python, and C++.

Java Implementation

import java.util.HashMap;

public class Solution {
    public int minCostTickets(int[] days, int[] costs) {
        HashMap<Integer, Integer> dp = new HashMap<>();
        return dfs(days, 0, costs, dp);
    }
    
    private int dfs(int[] days, int index, int[] costs, HashMap<Integer, Integer> dp) {
        if (index >= days.length) return 0;
        if (dp.containsKey(index)) return dp.get(index);

        int ans = Integer.MAX_VALUE;
        int[] durations = {1, 7, 30};
        
        for (int i = 0; i < durations.length; i++) {
            int nextIndex = index;
            while (nextIndex < days.length && days[nextIndex] < days[index] + durations[i]) {
                nextIndex++;
            }
            ans = Math.min(ans, costs[i] + dfs(days, nextIndex, costs, dp));
        }
        
        dp.put(index, ans);
        return ans;
    }
}

Python Implementation

class Solution:
    def mincostTickets(self, days: List[int], costs: List[int]) -> int:
        dp = {}

        def dfs(index):
            if index >= len(days):
                return 0
            if index in dp:
                return dp[index]

            ans = float('inf')
            durations = [1, 7, 30]
            
            for i in range(len(durations)):
                next_index = index
                while next_index < len(days) and days[next_index] < days[index] + durations[i]:
                    next_index += 1
                ans = min(ans, costs[i] + dfs(next_index))
            
            dp[index] = ans
            return ans

        return dfs(0)

C++ Implementation

#include <vector>
#include <unordered_map>
#include <algorithm>

using namespace std;

class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        unordered_map<int, int> dp;
        return dfs(days, 0, costs, dp);
    }

    int dfs(vector<int>& days, int index, vector<int>& costs, unordered_map<int, int>& dp) {
        if (index >= days.size()) return 0;
        if (dp.count(index)) return dp[index];

        int ans = INT_MAX;
        vector<int> durations = {1, 7, 30};
        
        for (int i = 0; i < durations.size(); i++) {
            int nextIndex = index;
            while (nextIndex < days.size() && days[nextIndex] < days[index] + durations[i]) {
                nextIndex++;
            }
            ans = min(ans, costs[i] + dfs(days, nextIndex, costs, dp));
        }
        
        dp[index] = ans;
        return ans;
    }
};

In each code, we use a method called depth-first search or DFS with memoization. This helps us avoid doing the same work over and over again. The main idea is to check all the ticket durations. Then we find the minimum cost for each option.

By using this method, we can solve the Minimum Cost For Tickets problem well in Java, Python, and C++. If we want to learn more about dynamic programming, we can look at dynamic programming approaches to coin change problems.

Frequently Asked Questions

What is the Minimum Cost for Tickets problem in dynamic programming?

The Minimum Cost for Tickets problem is about finding the least amount of money we need to pay for a series of travel days. We have different ticket options. Each ticket type lets us travel for a certain number of days. Our goal is to find the best combination of tickets so that we spend the least money while covering all travel days. This is a common problem in dynamic programming. It needs us to manage states well.

How do I implement the Minimum Cost for Tickets solution in Java?

To implement the Minimum Cost for Tickets solution in Java, we can use dynamic programming. We keep an array where each index shows a day. We will look at each day and see if it is within the time of each ticket type. Then, we update the cost for each day. We do this by using the minimum costs from the previous days. This way, we can find the total minimum cost for all the travel days.

What are the best practices for optimizing the space complexity in the Minimum Cost for Tickets problem?

To make space usage better for the Minimum Cost for Tickets problem, we can change from using a full array to using a rolling array or even just one variable. This works if we only need the latest states. By reusing space from past results, we keep our algorithm efficient. It helps us use less memory while still finding the minimum cost well.

Can I solve the Minimum Cost for Tickets problem using alternative approaches?

Yes, we can solve the Minimum Cost for Tickets problem in other ways. While dynamic programming is the best way, we can also use memoization. This means we store results we already calculated. This way, we do not do the same calculations again when we call functions. But usually, dynamic programming is faster for this problem.

How does the Minimum Cost for Tickets problem compare with similar dynamic programming problems?

The Minimum Cost for Tickets problem is like other dynamic programming problems, such as the Coin Change problem or the Minimum Path Sum in a Grid. All these problems try to optimize a total cost based on the choices we make at each step. Knowing these connections helps us understand dynamic programming better. It also improves our problem-solving skills. For more information, we can check related articles like Dynamic Programming Coin Change and Dynamic Programming Minimum Path Sum.