What is the time complexity of Minimum Cost For Tickets?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Minimum Cost For Tickets?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Cost For Tickets?

Insight 1: This is a dynamic programming problem where the optimal cost for day `i` depends on the optimal cost of previous days. Insight 2: We can build a DP array where `dp[i]` represents the minimum cost to travel up to day `i`.

Which companies ask LeetCode 983. Minimum Cost For Tickets in interviews?

This problem is frequently asked at Flipkart, Grab, Intuit, Snap, Turing and 1 more companies.

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Minimum Cost For Tickets - LeetCode 983 Solution

Minimum Cost For Tickets - Complete Solution Guide

Minimum Cost For Tickets is LeetCode problem 983, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have planned some train traveling one year in advance. The days of the year in which you will travel are given as an integer array days . Each day is an integer from 1 to 365 . Train tickets are sold in three different ways : a 1-day pass is sold for costs[0] dollars, a 7-day pass is sold for costs[1] dollars, and a 30-day pass is sold for costs[2] dollars. The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2 , then we can travel for 7 days: 2 ,

Detailed Explanation

The problem asks us to find the minimum cost to travel on specific days of the year, given three types of tickets: a 1-day pass costing `costs[0]`, a 7-day pass costing `costs[1]`, and a 30-day pass costing `costs[2]`. We are given a list of days `days` on which we need to travel. The goal is to minimize the total cost of tickets purchased to cover all travel days. `days` is a sorted array of integers from 1 to 365, and `costs` is an array of length 3 representing the costs of the three types of passes.

Solution Approach

The solution uses dynamic programming to calculate the minimum cost to travel up to each day. It iterates through each day from 1 to the last travel day. If the current day is not a travel day, the minimum cost is the same as the previous day. If the current day is a travel day, it considers the cost of buying a 1-day pass, a 7-day pass, and a 30-day pass, and chooses the minimum among these options. The DP array `dp[i]` stores the minimum cost to travel up to day `i`.

Step-by-Step Algorithm

Step 1: Create a set (or equivalent data structure) `travel_days` to store all the days on which we need to travel. This helps in O(1) lookup.
Step 2: Create a DP array `dp` of size `last_day + 1`, where `last_day` is the last day on which we travel. Initialize all elements to 0.
Step 3: Iterate from `i = 1` to `last_day`. For each `i`, check if `i` is a travel day (i.e., present in `travel_days`).
Step 4: If `i` is not a travel day, then `dp[i] = dp[i-1]` (no additional cost as we don't need to travel).
Step 5: If `i` is a travel day, calculate the cost of buying each type of pass and choose the minimum cost:
Step 6: - `cost1 = dp[i-1] + costs[0]` (buy a 1-day pass).
Step 7: - `cost7 = dp[max(0, i-7)] + costs[1]` (buy a 7-day pass, consider edge case where `i-7 < 0`).
Step 8: - `cost30 = dp[max(0, i-30)] + costs[2]` (buy a 30-day pass, consider edge case where `i-30 < 0`).
Step 9: - `dp[i] = min(cost1, cost7, cost30)`.
Step 10: Return `dp[last_day]`, which represents the minimum cost to travel up to the last travel day.

Key Insights

Insight 1: This is a dynamic programming problem where the optimal cost for day `i` depends on the optimal cost of previous days.
Insight 2: We can build a DP array where `dp[i]` represents the minimum cost to travel up to day `i`.
Insight 3: For each day, we need to consider three options: buy a 1-day pass, a 7-day pass, or a 30-day pass and choose the option that gives the minimum cost.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: Flipkart, Grab, Intuit, Snap, Turing, Zepto.

Minimum Cost For Tickets | LeetCode 983

Medium

Array Dynamic Programming

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.HashSet;
import java.util.Set;

class Solution {
    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];
            } else {
                // Cost with a 1-day pass on day i
                int cost1 = dp[i - 1] + costs[0];

                // Cost with a 7-day pass ending on or after day i
                int cost7 = dp[Math.max(0, i - 7)] + costs[1];

                // Cost with a 30-day pass ending on or after day i
                int cost30 = dp[Math.max(0, i - 30)] + costs[2];

                dp[i] = Math.min(cost1, Math.min(cost7, cost30));
            }
        }

        return dp[lastDay];
    }
}

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Problem Info

Description

Problem 983: Minimum Cost For Tickets - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

FlipkartGrabIntuitSnapTuringZepto

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:983

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Create a set (or equivalent data structure) `travel_days` to store all the days on which we need to travel. This helps in O(1) lookup.
2
Step 2: Create a DP array `dp` of size `last_day + 1`, where `last_day` is the last day on which we travel. Initialize all elements to 0.
3
Step 3: Iterate from `i = 1` to `last_day`. For each `i`, check if `i` is a travel day (i.e., present in `travel_days`).
4
Step 4: If `i` is not a travel day, then `dp[i] = dp[i-1]` (no additional cost as we don't need to travel).
5
Step 5: If `i` is a travel day, calculate the cost of buying each type of pass and choose the minimum cost:
6
Step 6: - `cost1 = dp[i-1] + costs[0]` (buy a 1-day pass).
7
Step 7: - `cost7 = dp[max(0, i-7)] + costs[1]` (buy a 7-day pass, consider edge case where `i-7 < 0`).
8
Step 8: - `cost30 = dp[max(0, i-30)] + costs[2]` (buy a 30-day pass, consider edge case where `i-30 < 0`).
9
Step 9: - `dp[i] = min(cost1, cost7, cost30)`.
10
Step 10: Return `dp[last_day]`, which represents the minimum cost to travel up to the last travel day.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the loop iterating up to `last_day`. The space complexity is determined by the size of the `dp` array which is proportional to `last_day`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursion with Memoization

Solve the problem using recursion with memoization. Define a recursive function that takes the current day as input and returns the minimum cost to travel from that day onwards. Memoize the results to avoid redundant calculations.

Time: O(n), where n is the last day.Space: O(n), due to the memoization table and recursion depth.

Trade-offs: Recursion with memoization can be more intuitive for some, but might have a larger overhead due to recursive function calls. The iterative DP approach might be slightly more efficient due to avoiding function call overhead.

✨

Key Insights

1Insight 1: This is a dynamic programming problem where the optimal cost for day `i` depends on the optimal cost of previous days.
2Insight 2: We can build a DP array where `dp[i]` represents the minimum cost to travel up to day `i`.
3Insight 3: For each day, we need to consider three options: buy a 1-day pass, a 7-day pass, or a 30-day pass and choose the option that gives the minimum cost.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: The first few days are travel days. The `i - 7` and `i - 30` indices might be negative. The solution uses `max(0, i - 7)` and `max(0, i - 30)` to handle this by accessing `dp[0]` which represents the initial cost (0).
Edge case 2: The input `days` array might be empty. While the constraints say there will always be at least one day, if the `days` array *were* empty, accessing `days[-1]` would throw an error. However, the prompt states 1 <= days.length <= 365

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge cases where `i - 7` or `i - 30` becomes negative. Forgetting to take the `max(0, i - 7)` or `max(0, i - 30)` will result in an index out of bounds error.

✗Common mistake 2: Not initializing the DP array correctly. Forgetting to set `dp[0] = 0` can lead to incorrect results.

✗Common mistake 3: Misunderstanding the problem statement. Not realizing the days array is sorted can cause inefficiencies if one attempts to search the array instead of utilizing its sorted nature.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing subscription costs for services with varying duration packages (e.g., monthly, quarterly, yearly subscriptions).

🏦Real-world scenario 2: Inventory management where items have different expiry dates and costs associated with discarding expired items. Finding the optimal purchasing strategy to minimize waste and cost.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 23, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Minimum Cost For Tickets"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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