Minimum Cost For Tickets
The “Minimum Cost For Tickets” problem asks you to minimize the total amount you spend on travel passes over a set of planned travel days. You are given a list of days in a year when you will travel, and you can buy passes that cover different durations, typically a one-day pass, a seven-day pass, and a thirty-day pass. Each pass has a cost, and once purchased, it covers travel for its full duration, starting from the day you buy it. The goal is to cover every travel day with at least one valid pass while spending as little as possible.
This problem looks like a pricing choice problem, but it is really an optimization over time. Every decision you make affects which future travel days are already covered, which means you cannot reliably choose passes greedily based on the current day alone.
Why greedy buying fails
A common instinct is to buy the cheapest pass whenever you travel, or to always buy the longest pass when you see a cluster of travel days. Both strategies can fail because the optimal decision depends on the pattern of travel days ahead.
For example, a seven-day pass can be wasteful if you travel only once in that week, while a one-day pass can be wasteful if you travel six times in that same week. The best choice depends on how coverage overlaps with future travel days, which is exactly the kind of dependency that calls for dynamic programming.
Recognizing the dynamic programming structure
The key insight is that the problem can be expressed as “minimum cost from day i onward.” If you know the minimum cost required to cover travel from a certain point in time, you can compute the minimum cost for earlier days by considering which pass you buy next.
This works because the travel calendar is sequential. Once you decide what pass to buy starting on a specific travel day, you immediately know how far into the future you are covered, and the remaining problem becomes the same question for the next uncovered travel day.
Want to explore more coding problem solutions? Check out Delete Nodes and Return Forest and Furthest Building You Can Reach.
Defining the state: the next travel day to cover
A clean way to model the problem is to define a state based on the position in the travel-days list. At each index, you are trying to cover that travel day and all later travel days as cheaply as possible.
From that state, you have three options: buy a one-day pass, buy a seven-day pass, or buy a thirty-day pass. Each option covers the current travel day and potentially several future travel days. The important part is determining which travel day becomes the next uncovered day after buying a pass.
Jumping to the next uncovered day
When you buy a pass, you do not move forward by a fixed number of travel days. You move forward in time by a fixed duration, and then you skip to the first travel day that falls outside that coverage window.
This “jump” is the heart of the optimization. It prevents you from paying repeatedly for days that are already covered. Once you can compute the next uncovered travel day for each pass choice, the rest of the solution becomes a straightforward minimum comparison across the three options.
Why this approach guarantees optimality
The dynamic programming solution works because of the optimal substructure. If you choose a particular pass today, the best total cost is the pass cost plus the best possible cost to cover everything after its coverage ends.
You are not guessing which pass is best globally. You are evaluating all valid choices locally while using cached optimal results for the remaining suffix of travel days. This combination ensures you always compute the true minimum.
Memoization or bottom-up DP both fit naturally
Because the same “minimum cost from this travel-day index onward” question will appear repeatedly, storing results is essential. Memoization makes the solution efficient by caching computed states. A bottom-up approach works as well by computing answers from the end of the travel schedule back to the beginning.
Both strategies rely on the same core recurrence. The difference is whether you compute on demand or precompute in reverse order.
Time and space complexity considerations
The number of states is proportional to the number of travel days, not the number of days in the year. That is an important optimization because travel days are often sparse compared to a full calendar.
Each state evaluates three pass options and advances to the next uncovered travel day, which keeps total computation manageable. Space usage is driven by storing the DP results and the travel days list.
Why this problem is common in interviews
Minimum Cost For Tickets appears often in interviews because it tests whether candidates can recognize when time-based decisions require dynamic programming. It’s easy to oversimplify this into a greedy choice problem, but the pass durations introduce forward dependencies that greedy methods cannot reliably handle.
The problem also checks whether you can model a calendar or timeline efficiently without iterating over every day of the year.
What this problem teaches beyond ticket pricing
Beyond travel passes, this problem teaches a general lesson about interval coverage optimization. Whenever you have items that cover a range of future requirements, the optimal plan usually involves comparing choices based on how far they advance you and what remains afterward.
If you can clearly explain why greedy fails, how the state is defined around the next uncovered travel day, and why jumping to the next uncovered day keeps the solution efficient, you demonstrate strong dynamic programming intuition. That depth of understanding makes “Minimum Cost For Tickets” an excellent exercise in time-based optimization and decision modeling.
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