Minimum Cost For Tickets

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Amazon

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Dynamic Programming

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, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total, you spent $11 and covered all the days of your travel.

Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total, you spent $17 and covered all the days of your travel.

Constraints:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

Try coding it in LeetCode

Minimum Cost For Tickets

Medium

Amazon

2 views

Topics:

Dynamic Programming

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, 3, 4, 5, 6, 7, and 8.

Return the minimum number of dollars you need to travel every day in the given list of days.

Example 1:

Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total, you spent $11 and covered all the days of your travel.

Example 2:

Input: days = [1,2,3,4,5,6,7,8,9,10,30,31], costs = [2,7,15]
Output: 17
Explanation: For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total, you spent $17 and covered all the days of your travel.

Constraints:

1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `days` array, and the range of values within `days` and `costs` arrays?
Are the days in the `days` array guaranteed to be in ascending order?
Can the `days` array be empty? If so, what should I return?
Are the costs in the `costs` array guaranteed to be non-negative?
If there are multiple ways to achieve the minimum cost, is any of them acceptable, or is there a specific way to determine which one to return?

Brute Force Solution

Approach

The brute force approach to finding the minimum cost for tickets involves checking every possible combination of ticket types we could buy for the given travel days. We consider all the combinations and pick the cheapest one. Basically, we're trying out every single way to buy tickets and finding the cheapest one.

Here's how the algorithm would work step-by-step:

Start from the first day we need a ticket.
Consider all the possible ticket options for that day: a one-day pass, a seven-day pass, or a thirty-day pass.
For each of these options, calculate the cost of the ticket and which days are now covered by that ticket.
Move on to the next day that still requires a ticket.
Again, consider all the ticket options for that day, taking into account any previous tickets that might already cover some of the days.
Repeat this process for every travel day, exploring all the possible combinations of tickets.
For each complete combination of tickets that covers all the travel days, calculate the total cost.
Compare the total cost of all the possible combinations, and select the combination with the lowest total cost.

Code Implementation

def mincostTicketsBruteForce(days, costs):
    number_of_days = len(days)
    minimum_cost = float('inf')

    def calculate_cost(current_day_index, current_cost):
        nonlocal minimum_cost

        # If all days are covered, update the minimum cost
        if current_day_index == number_of_days:
            minimum_cost = min(minimum_cost, current_cost)
            return

        # Try buying a 1-day pass
        calculate_cost(find_next_day(current_day_index, days[current_day_index] + 1), current_cost + costs[0])

        # Try buying a 7-day pass
        # Explore buying a 7-day pass
        calculate_cost(find_next_day(current_day_index, days[current_day_index] + 7), current_cost + costs[1])

        # Try buying a 30-day pass
        # Explore buying a 30-day pass
        calculate_cost(find_next_day(current_day_index, days[current_day_index] + 30), current_cost + costs[2])

    def find_next_day(current_day_index, covered_until):
        next_day_index = current_day_index + 1
        # Find the next travel day not covered by the current ticket
        while next_day_index < number_of_days and days[next_day_index] < covered_until:
            next_day_index += 1
        return next_day_index

    calculate_cost(0, 0)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach explores all possible combinations of ticket types (1-day, 7-day, and 30-day) for each travel day. In the worst-case scenario, we have to make a decision among these 3 ticket options for each of the 'n' travel days. This leads to a decision tree where each node has 3 branches, giving a total of approximately 3^n possible paths to explore. Therefore, the time complexity grows exponentially with the number of travel days 'n'.

Space Complexity

O(N) – The brute force approach explores all possible combinations of ticket purchases using recursion. In the worst-case scenario, the depth of the recursion can reach N, where N is the number of travel days. Each recursive call requires its own stack frame, leading to a call stack that can grow up to size N. Therefore, the auxiliary space used by the recursion stack is proportional to the number of travel days, N, resulting in a space complexity of O(N).

Optimal Solution

Approach

The best way to solve this problem is by figuring out the cheapest cost to travel up to each day. We use previous best costs to calculate the cost of traveling on any given day, picking the cheapest option at each step. This avoids recalculating costs and ensures we find the absolute minimum.

Here's how the algorithm would work step-by-step:

Create a plan to track the minimum cost to travel up to any given day. Start by filling in the cost to travel up to the day before any travel is needed, setting it to zero.
Work through each day, from the first day of travel onwards, and calculate the cheapest way to get there.
For each day you need to travel, consider three options: buying a one-day ticket, a seven-day ticket, or a thirty-day ticket.
For each option, figure out the cost of the ticket plus the cheapest way to get to the day *before* the ticket is effective. For example, the cost of a 7 day pass would be the pass cost + the cost of getting to 7 days before today.
Compare the costs of the three options, and save the cheapest one as the minimum cost to get to that day.
If you are on a day when you don't need to travel, the cheapest way to get there is to not buy a ticket, so the cost to get there is the same as the cost to get to the day before.
Keep going until you have calculated the minimum cost to travel up to the very last day you need to travel.
The minimum cost to travel up to the last travel day is your final answer.

Code Implementation

def minimum_cost_for_tickets(days, costs):
    last_day = days[-1]
    minimum_cost = [0] * (last_day + 1)

    days_set = set(days)

    for day in range(1, last_day + 1):
        if day not in days_set:
            minimum_cost[day] = minimum_cost[day - 1]
        else:
            # Consider all ticket options to find the min cost.
            minimum_cost[day] = costs[0] + minimum_cost[day - 1]

            minimum_cost[day] = min(minimum_cost[day],
                                  costs[1] + minimum_cost[max(0, day - 7)])

            # Thirty day pass
            minimum_cost[day] = min(minimum_cost[day],
                                  costs[2] + minimum_cost[max(0, day - 30)])

    return minimum_cost[last_day]

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each day from 1 to the last travel day. Let n be the largest day in the `days` array. Inside the loop, for each travel day, it considers three ticket options (1-day, 7-day, and 30-day). These options involve constant-time calculations (finding the minimum of three values and accessing previously computed costs). Therefore, the time complexity is directly proportional to n, resulting in O(n).

Space Complexity

–

Edge Cases

Case	How to Handle
days is null or empty	Return 0, as no travel is required.
costs is null or of incorrect size (not 3 elements)	Throw an IllegalArgumentException or return a predefined error code like -1.
days contains duplicate days	The dynamic programming solution correctly handles duplicate days since only the presence or absence of a day affects the result.
days contains non-positive integers (<=0)	Consider these invalid inputs and throw an IllegalArgumentException, as days should represent positive calendar days.
days is a very large array (e.g., 365 days)	Dynamic programming solution's space complexity is O(max(days)), so it might consume considerable memory but should still be solvable.
costs contains very large values (approaching Integer.MAX_VALUE)	Potential integer overflow during calculations, so use long data types to store intermediate cost values.
days is sorted in descending order	Sort the `days` array in ascending order as dynamic programming solution generally assumes that days are visited in increasing order.
days contains values beyond a reasonable time frame (e.g., year 10000)	Limit `days` value range to prevent excessive memory usage and/or integer overflow when indexing (e.g., maximum day being 365 for a year) and throw IllegalArgumentException if exceeded.