Minimum Costs Using the Train Line

Hard

Citadel

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Topics:

Dynamic Programming

A train line has n stations numbered from 1 to n. You are given a 0-indexed array regular of length n - 1 where regular[i] is the cost to travel from station i + 1 to station i + 2. You are also given a 0-indexed array express of length n - 1 where express[i] is the cost to travel from station i + 1 to station i + 2 using the express line.

You can only use the express line at most once. Return the minimum cost to travel from station 1 to station n.

Example 1:

Input: regular = [1,5,6,9], express = [8,4,2,1]
Output: 15
Explanation:
One possible way to minimize the cost is to take the regular line from station 1 to station 3, the express line from station 3 to station 4, and the regular line from station 4 to station 5.
The total cost is regular[0] + regular[1] + express[2] + regular[3] = 1 + 5 + 2 + 9 = 17.
However, a better way is to take the express line from station 2 to station 3, and the regular line from station 1 to station 2 and from station 3 to station 5.
The total cost is regular[0] + express[1] + regular[2] + regular[3] = 1 + 4 + 6 + 9 = 20.
But the optimal way is to take the regular line from station 1 to station 2, the express line from station 2 to station 4, and the regular line from station 4 to station 5.
The total cost is regular[0] + express[1] + express[2] + regular[3] = 1 + 4 + 2 + 9 = 16.
But the optimal way is to take the regular line from station 1 to station 3, the express line from station 3 to station 5.
The total cost is regular[0] + regular[1] + express[2] + express[3] = 1 + 5 + 2 + 1 = 9.

Example 2:

Input: regular = [1,1,1,1], express = [9,9,9,9]
Output: 4
Explanation:
It is optimal to use the regular line from station 1 to station 5.

Constraints:

2 <= n <= 10⁵
regular.length == n - 1
express.length == n - 1
1 <= regular[i], express[i] <= 10⁵

Try coding it in LeetCode

Minimum Costs Using the Train Line

Hard

Citadel

3 views

Topics:

Dynamic Programming

A train line has n stations numbered from 1 to n. You are given a 0-indexed array regular of length n - 1 where regular[i] is the cost to travel from station i + 1 to station i + 2. You are also given a 0-indexed array express of length n - 1 where express[i] is the cost to travel from station i + 1 to station i + 2 using the express line.

You can only use the express line at most once. Return the minimum cost to travel from station 1 to station n.

Example 1:

Input: regular = [1,5,6,9], express = [8,4,2,1]
Output: 15
Explanation:
One possible way to minimize the cost is to take the regular line from station 1 to station 3, the express line from station 3 to station 4, and the regular line from station 4 to station 5.
The total cost is regular[0] + regular[1] + express[2] + regular[3] = 1 + 5 + 2 + 9 = 17.
However, a better way is to take the express line from station 2 to station 3, and the regular line from station 1 to station 2 and from station 3 to station 5.
The total cost is regular[0] + express[1] + regular[2] + regular[3] = 1 + 4 + 6 + 9 = 20.
But the optimal way is to take the regular line from station 1 to station 2, the express line from station 2 to station 4, and the regular line from station 4 to station 5.
The total cost is regular[0] + express[1] + express[2] + regular[3] = 1 + 4 + 2 + 9 = 16.
But the optimal way is to take the regular line from station 1 to station 3, the express line from station 3 to station 5.
The total cost is regular[0] + regular[1] + express[2] + express[3] = 1 + 5 + 2 + 1 = 9.

Example 2:

Input: regular = [1,1,1,1], express = [9,9,9,9]
Output: 4
Explanation:
It is optimal to use the regular line from station 1 to station 5.

Constraints:

2 <= n <= 10⁵
regular.length == n - 1
express.length == n - 1
1 <= regular[i], express[i] <= 10⁵

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 possible values for the costs associated with traveling between stations, and the cost of skipping a station? Can these values be negative?
What are the constraints on the number of stations? Could the train line be empty (no stations), or have only one or two stations?
If it's impossible to reach the end of the line, what value should I return? Should I return a special value (e.g., -1 or infinity) or throw an exception?
Is skipping a station always allowed, or are there any constraints on when I can skip a station?
Are the station positions provided in a specific order (e.g., sorted along the train line), or can they be in any order? Does the cost of skipping a station depend on the position (index) of the station being skipped?

Brute Force Solution

Approach

The brute force way to solve this is to consider every possible travel plan along the train line. We'll explore absolutely every combination of stops to find the cheapest way to get to the end. It's like trying every possible route on a map until you find the one that costs the least.

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

Start at the beginning and consider going directly to the end. Calculate the cost of this direct trip.
Next, consider stopping at the first possible station along the way, and then going to the end from there. Calculate the total cost.
Then, consider stopping at the second station instead, and again calculate the total cost to the end.
Continue this process, trying every single possible station as your first stop, and calculating the cost for each of those options to the end.
Now, for each of those first stops, consider every possible second stop. Calculate the cost for each combination of first and second stops to the end.
Keep doing this, adding more and more stops into the mix, trying absolutely every possible combination of stations you could visit along the way.
Once you've tried every conceivable combination of stations, compare the total cost of all the different travel plans you've generated.
The travel plan with the lowest total cost is your answer.

Code Implementation

def minimum_costs_using_the_train_line_brute_force(costs):
    number_of_stations = len(costs)
    minimum_cost = float('inf')

    def calculate_cost(path):
        total_cost = 0
        for i in range(len(path) - 1):
            total_cost += costs[path[i]][path[i+1]]
        return total_cost

    def find_all_paths(current_path):
        nonlocal minimum_cost

        # If we've reached the end, calculate the cost
        if current_path[-1] == number_of_stations - 1:

            path_cost = calculate_cost(current_path)
            minimum_cost = min(minimum_cost, path_cost)
            return

        # Explore all possible next stops
        last_station = current_path[-1]

        for next_station in range(last_station + 1, number_of_stations):

            # Ensure we don't revisit stations
            if next_station not in current_path:

                find_all_paths(current_path + [next_station])

    # Start the search from the first station
    find_all_paths([0])

    return minimum_cost

Big(O) Analysis

Time Complexity

O(2^n) – The provided brute-force approach explores all possible subsets of stations along the train line. For each station, we either include it in our travel plan or exclude it. With n stations, this leads to 2^n possible combinations to evaluate. Therefore, the time complexity grows exponentially with the number of stations. Checking each combination involves calculating the cost, which takes O(n) in the worst case. Overall, the time complexity is dominated by the exploration of all subsets, resulting in O(2^n).

Space Complexity

O(N^N) – The brute force approach, as described, explores every possible travel plan, considering all combinations of stops. This involves implicitly building up many different paths to try, potentially storing the cost of each path. Since we're considering every combination of stops, and each stop can be present or absent in a path, the number of possible paths grows exponentially with the number of stations, N. In the worst case, the number of possible combinations to store approaches N^N, as each of the N possible stops could be considered at each stage. Therefore, the auxiliary space grows proportionally to the number of combinations of stops, resulting in a space complexity of O(N^N).

Optimal Solution

Approach

The goal is to find the cheapest way to travel between stations using a train line, where you can either travel directly between two stations or transfer to another train. We solve this by calculating the minimum cost to reach each station step-by-step, remembering the cheapest path we've found so far to each location.

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

Start at the beginning station, which has a cost of zero to reach.
For each station you can directly reach from the starting station, calculate the cost of going there.
If this calculated cost is less than the current cheapest cost to reach that station (which might be infinity at first), update the cheapest cost to that station.
Repeat this process for each station you've now reached: look at all the stations you can reach from it, calculate the cost, and update the cheapest costs if necessary.
Also consider transfers: if you can transfer at a station, see if transferring and then taking a different train line would give you a cheaper route to a new station. Update the cheapest costs accordingly.
Continue this process until you've considered all possible routes and transfers to all reachable stations.
The final cheapest cost to the destination station is the minimum cost to travel between the stations.

Code Implementation

def find_minimum_costs(train_lines, start_station, end_station, transfers):
    number_of_stations = len(set([station for line in train_lines for station in line] + [station1 for station1, station2, cost in transfers] + [station2 for station1, station2, cost in transfers]))
    minimum_costs = {station: float('inf') for station in range(number_of_stations)}
    minimum_costs[start_station] = 0
    visited_stations = {start_station}

    while visited_stations:
        current_station = visited_stations.pop()

        # Iterate through each train line.
        for train_line in train_lines:
            if current_station in train_line:
                current_station_index = train_line.index(current_station)

                # Check stations reachable in the forward direction
                for next_station_index in range(current_station_index + 1, len(train_line)):
                    next_station = train_line[next_station_index]
                    cost = next_station_index - current_station_index
                    new_cost = minimum_costs[current_station] + cost

                    # Update the cost if a cheaper path is found
                    if new_cost < minimum_costs[next_station]:
                        minimum_costs[next_station] = new_cost
                        visited_stations.add(next_station)

                # Check stations reachable in the backward direction
                for previous_station_index in range(current_station_index - 1, -1, -1):
                    previous_station = train_line[previous_station_index]
                    cost = current_station_index - previous_station_index
                    new_cost = minimum_costs[current_station] + cost

                    # Update the cost if a cheaper path is found
                    if new_cost < minimum_costs[previous_station]:
                        minimum_costs[previous_station] = new_cost
                        visited_stations.add(previous_station)

        # Check for possible transfers from the current station.
        for station1, station2, cost in transfers:
            if station1 == current_station:

                # Update the cost if a cheaper transfer is found.
                new_cost = minimum_costs[current_station] + cost
                if new_cost < minimum_costs[station2]:
                    minimum_costs[station2] = new_cost
                    visited_stations.add(station2)

    return minimum_costs[end_station]

Big(O) Analysis

Time Complexity

O(V*E) – The algorithm resembles Dijkstra's algorithm and iterates through each station (vertex) to find the minimum cost. For each station, it explores all possible connections to other stations (edges), including transfers, to update the minimum cost. In the worst-case scenario, we might visit each vertex V and for each vertex consider all of its outgoing edges E to potentially update costs. Therefore, the time complexity is proportional to the product of the number of vertices and edges, giving us O(V*E).

Space Complexity

O(N) – The algorithm maintains a data structure to store the cheapest cost to reach each station. In the worst-case scenario, the number of stations can be proportional to the input size, N, where N represents the number of stations. Therefore, the auxiliary space used to store these costs is proportional to N. This results in an auxiliary space complexity of O(N).

Edge Cases

Case	How to Handle
Null or empty price/time lists	Return 0 or throw an IllegalArgumentException as there are no stations to traverse.
Only one station (single element array)	Return 0, as there's no travel needed to reach the 'destination'.
Prices or times contain negative values	Throw an IllegalArgumentException, since prices and times can't be negative.
Prices or times contain very large values (potential integer overflow)	Use long data type to store intermediate cost calculations to prevent overflow.
Zero travel time between stations	Handle division by zero in the speed calculation if applicable; otherwise continue calculating minimum costs without issues
Maximum number of stations (large input size)	Ensure that the dynamic programming solution scales efficiently (e.g., O(n) space and time complexity) to avoid memory or time limit issues.
Prices or times are all zeros.	If times are all zero, either return 0 or throw an exception, since average speed is undefined; if prices are all zero, the total minimum cost will be 0
Prices or times are extremely skewed (one value much larger than others).	The algorithm should correctly handle scenarios where a single high-cost or long-duration segment dominates the overall cost.