Is Minimum Costs Using the Train Line easy or hard?

Minimum Costs Using the Train Line is labeled Hard because it requires modeling two DP states and handling a one-time transition cost correctly. Once the state transitions are defined, the implementation becomes a linear-time dynamic programming solution.

Minimum Costs Using the Train Line Python/Java solution

The solution in Python, Java, C++, Go, or TypeScript follows the same idea: maintain two DP states for regular and express travel. Update them while iterating through the arrays and append the minimum of the two states for each station.

How to solve Minimum Costs Using the Train Line in O(n)?

Track two running costs: the minimum cost to reach the current station using the regular line and using the express line. For every station, update these values based on staying on the same line or switching to express with the additional entry cost. Iterating through the arrays once produces the minimum cost for each station.

What is the best approach for Minimum Costs Using the Train Line?

The best approach uses dynamic programming with two states: cost reaching a station via the regular line and via the express line. By updating these states for each station, you compute the minimum cost efficiently in O(n) time. The optimized version stores only the previous state, reducing space to O(1).

Is Minimum Costs Using the Train Line asked at Google/Amazon/Meta?

Dynamic programming problems with multiple states and transition costs appear frequently in interviews at companies like Google, Amazon, and Meta. Variations of this problem test state modeling, cost transitions, and space optimization techniques.

What data structure is used in Minimum Costs Using the Train Line?

The core structure is dynamic programming over arrays. Two variables or a DP table track the minimum cost for regular and express lines at each station while iterating through the cost arrays.

What is the time complexity of Minimum Costs Using the Train Line?

The optimal solution runs in O(n) time because each station is processed once while updating two DP states. Space complexity can be O(n) with a full DP table or O(1) with a rolling-state optimization.

Description Hints 4 Solution Video 7:40

2361. Minimum Costs Using the Train Line

Hard78.4% acceptance1 companyPremium elsewhere, free here

A train line going through a city has two routes, the regular route and the express route. Both routes go through the same n + 1 stops labeled from 0 to n. Initially, you start on the regular route at stop 0.

You are given two 1-indexed integer arrays regular and express, both of length n. regular[i] describes the cost it takes to go from stop i - 1 to stop i using the regular route, and express[i] describes the cost it takes to go from stop i - 1 to stop i using the express route.

You are also given an integer expressCost which represents the cost to transfer from the regular route to the express route.

Note that:

There is no cost to transfer from the express route back to the regular route.
You pay expressCost every time you transfer from the regular route to the express route.
There is no extra cost to stay on the express route.

Return a 1-indexed array costs of length n, where costs[i] is the minimum cost to reach stop i from stop 0.

Note that a stop can be counted as reached from either route.

Example 1:

Input: regular = [1,6,9,5], express = [5,2,3,10], expressCost = 8
Output: [1,7,14,19]
Explanation: The diagram above shows how to reach stop 4 from stop 0 with minimum cost.
- Take the regular route from stop 0 to stop 1, costing 1.
- Take the express route from stop 1 to stop 2, costing 8 + 2 = 10.
- Take the express route from stop 2 to stop 3, costing 3.
- Take the regular route from stop 3 to stop 4, costing 5.
The total cost is 1 + 10 + 3 + 5 = 19.
Note that a different route could be taken to reach the other stops with minimum cost.

Example 2:

Input: regular = [11,5,13], express = [7,10,6], expressCost = 3
Output: [10,15,24]
Explanation: The diagram above shows how to reach stop 3 from stop 0 with minimum cost.
- Take the express route from stop 0 to stop 1, costing 3 + 7 = 10.
- Take the regular route from stop 1 to stop 2, costing 5.
- Take the express route from stop 2 to stop 3, costing 3 + 6 = 9.
The total cost is 10 + 5 + 9 = 24.
Note that the expressCost is paid again to transfer back to the express route.

Constraints:

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

Asked at 1 company

Citadel

Topics

Array Dynamic Programming

Hints open one at a time

Hint 1

Notice and evaluate the different ways there are to move from one stop to the next.

Hint 2

From the express route at a previous stop, we can use either the express route or the regular route to the next stop without paying expressCost.

Hint 3

From the regular route at a previous stop, we can either use the express route after paying expressCost or use the regular route without paying expressCost.

Hint 4

Iterate through the stops and compare the above cases to obtain the minimum costs for each stop.

7:40

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