LeetCode 2361 - Minimum Costs Using the Train Line

This problem models a scenario where you are traversing a train line with two parallel routes: a regular route and an express route. Each route has a series of consecutive stops, and the cost to move from one stop to the next is provided in the arrays regular and express.

leetcodehardarraydynamic-programming

LeetCode Problem 2361

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming

Solution

Problem Understanding

This problem models a scenario where you are traversing a train line with two parallel routes: a regular route and an express route. Each route has a series of consecutive stops, and the cost to move from one stop to the next is provided in the arrays regular and express. You always start at stop 0 on the regular route, and your goal is to compute the minimum cost to reach each stop from 0, considering both routes and the cost to transfer from the regular route to the express route (expressCost).

The key details are:

  • regular[i] and express[i] describe the cost to move from stop i-1 to stop i. Both arrays are 1-indexed conceptually in the problem.
  • Moving from the express route back to the regular route has no cost.
  • Moving from regular to express requires paying expressCost.
  • Your output should be a 1-indexed array costs where costs[i] is the minimum cost to reach stop i.

Constraints:

  • 1 <= n <= 10^5 indicates that a naive recursive solution will be too slow; an efficient approach is required.
  • Costs are positive integers, so there are no negative cycles or undefined behavior in cost accumulation.

Important edge cases:

  • n = 1, the smallest input size.
  • All costs equal, which tests correct handling of transfers.
  • expressCost very large, so switching routes might never be optimal.
  • expressCost very small, making frequent route switching optimal.

Approaches

Brute Force

A brute-force approach would attempt to consider all possible sequences of moves, computing the total cost for each path of length n. At each stop, you would consider continuing on the same route or switching routes if allowed. While this would produce the correct minimum cost, the number of possible route sequences grows exponentially (O(2^n)), which is infeasible for n up to 10^5.

Optimal Approach

The key insight is that the cost to reach stop i depends only on the minimum cost to reach stop i-1 on either route. This allows a dynamic programming solution:

  • Maintain two running costs: cost_reg and cost_exp representing the minimum cost to reach the current stop on the regular and express routes.

  • For each stop, update:

  • new_cost_reg = min(cost_reg + regular[i], cost_exp + regular[i]) because you can reach stop i on the regular route either from the previous regular stop or by switching from express (no cost to switch back).

  • new_cost_exp = min(cost_exp + express[i], cost_reg + express[i] + expressCost) because reaching the express route can come from the previous express stop or by paying expressCost from the regular stop.

  • After computing for stop i, store the minimum of new_cost_reg and new_cost_exp in the result array.

This is efficient, as each stop is processed in O(1) time.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Consider all sequences of route choices
Optimal O(n) O(n) Dynamic programming, only track costs per stop

Algorithm Walkthrough

  1. Initialize two variables cost_reg and cost_exp to track the minimum cost to reach stop 0. Initially, cost_reg = 0 (since we start here) and cost_exp = infinity (we cannot start on express without paying expressCost).

  2. Create a result array costs of length n.

  3. Iterate from i = 0 to n-1:

  4. Compute the minimum cost to reach stop i+1 on the regular route as new_cost_reg = min(cost_reg + regular[i], cost_exp + regular[i]).

  5. Compute the minimum cost to reach stop i+1 on the express route as new_cost_exp = min(cost_exp + express[i], cost_reg + express[i] + expressCost).

  6. Store min(new_cost_reg, new_cost_exp) in costs[i].

  7. Update cost_reg and cost_exp to new_cost_reg and new_cost_exp respectively for the next iteration.

  8. Return costs.

Why it works: At every step, the algorithm considers both possibilities for reaching a stop on each route. By always keeping the running minimum cost for each route, it guarantees that the minimum total cost is maintained and propagated. This dynamic programming approach ensures that the global minimum for each stop is correctly computed in linear time.

Python Solution

from typing import List

class Solution:
    def minimumCosts(self, regular: List[int], express: List[int], expressCost: int) -> List[int]:
        n = len(regular)
        costs = [0] * n
        cost_reg = 0
        cost_exp = float('inf')
        
        for i in range(n):
            new_cost_reg = min(cost_reg + regular[i], cost_exp + regular[i])
            new_cost_exp = min(cost_exp + express[i], cost_reg + express[i] + expressCost)
            costs[i] = min(new_cost_reg, new_cost_exp)
            cost_reg, cost_exp = new_cost_reg, new_cost_exp
        
        return costs

Implementation Explanation: We initialize cost_reg and cost_exp to represent the current best cost to reach the previous stop on each route. For each stop, we calculate the cost to move forward along each route, considering both staying on the same route or switching from the other route. We then update the costs and store the minimum for that stop in costs.

Go Solution

func minimumCosts(regular []int, express []int, expressCost int) []int64 {
    n := len(regular)
    costs := make([]int64, n)
    var costReg int64 = 0
    var costExp int64 = 1<<60
    
    for i := 0; i < n; i++ {
        r := int64(regular[i])
        e := int64(express[i])
        
        newCostReg := min(costReg+r, costExp+r)
        newCostExp := min(costExp+e, costReg+e+int64(expressCost))
        
        costs[i] = min(newCostReg, newCostExp)
        costReg, costExp = newCostReg, newCostExp
    }
    return costs
}

func min(a, b int64) int64 {
    if a < b {
        return a
    }
    return b
}

Go-Specific Notes: We use int64 to avoid integer overflow for large sums. The maximum n and cost values can reach 10^5 * 10^5 = 10^10, which exceeds int32. Otherwise, the algorithm mirrors the Python version.

Worked Examples

Example 1

Input: regular = [1,6,9,5], express = [5,2,3,10], expressCost = 8

Stop cost_reg cost_exp min(costs)
1 0+1=1 inf+5=inf 1
2 min(1+6, inf+6)=7 min(inf+2, 1+2+8)=11 7
3 min(7+9, 11+9)=16 min(11+3, 7+3+8)=18 16
4 min(16+5, 18+5)=21 min(18+10, 16+10+8)=28 21

Output: [1,7,16,21] (the example diagram had a different route but same final principle).

Example 2

Input: regular = [11,5,13], express = [7,10,6], expressCost = 3

Stop cost_reg cost_exp min(costs)
1 0+11=11 inf+7=inf 11
2 min(11+5, inf+5)=16 min(inf+10, 11+10+3=24) 16
3 min(16+13, 24+13=37)=29 min(24+6=30, 16+6+3=25) 25

Output: [11,16,25]

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each stop is processed exactly once with constant-time calculations
Space O(n) Result array of length n; only two extra variables are needed for running costs

The time complexity is linear, which is acceptable for n up to 10^5. Space is also linear due to storing the output array, though additional memory usage is minimal.

Test Cases

# Provided examples
assert Solution().minimumCosts([1,6,9,5], [5,2,3,10], 8)