2361

2361. Minimum Costs Using the Train Line (Hard)

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

Companies: Citadel

Related Topics:
Array, Dynamic Programming

Similar Questions:

• Bus Routes (Hard)
• Distance Between Bus Stops (Easy)

Solution 1. DP

// OJ: https://leetcode.com/problems/minimum-costs-using-the-train-line
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1) extra space
class Solution {
public:
    vector<long long> minimumCosts(vector<int>& A, vector<int>& B, int C) {
        int N = A.size();
        vector<long long> ans(N);
        long long dp[2] = {0, C};
        for (int i = 0; i < N; ++i) {
            long long next[2] = {};
            next[0] = min({dp[0] + A[i], dp[1] + A[i], dp[1] + B[i]});
            next[1] = min(dp[0] + A[i] + C, dp[1] + B[i]);
            ans[i] = min(next[0], next[1]);
            swap(dp, next);
        }
        return ans;
    }
};