Minimum Costs Using the Train Line

Minimum Costs Using the Train Line - Problem

🚆 Minimum Costs Using the Train Line

Imagine you're planning a journey through a city with two train routes: a regular route and an express route. Both routes connect the same n + 1 stops (labeled 0 to n), but they have different costs and speeds.

Your Journey: You start at stop 0 on the regular route and want to find the cheapest way to reach each subsequent stop.

Route Rules:

🚂 Regular Route: regular[i] = cost from stop i-1 to stop i
🚄 Express Route: express[i] = cost from stop i-1 to stop i
💰 Transfer Cost: Pay expressCost each time you switch from regular to express
🆓 Free Transfer: Switch from express back to regular anytime (no cost)

Goal: Return an array costs where costs[i] is the minimum cost to reach stop i from stop 0.

Example: If regular = [1, 3, 2], express = [4, 1, 1], and expressCost = 2, you might take the regular route to stop 1 (cost: 1), then transfer to express for stops 2-3 (total additional cost: 2 + 1 + 1 = 4).

Input & Output

example_1.py — Basic Route Selection

$ Input: regular = [1, 3, 2], express = [4, 1, 1], expressCost = 2

› Output: [1, 4, 5]

💡 Note: Stop 1: Take regular route (cost: 1). Stop 2: Stay on regular (cost: 1+3=4). Stop 3: Transfer to express at stop 1 is better: regular[0] + expressCost + express[1] + express[2] = 1 + 2 + 1 + 1 = 5, vs staying regular = 1+3+2 = 6.

example_2.py — Express Always Better

$ Input: regular = [10, 10, 10], express = [1, 1, 1], expressCost = 1

› Output: [2, 3, 4]

💡 Note: Express route is much cheaper despite transfer cost. Stop 1: Transfer immediately (cost: 1+1=2). Stop 2: Continue on express (cost: 2+1=3). Stop 3: Continue on express (cost: 3+1=4).

example_3.py — High Transfer Cost

$ Input: regular = [2, 2, 2], express = [1, 1, 1], expressCost = 10

› Output: [2, 4, 6]

💡 Note: Transfer cost is too high, so regular route is always better. Stop 1: Regular (cost: 2). Stop 2: Regular (cost: 4). Stop 3: Regular (cost: 6). Even though express is cheaper per segment, the 10-unit transfer cost makes it uneconomical.

Constraints

1 ≤ n ≤ 1000
1 ≤ regular[i], express[i] ≤ 1000
1 ≤ expressCost ≤ 1000
Important: You always start on the regular route at stop 0

Visualization

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Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28 Apple 22

The optimal solution uses dynamic programming to track the minimum cost to reach each stop via both the regular and express routes. At each stop, we calculate: (1) Regular cost = min(previous regular, previous express) + regular route cost (free transfer from express), (2) Express cost = min(previous regular + transfer cost, previous express) + express route cost. The answer for each stop is the minimum of these two costs. This approach runs in O(n) time and O(1) extra space.

Common Approaches

✓ Brute Force (Try All Route Combinations)

⏱️ Time: O(2^n) Space: O(n)

Generate all possible ways to traverse from stop 0 to each destination, considering every possible point where we could switch from regular to express route. For each combination, calculate the total cost and keep track of the minimum.

Dynamic Programming (Optimal)

⏱️ Time: O(n) Space: O(1)

Use dynamic programming to efficiently compute the minimum cost to reach each stop. At each stop, we maintain two values: the minimum cost to reach that stop via the regular route, and the minimum cost via the express route. We build the solution incrementally, using the optimal costs from the previous stop.

Brute Force (Try All Route Combinations) — Algorithm Steps

For each destination stop, generate all possible route combinations
For each combination, calculate total cost including transfer fees
Track the minimum cost among all combinations
Repeat for each destination stop

Visualization

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Step-by-Step Walkthrough

Generate Combinations

Create all possible ways to switch between regular and express routes

Calculate Costs

For each combination, sum up route costs and transfer fees

Find Minimum

Compare all combinations and select the one with minimum cost

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string.h>

int* minimumCosts(int* regular, int regularSize, int* express, int expressSize, int expressCost, int* returnSize) {
    int n = regularSize;
    int* result = (int*)malloc(n * sizeof(int));
    *returnSize = n;
    
    for (int target = 1; target <= n; target++) {
        int minCost = INT_MAX;
        
        // Try all possible route combinations
        for (int mask = 0; mask < (1 << target); mask++) {
            int cost = 0;
            int onExpress = 0;
            
            for (int i = 0; i < target; i++) {
                // Check if we should use express for this segment
                int shouldUseExpress = (mask & (1 << i)) != 0;
                
                // If switching from regular to express, pay transfer cost
                if (shouldUseExpress && !onExpress) {
                    cost += expressCost;
                    onExpress = 1;
                } else if (!shouldUseExpress) {
                    onExpress = 0;
                }
                
                // Add segment cost
                if (onExpress) {
                    cost += express[i];
                } else {
                    cost += regular[i];
                }
            }
            
            if (cost < minCost) {
                minCost = cost;
            }
        }
        
        result[target - 1] = minCost;
    }
    
    return result;
}

int main() {
    char line[10000];
    
    // Read regular array
    fgets(line, sizeof(line), stdin);
    int regular[1000];
    int regularSize = 0;
    char* token = strtok(line, ",\n");
    while (token != NULL) {
        regular[regularSize++] = atoi(token);
        token = strtok(NULL, ",\n");
    }
    
    // Read express array
    fgets(line, sizeof(line), stdin);
    int express[1000];
    int expressSize = 0;
    token = strtok(line, ",\n");
    while (token != NULL) {
        express[expressSize++] = atoi(token);
        token = strtok(NULL, ",\n");
    }
    
    // Read express cost
    int expressCost;
    scanf("%d", &expressCost);
    
    int returnSize;
    int* result = minimumCosts(regular, regularSize, express, expressSize, expressCost, &returnSize);
    
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

For each stop, we have 2 choices (regular or express), leading to 2^n total combinations

⚠ Quadratic Growth

Space Complexity

O(n)

Space for recursion stack and storing the result array

⚡ Linearithmic Space

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🚆 Minimum Costs Using the Train Line

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Route Combinations) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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