Maximum Cost of Trip With K Highways - Problem

A series of highways connect n cities numbered from 0 to n - 1. You are given a 2D integer array highways where highways[i] = [city1_i, city2_i, toll_i] indicates that there is a highway that connects city1_i and city2_i, allowing a car to go from city1_i to city2_i and vice versa for a cost of toll_i.

You are also given an integer k. You are going on a trip that crosses exactly k highways. You may start at any city, but you may only visit each city at most once during your trip.

Return the maximum cost of your trip. If there is no trip that meets the requirements, return -1.

Input & Output

Example 1 — Basic Case

$ Input: highways = [[0,1,15],[0,3,8],[1,2,10],[1,3,20],[2,3,5]], k = 2

› Output: 35

💡 Note: Start at city 0, take highway to city 1 (cost 15), then take highway to city 3 (cost 20). Total cost = 15 + 20 = 35, using exactly 2 highways.

Example 2 — No Valid Path

$ Input: highways = [[0,1,10],[1,2,5]], k = 3

› Output: -1

💡 Note: Only 3 cities connected in a line, but we need exactly 3 highways. Maximum possible is 2 highways (0→1→2), so return -1.

Example 3 — Single Highway

$ Input: highways = [[0,1,100]], k = 1

› Output: 100

💡 Note: Only one highway available with cost 100. We need exactly 1 highway, so the answer is 100.

Constraints

1 ≤ highways.length ≤ 15
highways[i].length == 3
0 ≤ city1_i, city2_i ≤ n - 1
city1_i ≠ city2_i
0 ≤ toll_i ≤ 100
1 ≤ k ≤ 50
There are no duplicate highways.

Visualization

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Asked in

G Google 12 a Amazon 8 M Microsoft 6 f Meta 4

The key insight is to use DFS with bitmask memoization to efficiently track visited cities and avoid recomputing the same states. The optimal approach combines dynamic programming with graph traversal. Time: O(n × 2ⁿ × k), Space: O(n × 2ⁿ × k).

Common Approaches

✓ Brute Force DFS

⏱️ Time: O(n × k!) Space: O(k)

For each starting city, use DFS to explore all possible paths of exactly k highways. Track visited cities to avoid revisiting and keep track of the maximum cost found.

DFS with Bitmask Memoization

⏱️ Time: O(n × 2ⁿ × k) Space: O(n × 2ⁿ × k)

Combine DFS with dynamic programming using bitmasks to represent visited cities. Memoize results to avoid recomputing the same state (current city, visited cities, remaining highways).

Brute Force DFS — Algorithm Steps

Build adjacency list from highways
For each city, start DFS with k remaining highways
Recursively explore unvisited neighbors
Track maximum cost among all valid paths

Visualization

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Step-by-Step Walkthrough

Build Graph

Create adjacency list from highways

Try Each Start

Start DFS from each city

Explore Paths

Recursively visit unvisited neighbors

Track Maximum

Keep track of maximum cost found

Code -

solution.c — C

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

int maxCost = -1;
int graph[20][20];
int n = 0;

void dfs(int city, int* visited, int remaining, int currentCost) {
    if (remaining == 0) {
        if (currentCost > maxCost) {
            maxCost = currentCost;
        }
        return;
    }
    
    for (int neighbor = 0; neighbor < n; neighbor++) {
        if (graph[city][neighbor] > 0 && !visited[neighbor]) {
            visited[neighbor] = 1;
            dfs(neighbor, visited, remaining - 1, currentCost + graph[city][neighbor]);
            visited[neighbor] = 0;
        }
    }
}

int solution(int highways[][3], int highwaysSize, int k) {
    // Initialize graph
    memset(graph, 0, sizeof(graph));
    n = 0;
    
    // Build adjacency matrix
    for (int i = 0; i < highwaysSize; i++) {
        int u = highways[i][0], v = highways[i][1], toll = highways[i][2];
        graph[u][v] = toll;
        graph[v][u] = toll;
        if (u + 1 > n) n = u + 1;
        if (v + 1 > n) n = v + 1;
    }
    
    maxCost = -1;
    
    // Try starting from each city
    for (int start = 0; start < n; start++) {
        int visited[20] = {0};
        visited[start] = 1;
        dfs(start, visited, k, 0);
    }
    
    return maxCost;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for small inputs
    int highways[50][3];
    int highwaysSize = 0;
    
    // Parse highways (simplified for demonstration)
    char* ptr = line + 1; // Skip first [
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            ptr++;
            highways[highwaysSize][0] = atoi(ptr);
            while (*ptr && *ptr != ',') ptr++;
            ptr++;
            highways[highwaysSize][1] = atoi(ptr);
            while (*ptr && *ptr != ',') ptr++;
            ptr++;
            highways[highwaysSize][2] = atoi(ptr);
            while (*ptr && *ptr != ']') ptr++;
            ptr++;
            highwaysSize++;
        }
        ptr++;
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(highways, highwaysSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × k!)

For each starting city, we explore paths of length k with factorial branching

✓ Linear Growth

Space Complexity

O(k)

Recursion stack depth up to k levels plus visited array

✓ Linear Space

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Maximum Cost of Trip With K Highways - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force DFS — Algorithm Steps

Visualization

Code -

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