Is Maximum Cost of Trip With K Highways easy or hard?

Maximum Cost of Trip With K Highways is classified as Hard. The challenge comes from recognizing that the path constraint requires subset dynamic programming with bitmasking, a technique often used when the number of nodes is small but all combinations must be explored efficiently.

Maximum Cost of Trip With K Highways Python/Java solution

Python and Java implementations both follow the same idea: iterate through all bitmasks, track dp[mask][u], and transition to neighbors not yet visited. Bit operations like mask | (1 << v) update the visited set. The algorithm runs in O(2^n * n^2) regardless of language.

How to solve Maximum Cost of Trip With K Highways in O(2^n * n^2)?

Represent visited cities with a bitmask and use dynamic programming. Maintain dp[mask][u] as the maximum cost for visiting the cities in mask and ending at u. For each state, try extending the trip to an unvisited neighbor v if a highway exists. Only consider states where the number of edges used equals k when computing the final answer.

What is the best approach for Maximum Cost of Trip With K Highways?

The most effective solution uses state compression dynamic programming with a bitmask. The DP state dp[mask][u] stores the maximum trip cost for visiting the set of cities in mask and ending at city u. By expanding the mask to unvisited neighbors and counting edges using bitcount(mask) - 1, you compute the best path with exactly k highways. The complexity is O(2^n * n^2).

Is Maximum Cost of Trip With K Highways asked at Google/Amazon/Meta?

Graph path optimization combined with bitmask dynamic programming appears frequently in interviews at companies like Google, Amazon, and Meta. Variants include traveling with constraints, visiting subsets of nodes, or maximizing path value under length limits. This problem tests subset DP and graph traversal together.

What data structure is used in Maximum Cost of Trip With K Highways?

The solution relies on bitmasks to represent visited city sets and a 2D DP table dp[mask][node]. The graph itself is typically stored as an adjacency matrix or adjacency list for fast edge lookups during transitions.

What is the time complexity of Maximum Cost of Trip With K Highways?

The optimal bitmask DP solution runs in O(2^n * n^2) time and uses O(2^n * n) space. Each subset of cities (mask) is processed, and transitions check possible neighboring cities. Because n is small (typically ≤15), this state compression approach is efficient enough.

2247. Maximum Cost of Trip With K Highways - Practice Online

Q: How to solve Maximum Cost of Trip With K Highways in O(2^n * n^2)?

Represent visited cities with a bitmask and use dynamic programming. Maintain dp[mask][u] as the maximum cost for visiting the cities in mask and ending at u. For each state, try extending the trip to an unvisited neighbor v if a highway exists. Only consider states where the number of edges used equals k when computing the final answer.

Q: What is the best approach for Maximum Cost of Trip With K Highways?

The most effective solution uses state compression dynamic programming with a bitmask. The DP state dp[mask][u] stores the maximum trip cost for visiting the set of cities in mask and ending at city u. By expanding the mask to unvisited neighbors and counting edges using bitcount(mask) - 1, you compute the best path with exactly k highways. The complexity is O(2^n * n^2).

Q: Is Maximum Cost of Trip With K Highways asked at Google/Amazon/Meta?

Graph path optimization combined with bitmask dynamic programming appears frequently in interviews at companies like Google, Amazon, and Meta. Variants include traveling with constraints, visiting subsets of nodes, or maximizing path value under length limits. This problem tests subset DP and graph traversal together.

Q: What data structure is used in Maximum Cost of Trip With K Highways?

The solution relies on bitmasks to represent visited city sets and a 2D DP table dp[mask][node]. The graph itself is typically stored as an adjacency matrix or adjacency list for fast edge lookups during transitions.

Q: What is the time complexity of Maximum Cost of Trip With K Highways?

The optimal bitmask DP solution runs in O(2^n * n^2) time and uses O(2^n * n) space. Each subset of cities (mask) is processed, and transitions check possible neighboring cities. Because n is small (typically ≤15), this state compression approach is efficient enough.

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.

Example 1:

Input: n = 5, highways = [[0,1,4],[2,1,3],[1,4,11],[3,2,3],[3,4,2]], k = 3
Output: 17
Explanation:
One possible trip is to go from 0 -> 1 -> 4 -> 3. The cost of this trip is 4 + 11 + 2 = 17.
Another possible trip is to go from 4 -> 1 -> 2 -> 3. The cost of this trip is 11 + 3 + 3 = 17.
It can be proven that 17 is the maximum possible cost of any valid trip.

Note that the trip 4 -> 1 -> 0 -> 1 is not allowed because you visit the city 1 twice.

Example 2:

Input: n = 4, highways = [[0,1,3],[2,3,2]], k = 2
Output: -1
Explanation: There are no valid trips of length 2, so return -1.

Constraints:

2 <= n <= 15
1 <= highways.length <= 50
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.