Problem Description

You are given n cities numbered from 0 to n - 1 that are connected by highways. Each highway is described by highways[i] = [city1_i, city2_i, toll_i], which means there is a bidirectional highway between city1_i and city2_i with a toll cost of toll_i.

You need to plan a trip that:

• Crosses exactly k highways
• Can start from any city
• Can visit each city at most once during the trip

Your goal is to find the maximum total toll cost for such a trip. If no valid trip exists that meets these requirements, return -1.

For example, if you have 4 cities and need to cross exactly 2 highways, you might start at city 0, take a highway to city 1 (paying toll₁), then take another highway to city 3 (paying toll₂). The total cost would be toll₁ + toll₂. You want to find the path that maximizes this total cost while meeting all constraints.

The key constraints are:

• You must use exactly k highways (not fewer, not more)
• Each city can be visited at most once (no revisiting cities)
• The highways are bidirectional (you can travel in either direction)

Intuition

The first observation is that if we need to cross exactly k highways and can visit each city at most once, then we can visit at most k + 1 cities (since each highway connects two cities). With n cities total, the maximum number of highways we can cross is n - 1. So if k >= n, it's impossible to meet the requirements.

The second key observation is that n is at most 15, which is a small number. This suggests we can use bitmask dynamic programming to track which cities we've visited. We can represent the set of visited cities as a binary number where bit i is 1 if city i has been visited.

The core insight is that to find the maximum cost path, we need to know:

1. Which cities we've visited so far (represented by a bitmask)
2. Which city we're currently at (since we need to know what highways are available from here)

This leads us to define f[mask][j] as the maximum cost to reach city j when we've visited exactly the cities represented by mask.

Starting from any city (with cost 0), we can build up paths by:

• Looking at each possible set of visited cities (mask)
• For each city j in that set, considering how we could have arrived there
• We could have come from any neighboring city h that's also in the visited set
• The cost would be the maximum cost to reach h with j removed from the visited set, plus the toll from h to j

By tracking the number of 1s in our bitmask (which equals the number of visited cities), we can identify when we've visited exactly k + 1 cities (meaning we've crossed exactly k highways). Among all such valid states, we take the maximum cost.

The beauty of this approach is that it systematically explores all possible paths while avoiding revisiting cities, and the small value of n makes the 2^n state space manageable.

Learn more about Graph, Dynamic Programming and Bitmask patterns.

Solution Approach

We implement the solution using state compression dynamic programming with the following steps:

1. Initial Check: First, we check if k >= n. If true, return -1 immediately since we can cross at most n - 1 highways when visiting each city at most once.

2. Build the Graph: We create an adjacency list g using a defaultdict where g[a] stores all neighbors of city a along with their toll costs. For each highway [a, b, cost], we add (b, cost) to g[a] and (a, cost) to g[b] since highways are bidirectional.

3. Initialize DP Table: We create a 2D array f where f[i][j] represents the maximum cost when:

• i is a bitmask representing visited cities
• j is the last city we're currently at

Initialize f[1 << i][i] = 0 for all cities i, meaning we start at city i with cost 0. All other states are initialized to negative infinity.

4. State Transitions: For each possible state i (bitmask of visited cities):

• For each city j, check if it's in the visited set by testing i >> j & 1
• If city j is visited, look at all its neighbors h with toll cost
• If neighbor h is also in the visited set, we could have reached j from h
• Update: f[i][j] = max(f[i][j], f[i ^ (1 << j)][h] + cost)
  • Here i ^ (1 << j) removes city j from the visited set, giving us the state before visiting j

5. Track the Answer: After updating f[i][j], check if the number of visited cities equals k + 1 using i.bit_count(). If so, update the answer: ans = max(ans, f[i][j]).

6. Return Result: Return the maximum cost found. If no valid path exists, ans remains -1.

The time complexity is O(2^n × n^2) where we iterate through all 2^n states, and for each state and city, we check all neighboring cities. The space complexity is O(2^n × n) for the DP table.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example Input:

• n = 4 cities (numbered 0, 1, 2, 3)
• k = 2 (we need to cross exactly 2 highways)
• highways = [[0,1,4], [1,2,3], [2,3,5], [0,3,2]]

This creates the following graph:

    0 ---4--- 1
    |         |
    2         3
    |         |
    3 ---5--- 2

Step 1: Initial Check

• Is k >= n? No, 2 < 4, so we can proceed.

Step 2: Build Graph

g[0] = [(1,4), (3,2)]
g[1] = [(0,4), (2,3)]
g[2] = [(1,3), (3,5)]
g[3] = [(2,5), (0,2)]

Step 3: Initialize DP Table We initialize f[mask][city] where mask represents visited cities:

• f[0001][0] = 0 (start at city 0, mask = 1)
• f[0010][1] = 0 (start at city 1, mask = 2)
• f[0100][2] = 0 (start at city 2, mask = 4)
• f[1000][3] = 0 (start at city 3, mask = 8)

Step 4: State Transitions Let's trace one successful path: Start at city 2 → city 3 → city 0

First transition (city 2 to city 3):

• Current state: mask = 0100 (only city 2 visited)
• We want to visit city 3 next
• New mask = 1100 (cities 2 and 3 visited)
• From city 2, we can reach city 3 with toll 5
• Update: f[1100][3] = f[0100][2] + 5 = 0 + 5 = 5

Second transition (city 3 to city 0):

• Current state: mask = 1100 (cities 2 and 3 visited)
• We want to visit city 0 next
• New mask = 1101 (cities 0, 2, and 3 visited)
• From city 3, we can reach city 0 with toll 2
• Update: f[1101][0] = f[1100][3] + 2 = 5 + 2 = 7

Step 5: Track Answer When we have mask = 1101:

• bit_count(1101) = 3 cities visited
• Since we need k + 1 = 3 cities for exactly 2 highways, this is valid
• Update answer: ans = max(-1, 7) = 7

Other Valid Paths: The algorithm explores all possibilities:

• Path 0→1→2: cost = 4 + 3 = 7
• Path 1→0→3: cost = 4 + 2 = 6
• Path 3→2→1: cost = 5 + 3 = 8 ✓ (maximum)

Step 6: Return Result The maximum toll cost found is 8 (from path 3→2→1).

The algorithm systematically explores all possible combinations of visited cities and tracks the maximum cost for each valid configuration where exactly k highways are crossed.

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n × n^2)

The algorithm uses dynamic programming with bitmask to track visited cities. The main computation involves:

• Iterating through all possible subsets of cities: 2^n iterations (outer loop over i)
• For each subset, iterating through all cities: n iterations (loop over j)
• For each city in the subset, iterating through its neighbors: up to n-1 neighbors in the worst case (loop over g[j])

Therefore, the total time complexity is O(2^n × n × n) = O(2^n × n^2).

Space Complexity: O(2^n × n)

The space is dominated by:

• The DP table f: a 2D array with dimensions 2^n × n, storing the maximum cost for each combination of visited cities (represented by bitmask) and current city
• The adjacency list g: stores all edges, which takes O(E) space where E is the number of highways, but this is bounded by O(n^2) in the worst case

Since the DP table dominates the space usage, the overall space complexity is O(2^n × n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect State Transition Logic

A critical pitfall in this solution is the state transition logic when updating the DP table. The current implementation has a subtle bug in how it processes state transitions.

The Problem: When iterating through states with for mask in range(1 << n), we're processing masks in increasing order. However, when we update dp[mask][current_city] based on dp[prev_mask][neighbor], we're trying to use a state (prev_mask) that hasn't been fully computed yet if prev_mask < mask.

Example of the Issue:

• Suppose mask = 0b111 (cities 0, 1, 2 visited) and current_city = 2
• We want to update based on prev_mask = 0b011 (cities 0, 1 visited)
• But when we're at mask = 0b111, we haven't finished computing all possible ways to reach states with mask = 0b011

The Solution: Process states by the number of cities visited (in increasing order) to ensure all smaller states are fully computed before using them:

def maximumCost(self, n: int, highways: List[List[int]], k: int) -> int:
    if k >= n:
        return -1
  
    graph = defaultdict(list)
    for city1, city2, cost in highways:
        graph[city1].append((city2, cost))
        graph[city2].append((city1, cost))
  
    dp = [[float('-inf')] * n for _ in range(1 << n)]
  
    # Base case
    for city in range(n):
        dp[1 << city][city] = 0
  
    max_cost = -1
  
    # Process states by number of cities visited
    for num_cities in range(1, k + 2):  # Need k+1 cities for k highways
        for mask in range(1 << n):
            if bin(mask).count('1') != num_cities:
                continue
              
            for current_city in range(n):
                if not ((mask >> current_city) & 1):
                    continue
              
                # Try to arrive at current_city from any neighbor
                prev_mask = mask ^ (1 << current_city)
                for neighbor, edge_cost in graph[current_city]:
                    if (prev_mask >> neighbor) & 1:
                        dp[mask][current_city] = max(
                            dp[mask][current_city],
                            dp[prev_mask][neighbor] + edge_cost
                        )
              
                # Check if we have exactly k+1 cities
                if num_cities == k + 1:
                    max_cost = max(max_cost, dp[mask][current_city])
  
    return max_cost

2. Edge Case: No Valid Path Exists

Another pitfall is not properly handling the case where no valid path with exactly k highways exists, even when k < n.

Example Scenario:

• n = 4 cities, k = 2 (need exactly 2 highways)
• highways = [[0, 1, 10], [2, 3, 20]]
• The graph has two disconnected components, so you can't traverse 2 highways in a single path

The current solution handles this correctly by initializing max_cost = -1 and only updating it when valid paths are found. However, developers might accidentally initialize it to 0, which would incorrectly return 0 instead of -1 for impossible cases.

Best Practice: Always initialize the result variable to -1 (or the specified "impossible" value) and only update it when valid solutions are found.