2247. Maximum Cost of Trip With K Highways

Problem Description

The problem presents a scenario where we have n cities and a series of highways that connect certain pairs of these cities. Each highway has an associated toll cost. We are given a map of these connections in the form of a two-dimensional array highways, where each element is a triplet containing two cities and the cost to travel between them via the highway that connects them. Additionally, we have a specific requirement: to plan a trip that crosses exactly k highways, starting from any city but without visiting any city more than once during this trip. The goal is to calculate the maximum possible cost of such a trip, given these constraints. If no such trip exists (i.e., it's not possible to cross exactly k highways without revisiting cities), the function should return -1.

Intuition

The key to solving this problem is recognizing that it is a variation of the classic traveling salesman problem (TSP), which is a well-known problem in computer science. TSP is concerned with finding the shortest possible route that visits each city exactly once and returns to the origin city. In our case, however, we do not need to return to the starting city, and instead of minimizing cost, we are aiming to maximize it, and we only need to cross exactly k highways.

The solution uses a dynamic programming (DP) approach in which we keep track of the maximum cost for each subset of cities visited and ending in a particular city, considering exactly k highways. The DP state is represented by a two-dimensional array f, where f[i][j] keeps the maximum cost to reach city j after visiting a particular subset of cities represented by the bitmask i.

The logic includes these steps:

• First, initialize all DP states to -infinity except f[1 << i][i], which are set to 0 since it costs nothing to start from any city (without having traveled any highway).
• Go through all possible subsets of cities represented as bitmasks. For each subset and each city j on the current subset, check if there's a highway from any other city h to j. If so, update the DP state f[i][j] with the maximum cost by comparing the existing cost with the cost to reach h plus the toll cost between h and j. This represents taking a highway from h to j.
• Keep track of the maximum cost ans when exactly k+1 cities have been visited, which represents k highways crossed.

With this DP approach, we consider all possible paths and each path's accumulated cost to achieve the maximum toll cost for k highway crossings without revisiting cities, which satisfies the constraints of the problem.

Learn more about Graph, Dynamic Programming and Bitmask patterns.

Solution Approach

The implementation utilizes several programming concepts and data structures such as a graph, dynamic programming, bitmasking, and recursion.

• Dynamic Programming (DP): This approach helps us to solve complex problems by breaking them down into simpler subproblems. In this case, DP is used to keep track of the maximum cost for reaching each city through a specific subset of cities.

• Bitmasking: This is a technique used to represent a subset of cities. Each bit in an integer value represents whether a particular city is included in the subset (1 means included, 0 means not included). This is a compact way to handle all combinations of cities.

• Graph Representation: Each city and its connected highways are represented using a graph data structure. In this implementation, a dictionary of lists (defaultdict(list)) is used where each key corresponds to a city, and the value is a list of tuples representing the connected cities and the associated cost.

Here is how the solution translates these concepts into code:

1. Initialize the graph g to store the connections and tolls between the cities.

2. Prepare the dynamic programming table f such that f[i][j] will hold the maximum cost achieved when reaching city j after traveling through cities represented by the bitmask i. Initially, all values except the starting conditions (f[1 << i][i] = 0) are set to -infinity to indicate that those states have not been reached.

3. Iterate over all possible subsets of cities (i is the loop variable used as a bitmask) and their respective cities (j represents each city in the subset).

4. For each subset-city pair (i, j), iterate through all connecting highways (neighbors h) to see if it is possible to come to city j from city h. If so, take the maximum between the existing cost in f[i][j] and the cost of reaching city h plus the toll cost between h and j.

5. Use the bit_count method on the bitmask i to check if exactly k+1 cities have been included in the subset (which means k highways crossed). At this stage, update the answer ans with the maximum value found in the current subset for exactly k highways crossed.

6. After completing the iterations, return the maximum cost ans. If no valid trip meets the requirements, ans remains -1, which is returned as the result.

In summary, the key algorithmic steps in the solution involve initializing DP states, updating states based on the graph of cities and highways, and tracking the maximum cost for trips that span exactly k highways.

Example Walkthrough

Let's consider a small example to illustrate the solution approach:

Suppose we have 3 cities and 3 highways between them with the following toll costs:

1highway[0] = [0, 1, 4] // highway from city 0 to 1 with a toll cost of 4
2highway[1] = [1, 2, 5] // highway from city 1 to 2 with a toll cost of 5
3highway[2] = [0, 2, 1] // highway from city 0 to 2 with a toll cost of 1

And we need to find the maximum possible toll cost for a trip that crosses exactly k = 2 highways.

Step-by-Step Solution:

1. Initialize a graph g to represent the cities and highways. For our example, g will look like this after initialization:

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

2. Prepare the DP table f with 1 << total_cities rows (total_cities being 3 here) and total_cities columns, initialized to -infinity except for f[1 << i][i], which are set to 0. The table now looks like:

   1f = [
   2   [-inf,   -inf, -inf], // bitmask 00
   3   [0,       -inf, -inf], // bitmask 01
   4   [-inf,    0,    -inf], // bitmask 10
   5   [-inf,   -inf,    0], // bitmask 11
   6]

3. Iterate over subsets i (from 1 to (1 << 3) - 1) which are 001, 010, 011, 100, 101, 110, and 111 as bitmasks, and for each subset, iterate over each city j (from 0 to 2).

4. For each subset-city pair (i, j), iterate through all the highways from cities h to j that are not already included in i. Update f[i][j] accordingly. This is how things evolve:

   • When i = 001 (city 0) and j = 1, we can go from 0 to 1 with a cost of 4. We update f[001 | (1 << 1)][1] = max(f[001 | (1 << 1)][1], f[001][0] + 4) which translates to f[011][1] = max(-inf, 0 + 4) thus f[011][1] = 4.

   • When i = 010 (city 1) and j = 2, we can travel from 1 to 2 with a cost of 5. We update f[010 | (1 << 2)][2] = max(f[010 | (1 << 2)][2], f[010][1] + 5) which translates to f[110][2] = max(-inf, 0 + 5) thus f[110][2] = 5.

   • When i = 100 (city 2) no updates occur since there are no outgoing highways.

   • Continue this process for other subsets.

5. After examining all routes and updating f, we look for the answer in f where bit_count(i) == k + 1. In our case, k = 2, so we are looking for 3 visited cities (k+1). For every bitmask i that ends on a city j, we check if bit_count(i) == 3. If it is, we consider f[i][j] as a candidate for ans.

6. For i = 011, we have f[011][1] = 4 and for i = 110, we have f[110][2] = 5. Since bit_count(011) = 2 and bit_count(110) = 2, neither fit our requirement of 3, so we skip them. Thus, we never update ans from its initial value of -1.

7. In this example, there's no way to travel across exactly 2 highways without visiting a city twice, so the answer remains -1.

Using this approach, the algorithm would evaluate all possible trips that can be made by crossing exactly k highways and thus determine the maximum toll cost for such a trip or return -1 if no such trip exists.

Solution Implementation

Time and Space Complexity

The given Python code is an implementation of the bitmask dynamic programming algorithm for the problem of finding the maximum total cost of servicing k highways among n different cities. Here's an analysis of the time complexity and space complexity of the code:

Time Complexity

The time complexity is determined by the nested loops in the code and the operations within those loops.

• The outer loop runs for every subset of the n cities, hence it goes through 2^n iterations as it considers every possible combination of cities.
• Inside the outer loop, there is another loop that iterates over n cities.
• Inside the second loop, there is an inner loop traversing the adjacency list of each city. In the worst case, this could be n-1 edges for a complete graph.

Putting it together, the worst-case time complexity is O(n * 2^n * n) = O(n^2 * 2^n).

Space Complexity

The space complexity is determined by the space used to store all the information needed during computation.

• The 2D array f has a size of (2^n) * n, used to store the maximum costs of traveling through subsets of cities, resulting in a space complexity of O(n * 2^n).
• The graph g stores the edges between cities. In the worst case, it can store up to n * (n - 1) / 2 edges for a complete graph. However, the space used for g is not dominant compared to f.

Therefore, the overall space complexity is O(n * 2^n) as it is the largest memory allocation in the code.

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