2662. Minimum Cost of a Path With Special Roads

Problem Description

You are placed on a grid at a starting position defined by coordinates [startX, startY]. Your goal is to move to a target position on the grid, specified by [targetX, targetY]. The cost of moving from any position (x1, y1) to another (x2, y2) is equal to the sum of the absolute differences of their x-coordinates and y-coordinates, i.e., |x2 - x1| + |y2 - y1|.

In addition to the regular cost of moving on the grid, there are special roads available. These are described in a list where each element is another list of the form [x1_i, y1_i, x2_i, y2_i, cost_i], representing a special road that connects (x1_i, y1_i) to (x2_i, y2_i) at a given cost_i. You can use these special roads multiple times if needed.

The challenge is to determine the minimum cost to move from the start position to the target position, considering the regular grid movement cost and any cost savings that can be achieved by using the special roads.

Intuition

To find the least costly path, we need an algorithm that can process the grid and special roads systematically while keeping track of the costs accumulated along the way. The most suitable algorithm for this problem is Dijkstra's algorithm. It’s a well-known algorithm designed to find the shortest path in a weighted graph with non-negative edge weights.

Dijkstra's algorithm fits our needs well because it uses a priority queue to explore nodes (positions) that have the smallest known cost to reach from the start. It updates the cost to reach neighboring nodes as it progresses, and it's greedy in nature, always choosing the least costly option to explore next.

The default movement cost across the grid can act as the edges with a regular weight, while the paths provided by the special roads can be seen as edges with potentially lower weights. The main difference here is that Dijkstra's algorithm typically operates on graphs with fixed nodes and edges, whereas in this problem, the edges are dynamic, as movement is not restricted to predefined connections between points.

By using a modified version of Dijkstra's algorithm, we can push into the priority queue the cost to move from the current position to every position that can be reached by a special road, while also considering the cost to move directly towards the target position from our current position. This approach systematically evaluates all possible moves (both regular and special), keeping track of visited positions to prevent circular routes and redundant calculations.

In summary, the intuition is to use a priority queue to greedily accumulate the least cost of reaching the adjacent positions or the target, considering the cost of normal moves and special roads, until we can determine the minimum cost required to reach the target.

Learn more about Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The solution is implemented using Dijkstra's algorithm, a classical algorithm used in computing the shortest paths in a weighted graph. The specific data structures and patterns used in this implementation are:

• A Priority Queue: This is used to maintain a set of all unvisited nodes, allowing for nodes to be processed in order based on their current known cost from the start. In Python, this is typically implemented with a min-heap using the heapq library.

• A Set for Visited Nodes: To keep track of all visited positions to prevent revisiting and recalculating costs for those positions, a set is used.

In Dijkstra's algorithm, nodes (in this case, grid positions) are visited in order based on their current lowest cost from a start position. For each visited node, we update the costs to its adjacent nodes, which, in the context of the grid problem, are the other nodes that can be reached either directly (one step away in any of the four cardinal directions) or through a special road. The algorithm typically checks if the new path to a node is better than the previously known path, and updates it accordingly.

Here, the dist function is defined to calculate the cost of moving between two positions on the grid. The priority queue q is initialized with a tuple containing the starting cost (0), and the x and y coordinates of the starting position.

1def dist(x1: int, y1: int, x2: int, y2: int) -> int:
2    return abs(x1 - x2) + abs(y1 - y2)
3
4q = [(0, start[0], start[1])]

The solution maintains a loop iterating over the priority queue until it is empty, meaning all reachable positions have been considered.

1while q:
2    d, x, y = heappop(q)

Then it checks if the current position has been visited before:

1if (x, y) in vis:
2    continue

If not, the current position is added to the visited set, and we calculate the cost to reach the target from here:

1vis.add((x, y))
2ans = min(ans, d + dist(x, y, *target))

The algorithm then considers all special roads, pushing the cost of traveling via each special road from the current node into the priority queue. This cost includes the distance from the current node to the start of the special road plus the special road's cost:

1for x1, y1, x2, y2, cost in specialRoads:
2    heappush(q, (d + dist(x, y, x1, y1) + cost, x2, y2))

Once the queue q is empty, the variable ans will contain the minimum cost to reach the target position, which is then returned.

Overall, the algorithm uses Dijkstra's pattern for pathfinding in an efficient manner, yielding the minimum cost considering both normal grid costs and special roads.

Example Walkthrough

Let's assume we start at position [0, 0] and aim to reach the target at position [2, 2]. The grid allows us to move at the cost of the Manhattan distance between points, which is the sum of the absolute differences in the x and y coordinates. On top of the regular movement, we have special roads that connect certain points at a potentially reduced cost. Let's consider one special road for our example that connects [1, 1] to [2, 2] with a cost of 1.

Now let's walk through how the algorithm works using Dijkstra's algorithm:

1. We initialize our priority queue q with the starting position [(0, 0, 0)] which represents a 0 cost to reach [0, 0].

2. We also initialize a visited set vis to keep track of the positions we've already considered to avoid calculating them again.

3. We start by popping the first item from our priority queue: the starting position (0, 0, 0), with 0 being the cost so far.

4. Since [0, 0] isn't in our visited set, we process it. We calculate the Manhattan distance to our target [2, 2], which is 4, and we add the current position to the visited set.

5. Now, we consider the available special roads. In our case, there's one that goes from [1, 1] to [2, 2]. We calculate the cost to get to the starting point of the special road, which is 2 (from [0, 0] to [1, 1]), and then add the cost of the special road (1).

6. We put the endpoint of the special road [2, 2] into our priority queue with the calculated cost to get there (a total of 3).

7. Now the priority queue has one item, [3, 2, 2], which we then pop. Since this is the target and we haven't visited it before, we check the total cost required to reach here and update our answer ans if it's lower than any previous one.

8. As there are no other positions to process (the priority queue is empty, and there is only one special road in our example), the algorithm ends. The minimum cost is 3, which is lower than the straight Manhattan distance to the target, demonstrating the use of the special road to reduce costs.

In a more complex scenario with multiple special roads and a larger grid, the algorithm would work similarly, considering all possible moves and special roads from each position visited, updating the priority queue and visited set accordingly until the minimum cost to reach the target is found.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n^2 \times \log n), where n is the number of special roads. This complexity arises because, in the worst case, each road is visited once and every time a road is visited, it could potentially lead to all other roads being considered as next steps. For each road, the distance to the next road is computed, which is an O(1) operation, but then a point is added to the priority queue that has a size up to O(n^2) (since every special road could be pushed to the queue with different starting points), and hence, the insertion time will be O(\log n^2) which simplifies to O(2\log n) = O(\log n). Since we could have O(n^2) insertions, the total time complexity is O(n^2 \times \log n).

The space complexity of the code is O(n^2) mainly due to the storage requirements of the priority queue and the visited set. In the worst case, each point in the grid could be added to the visited set and at most, every entry of the special roads could be stored in the priority queue simultaneously, if the points are all unique.

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