Minimum Cost of a Path With Special Roads

The approach involves using Dijkstra’s algorithm to find the minimum cost path from the starting point to the target point. Initially, we set the distances to all the special roads from the starting point, and push them into a priority queue with the current distance as the priority. We then start extracting the minimum element from the priority queue and check if its distance is equal to the distance stored in the vector. If not, we ignore it and continue with the next element.

We calculate the distance to the target point via the current element and update the answer if it is less than the current answer. We then iterate over all the special roads and update their distance if it is less than the previously stored distance. We then push this special road onto the priority queue with the new distance as the priority.

We repeat this process until the priority queue is empty and return the final answer.

Approach

The given problem can be solved using Dijkstra’s algorithm. We can think of each special road as a node in the graph and the weight of each edge as the cost of traveling between the two special roads. To compute the cost of traveling from start to target, we can first calculate the cost of traveling from the start to each special road and then use these costs to update the cost of traveling to other special roads. Finally, we can add the cost of traveling from the target to the special road with minimum total cost.

Specifically, we can first initialize a vector d of distances from the start to each special road. For each special road i, we can calculate d[i] as the sum of the Manhattan distance between the start and the special road and the cost of the special road. We can then use a min-heap to store the distances and the indices of the special roads. We can start with the distances and indices of the special roads as the source and use Dijkstra’s algorithm to update the distances and indices of the special roads. To update the distance of a special road i, we can consider all the special roads that are connected to it and update their distances if the distance to i plus the weight of the edge connecting i and another special road is less than their current distances. Finally, we can add the distance from the target to the special road with minimum total cost to the total cost.

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import heapq

class Solution:
    def minimumCost(self, start: [int], target: [int], specialRoads: [[int]]) -> int:
        INF = 1e9 + 10
        n = len(specialRoads)

        # Initialize the distance of each special road to infinity
        d = [INF] * n

        # Create a priority queue and push the distance from start to each special road
        pq = []
        for i in range(n):
            d[i] = abs(start[0] - specialRoads[i][0]) + abs(start[1] - specialRoads[i][1]) + specialRoads[i][4]
            heapq.heappush(pq, (d[i], i))

        # Initialize the answer with the manhattan distance between start and target
        ans = abs(start[0] - target[0]) + abs(start[1] - target[1])

        # Continue to search for the shortest path until the priority queue is empty
        while pq:
            # Pop the pair with smallest distance
            d_c, c = heapq.heappop(pq)

            # If the distance stored in d is not equal to the current distance d_c, skip this node
            if d_c != d[c]:
                continue

            # Update the answer by finding the distance from the current special road to the target
            ans = min(ans, d_c + abs(target[0] - specialRoads[c][2]) + abs(target[1] - specialRoads[c][3]))

            # For each special road that can be reached from the current special road, update its distance
            for nxt in range(n):
                w = abs(specialRoads[c][2] - specialRoads[nxt][0]) + abs(specialRoads[c][3] - specialRoads[nxt][1]) + specialRoads[nxt][4]
                if d_c + w < d[nxt]:
                    d[nxt] = d_c + w
                    heapq.heappush(pq, (d[nxt], nxt))

        # Return the minimum cost of reaching the target
        return ans

10 Prerequisite LeetCode Problems

For this, the following are a good preparation:

1. “1135. Connecting Cities With Minimum Cost” - This problem also involves finding a path with minimum cost, similar to the main problem.

2. “743. Network Delay Time” - This problem requires understanding of shortest paths in a graph which is useful in the given problem.

3. “787. Cheapest Flights Within K Stops” - Similar to the given problem, this problem requires determining the least cost path with additional conditions.

4. “1029. Two City Scheduling” - This problem involves determining a minimum cost, though the context is different.

5. “505. The Maze II” - This problem deals with finding the shortest path in a 2D grid, similar to the main problem.

6. “1202. Smallest String With Swaps” - This problem is about graph connectivity and minimum cost path, similar to the main problem.

7. “207. Course Schedule” - This problem helps in understanding topological sorting, which can be beneficial to solve the given problem.

8. “210. Course Schedule II” - This problem extends on the previous one, further reinforcing the concepts required for the main problem.

9. “399. Evaluate Division” - This problem involves graph traversal and minimum path which are essential for the main problem.

10. “332. Reconstruct Itinerary” - This problem is about determining a path in a graph. The graph traversal aspect is common with the given problem.

These cover graph traversal, shortest paths, determining minimum cost and topological sorting, which will be helpful for the problem “Minimum Cost of a Path With Special Roads”.

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import heapq

def minimumCost(start, target, specialRoads):
    filteredRoads = []
    for road in specialRoads:
        a, b, c, d, cost = road
        if cost < abs(a - c) + abs(b - d):
            filteredRoads.append([a, b, c, d, cost])
    dist = {(start[0], start[1]): 0}
    heap = [(0, start[0], start[1])]
    while heap:
        currdist, x, y = heapq.heappop(heap)
        for road in filteredRoads:
            a, b, c, d, cost = road
            if dist.get((c, d), float('inf')) > currdist + abs(x - a) + abs(y - b) + cost:
                dist[(c, d)] = currdist + abs(x - a) + abs(y - b) + cost
                heapq.heappush(heap, (dist[(c, d)], c, d))
    res = abs(target[0] - start[0]) + abs(target[1] - start[1])
    for road in filteredRoads:
        a, b, c, d, cost = road
        res = min(res, dist.get((c, d), float('inf')) + abs(target[0] - c) + abs(target[1] - d))
    return res

Problem Classification

This problem is a classic example from the Computer Science domain, particularly the sub-domain of Graph Theory and Path Finding Algorithms. The essence of the problem lies in finding the minimum cost path in a 2D space given certain rules.

The ‘What’ components of the problem are:

1. A 2D space defined by x and y coordinates.
2. An initial position in the 2D space (start).
3. A target position in the 2D space (target).
4. A cost function which is the sum of absolute differences of x and y coordinates (|x2 - x1| + |y2 - y1|).
5. Special roads defined by their start and end coordinates along with their specific cost.
6. An objective to find the minimum cost to reach the target from the start using regular paths and special roads.

This can be further classified as a ‘Shortest Path Finding’ problem under ‘Graph Theory’. This is because we are essentially dealing with a graph where the nodes represent coordinates in the 2D space, and the edges represent either a standard path or a special road, each with their respective costs. The task is to navigate this graph from the start node to the target node in a way that minimizes the total cost.

Language Agnostic Coding Drills

1. Dissect the code and identify each distinct concept it contains:

   a. Data Structures: The code uses basic data structures like arrays and maps/dictionaries.

   b. Control Flow: The code uses loops and conditionals.

   c. Priority Queue operations: The code uses operations such as push and pop on a priority queue, where the smallest distance is always at the top of the queue.

   d. Distance calculation: The code uses the concept of Manhattan distance for calculating distances in a 2D space.

   e. Shortest path algorithm: The implementation of a variant of the shortest path algorithm for finding the minimum distance from start to end.

2. Order and description of each concept:

   a. Data Structures (Easy): Understanding how to work with arrays and maps/dictionaries is fundamental in most languages.

   b. Control Flow (Easy): Basic understanding of how to control the flow of code using loops and conditional statements is a prerequisite for problem-solving in any language.

   c. Distance calculation (Medium): Understanding the concept of Manhattan distance and its application in a 2D space. This requires some basic mathematical knowledge.

   d. Priority Queue operations (Medium): Understanding how to use a priority queue and its operations like push (insertion in the queue) and pop (removal from the queue). This requires an understanding of the queue data structure and its properties.

   e. Shortest path algorithm (Hard): Understanding and implementing a variant of the shortest path algorithm for finding the minimum distance in a graph. This is a complex algorithm that requires knowledge of graph theory and advanced problem-solving skills.

3. Problem-solving approach:

   The code uses a variant of the shortest path algorithm (like Dijkstra’s algorithm) to solve the problem. It starts by filtering out all special roads that have a lower cost than the Manhattan distance. It then calculates the shortest distance from the start to all points reachable via special roads. This is done using a priority queue where the smallest distance is always at the top of the queue.

   The algorithm then considers two possibilities to reach the target: either directly using Manhattan distance or via a special road plus the Manhattan distance from the end of the road to the target. The minimum of these two possibilities is the final result.

   Each concept or ‘drill’ contributes to the solution in the following way:

   • Data structures store the special roads, the distances, and the queue.
   • Control flow allows the algorithm to iterate over all special roads and to stop when all roads have been processed.
   • The priority queue and its operations enable efficient selection of the smallest distance at each step.
   • The Manhattan distance is used to calculate the cost of moving from one point to another in the 2D space.
   • Finally, the shortest path algorithm brings all these concepts together to solve the problem.

Targeted Drills in Python

1. Data Structures:

   a. Arrays:

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   # Declaring an array arr = [1, 2, 3, 4, 5] # Accessing an array element print(arr[2]) # Output: 3

   b. Dictionaries:

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   # Declaring a dictionary dist = {'start': 0, 'end': 10} # Accessing a dictionary element print(dist['start']) # Output: 0

2. Control Flow:

   a. Loops:

   1 2	for i in range(5): print(i) # Output: 0 1 2 3 4

   b. Conditionals:

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   a = 5 if a > 0: print("Positive") # Output: Positive

3. Distance calculation:

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   def manhattan_distance(x1, y1, x2, y2): return abs(x2 - x1) + abs(y2 - y1) print(manhattan_distance(0, 0, 1, 1)) # Output: 2

4. Priority Queue operations:

   a. Insertion:

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   import heapq heap = [] heapq.heappush(heap, (1, 'one')) heapq.heappush(heap, (10, 'ten')) heapq.heappush(heap, (5, 'five')) print(heap) # Output: [(1, 'one'), (10, 'ten'), (5, 'five')]

   b. Removal:

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   smallest = heapq.heappop(heap) print(smallest) # Output: (1, 'one') print(heap) # Output: [(5, 'five'), (10, 'ten')]

5. Shortest path algorithm (Dijkstra’s algorithm):

   This concept is more complex and requires understanding of the above concepts. It involves creating a priority queue, and continually popping the smallest element and updating the distances to its neighboring nodes.

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   # Assume we have a graph represented as an adjacency list graph = {1: {2: 1, 3: 4}, 2: {3: 2, 4: 5}, 3: {4: 1}, 4: {}} start_node = 1 distances = {node: float('infinity') for node in graph} distances[start_node] = 0 queue = [(0, start_node)] while queue: current_distance, current_node = heapq.heappop(queue) for neighbor, weight in graph[current_node].items(): distance = current_distance + weight if distance < distances[neighbor]: distances[neighbor] = distance heapq.heappush(queue, (distance, neighbor)) print(distances) # Output: {1: 0, 2: 1, 3: 3, 4: 4}

These drills focus on the fundamental concepts required for the problem at hand. In addition, understanding the application of Dijkstra’s algorithm is critical for this problem, as it allows us to efficiently calculate the shortest distance from the start to the target.

To solve the initial problem, we would first filter out any special roads that don’t provide a cost benefit over the normal Manhattan distance. We would then use a similar approach to the Dijkstra’s algorithm drill above, but with modifications to accommodate the special roads. For each node (or special road), we calculate the cost to reach it from the start or another special road, always ensuring to take the path with the smallest cost. Finally, we calculate the cost to reach the target, either directly or via a special road, and return the smallest cost.