Introduction

In this blog post, we will explore how to solve the LeetCode problem "2662" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an array start where start = [startX, startY] represents your initial position (startX, startY) in a 2D space. You are also given the array target where target = [targetX, targetY] represents your target position (targetX, targetY). The cost of going from a position (x1, y1) to any other position in the space (x2, y2) is |x2 - x1| + |y2 - y1|. There are also some special roads. You are given a 2D array specialRoads where specialRoads[i] = [x1i, y1i, x2i, y2i, costi] indicates that the ith special road goes in one direction from (x1i, y1i) to (x2i, y2i) with a cost equal to costi. You can use each special road any number of times. Return the minimum cost required to go from (startX, startY) to (targetX, targetY). Example 1: Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]] Output: 5 Explanation: (1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1. (1,2) to (3,3). Use specialRoads[0] with the cost 2. (3,3) to (3,4) with a cost of |3 - 3| + |4 - 3| = 1. (3,4) to (4,5). Use specialRoads[1] with the cost 1. So the total cost is 1 + 2 + 1 + 1 = 5. Example 2: Input: start = [3,2], target = [5,7], specialRoads = [[5,7,3,2,1],[3,2,3,4,4],[3,3,5,5,5],[3,4,5,6,6]] Output: 7 Explanation: It is optimal not to use any special edges and go directly from the starting to the ending position with a cost |5 - 3| + |7 - 2| = 7. Note that the specialRoads[0] is directed from (5,7) to (3,2). Example 3: Input: start = [1,1], target = [10,4], specialRoads = [[4,2,1,1,3],[1,2,7,4,4],[10,3,6,1,2],[6,1,1,2,3]] Output: 8 Explanation: (1,1) to (1,2) with a cost of |1 - 1| + |2 - 1| = 1. (1,2) to (7,4). Use specialRoads[1] with the cost 4. (7,4) to (10,4) with a cost of |10 - 7| + |4 - 4| = 3. Constraints: start.length == target.length == 2 1 <= startX <= targetX <= 105 1 <= startY <= targetY <= 105 1 <= specialRoads.length <= 200 specialRoads[i].length == 5 startX <= x1i, x2i <= targetX startY <= y1i, y2i <= targetY 1 <= costi <= 105

Explanation

Here's the breakdown of the solution:

• Dijkstra's Algorithm: Treat the problem as a graph where nodes are locations (x, y) and edges are either direct paths (Manhattan distance) or special roads. Use Dijkstra's algorithm to find the shortest path from the start to the target.
• Graph Representation: Implicitly represent the graph. We don't need to create an explicit adjacency list. Instead, when calculating distances from a node, iterate through special roads and consider the direct path to the target.

• Optimization: Use a min-heap (priority queue) to efficiently select the node with the smallest distance during the Dijkstra's algorithm.

• Runtime Complexity: O(E log V), where E is the number of edges (dominated by the number of special roads) and V is the number of vertices (which in the worst case could involve all the special road endpoints plus start and target). In practical terms, since we're not explicitly building the full graph, it's closer to O(S log S), where S is the number of special roads. Storage Complexity: O(S), to store the distances in the dist dictionary.

Code

    import heapq

def minimumCost(start, target, specialRoads):
    """
    Calculates the minimum cost to travel from start to target using special roads.

    Args:
        start: The starting coordinates [startX, startY].
        target: The target coordinates [targetX, targetY].
        specialRoads: A list of special roads, where each road is [x1, y1, x2, y2, cost].

    Returns:
        The minimum cost to travel from start to target.
    """

    start_x, start_y = start
    target_x, target_y = target

    dist = {}  # Store minimum distances to each point
    pq = [(0, start_x, start_y)]  # Priority queue (cost, x, y)
    dist[(start_x, start_y)] = 0

    while pq:
        cost, x, y = heapq.heappop(pq)

        if (x, y) == (target_x, target_y):
            return cost

        if cost > dist.get((x, y), float('inf')):
            continue  # Skip if a shorter path to (x, y) is already found

        # Option 1: Direct path to target
        direct_cost = abs(target_x - x) + abs(target_y - y)
        if dist.get((target_x, target_y), float('inf')) > cost + direct_cost:
            dist[(target_x, target_y)] = cost + direct_cost
            heapq.heappush(pq, (cost + direct_cost, target_x, target_y))

        # Option 2: Use special roads
        for x1, y1, x2, y2, special_cost in specialRoads:
            if x == x1 and y == y1:
                if dist.get((x2, y2), float('inf')) > cost + special_cost:
                    dist[(x2, y2)] = cost + special_cost
                    heapq.heappush(pq, (cost + special_cost, x2, y2))
            else:
                # Consider direct path to the START of the special road
                direct_cost_to_start = abs(x1 - x) + abs(y1 - y)

                if dist.get((x1, y1), float('inf')) > cost + direct_cost_to_start:
                    dist[(x1, y1)] = cost + direct_cost_to_start
                    heapq.heappush(pq, (cost + direct_cost_to_start, x1, y1))


    return dist.get((target_x, target_y), float('inf')) #should theoretically never get here, but included for safety.