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Minimum Cost of a Path With Special Roads

Difficulty: Medium

Problem Description

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 represented by a 2D array specialRoads where specialRoads[i] = [x1_i, y1_i, x2_i, y2_i, cost_i] indicates that the i-th special road goes in one direction from (x1_i, y1_i) to (x2_i, y2_i) with a cost equal to cost_i. You can use each special road any number of times.

Return the minimum cost required to go from (startX, startY) to (targetX, targetY).

Key Insights

  • The cost to move between points is based on the Manhattan distance.
  • Special roads provide a potentially cheaper alternative route between specific points.
  • A graph-like approach can be used, where nodes represent positions in the 2D space and edges represent the cost of moving between them.
  • Dijkstra's algorithm is suitable since we need to find the minimum cost path in a weighted graph.

Space and Time Complexity

Time Complexity: O((N + M) log N) where N is the number of nodes (positions you traverse) and M is the number of special roads (edges). Space Complexity: O(N + M) to store the graph representation and the priority queue.

Solution

To solve the problem, we can model the positions in 2D space as a graph where each node represents a point (x, y). The edges will represent either the direct cost of moving from one point to another or the cost of using special roads.

  1. Initialization: Use a priority queue (min-heap) to store the current position and its cost. Start with the initial position and a cost of zero.
  2. Graph Traversal: Use Dijkstra's algorithm to explore all possible positions. For each position, calculate the cost to move directly to the target position and check if it's cheaper than the current known cost.
  3. Special Roads: For each special road, if the starting point of the road is reachable, add the end point of the road along with the associated cost to the priority queue.
  4. Termination: Stop when the target position is reached with the minimum cost.

Code Solutions

import heapq

def minimumCost(start, target, specialRoads):
    startX, startY = start
    targetX, targetY = target
    min_cost = float('inf')
    
    # Priority queue for Dijkstra's algorithm
    pq = [(0, startX, startY)]  # (cost, x, y)
    visited = set()

    while pq:
        cost, x, y = heapq.heappop(pq)
        
        # If we reach the target, update the minimum cost
        if (x, y) == (targetX, targetY):
            min_cost = min(min_cost, cost)
            continue
        
        if (x, y) in visited:
            continue
        visited.add((x, y))
        
        # Calculate direct cost to target
        direct_cost = abs(targetX - x) + abs(targetY - y)
        heapq.heappush(pq, (cost + direct_cost, targetX, targetY))

        # Explore special roads
        for x1, y1, x2, y2, special_cost in specialRoads:
            if (x, y) == (x1, y1):  # Can use special road
                heapq.heappush(pq, (cost + special_cost, x2, y2))
    
    return min_cost
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