Minimum Cost of a Path With Special Roads

Medium

Samsung

1 view

Topics:

GraphsGreedy Algorithms

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] = [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).

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 <= x1_i, x2_i <= targetX
startY <= y1_i, y2_i <= targetY
1 <= cost_i <= 10⁵

Minimum Cost of a Path With Special Roads

Medium

Samsung

1 view

Topics:

GraphsGreedy Algorithms

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] = [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).

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 <= x1_i, x2_i <= targetX
startY <= y1_i, y2_i <= targetY
1 <= cost_i <= 10⁵

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the input values startX, startY, finishX, finishY, and the coordinates and costs within specialRoads? Specifically, what are the maximum and minimum possible values for each?
Can special roads overlap, and if so, does the cost of using an overlapping section of multiple special roads combine, or can I only use one?
Is it possible for a special road to have a cost of 0, or a negative cost?
If there's no possible path from (startX, startY) to (finishX, finishY) using the given constraints, what value should I return?
Are the start and end points of a special road guaranteed to be distinct? Can x1i equal x2i and y1i equal y2i in any specialRoads[i]?

Brute Force Solution

Approach

We want to find the cheapest way to get from a starting point to a destination, with the option of taking special roads. The brute force method simply considers every single possible route we could take.

Here's how the algorithm would work step-by-step:

Begin at the starting point.
Consider all possible routes we can take from the current location: either take a regular path directly to the destination, or take one of the special roads to a new location.
If we took a special road, repeat the process from that new location: consider all possible routes from there to the destination, including taking another special road or going directly to the destination.
Continue exploring all possible combinations of special roads and direct paths until we eventually reach the destination.
For each complete route we find, calculate the total cost of that route by adding up the cost of each special road and the cost of the direct paths taken.
Compare the total costs of all the routes we found.
Select the route with the lowest total cost as the cheapest way to get to the destination.

Code Implementation

def minimum_cost_brute_force(start, target, special_roads):
    def calculate_distance(point1, point2):
        return abs(point1[0] - point2[0]) + abs(point1[1] - point2[1])

    def find_minimum_cost(current_location):
        if current_location == target:
            return 0

        minimum_cost = calculate_distance(current_location, target)

        # Iterate through all possible special roads.
        for road_start, road_end, road_cost in special_roads:
            if current_location == road_start:

                # Explore path with current special road
                cost_with_road = road_cost + find_minimum_cost(road_end)
                minimum_cost = min(minimum_cost, cost_with_road)

        return minimum_cost

    return find_minimum_cost(start)

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m!) – The brute force approach explores all possible combinations of special roads, where m is the number of special roads. In the worst case, the algorithm could try every possible permutation of special roads before considering direct paths. This is because, from the starting point, it might choose any of the m special roads, then from that new point, any of the remaining (m-1) special roads, and so on. Therefore, the number of paths explored grows factorially with the number of special roads, leading to a time complexity of O(m!).

Space Complexity

O(N) – The brute force approach implicitly uses a recursion stack to explore all possible paths. In the worst-case scenario, the algorithm might explore a path that includes every special road. If there are N special roads, the maximum depth of the recursion stack can be N. Therefore, the space complexity is O(N) due to the recursion stack, where N is the number of special roads.

Optimal Solution

Approach

The core idea is to find the cheapest path from the start point to the end point. We use a clever technique to efficiently explore possible paths, considering both normal movement and special road shortcuts.

Here's how the algorithm would work step-by-step:

Imagine the problem as finding the shortest distance between two cities on a map, where some special roads offer quicker routes.
First, figure out the normal cost of traveling directly between any two points (including the start, the end, and the special road endpoints) using the standard distance formula.
Then, consider the special roads. For each special road, calculate the cost of using that road plus the normal cost of getting to its start and from its end to all other points.
Use a tool (like Dijkstra's algorithm) to keep track of the shortest known distance to each point. Start with the starting point, and gradually update the shortest distances to other points as you explore.
Repeat the process of checking normal distances and special road combinations, always choosing the path that gives you the lowest cost to reach each point.
The final shortest distance to the end point is the minimum cost of the path.

Code Implementation

import heapq

def minimum_cost(start_point, end_point, special_roads):
    points = [start_point, end_point]
    for road_start, road_end, _ in special_roads:
        points.append(road_start)
        points.append(road_end)

    number_of_points = len(points)
    adjacency_list = [[] for _ in range(number_of_points)]

    # Calculate normal distances between all points
    def calculate_distance(point1, point2):
        return abs(point1[0] - point2[0]) + abs(point1[1] - point2[1])

    for i in range(number_of_points):
        for j in range(number_of_points):
            if i != j:
                distance = calculate_distance(points[i], points[j])
                adjacency_list[i].append((j, distance))

    # Add special roads as edges
    for road_start, road_end, road_cost in special_roads:
        start_index = points.index(road_start)
        end_index = points.index(road_end)
        adjacency_list[start_index].append((end_index, road_cost))

    # Dijkstra's algorithm to find shortest paths
    distance_to_point = {i: float('inf') for i in range(number_of_points)}
    start_index = 0  # Start point is always the first point
    distance_to_point[start_index] = 0
    priority_queue = [(0, start_index)]

    while priority_queue:
        current_distance, current_point_index = heapq.heappop(priority_queue)

        if current_distance > distance_to_point[current_point_index]:
            continue

        # Iterate through neighbors and update distances
        for neighbor_index, edge_weight in adjacency_list[current_point_index]:
            new_distance = current_distance + edge_weight

            # We need to update if a shorter path is found
            if new_distance < distance_to_point[neighbor_index]:
                distance_to_point[neighbor_index] = new_distance
                heapq.heappush(priority_queue, (new_distance, neighbor_index))

    end_index = points.index(end_point)

    # Return shortest distance to the end point
    return distance_to_point[end_index]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(N^2 log N) – The algorithm involves calculating distances between all pairs of points (start, end, special road endpoints), resulting in O(N^2) distance calculations initially where N is the number of special road endpoints plus 2 (for start and end). Dijkstra's algorithm, used to find the shortest path, typically has a time complexity of O(E log V), where E is the number of edges and V is the number of vertices. In this case, V is the number of points (N), and E can be at most N^2 (representing all possible paths including special roads), therefore Dijkstra's has a complexity of O(N^2 log N). Since the initial distance calculation is dominated by Dijkstra's time complexity, the overall time complexity is O(N^2 log N).

Space Complexity

O(N) – The algorithm uses Dijkstra's algorithm which typically involves a priority queue (or min-heap) and a distance array to store shortest distances from the starting point to all other points. In this problem, 'points' include the start, end, and endpoints of special roads. If we represent the number of these points as N (which depends on the number of special roads), the distance array will have a size of N, and the priority queue could potentially hold all N points. Therefore, the auxiliary space is proportional to N, leading to a space complexity of O(N).

Edge Cases

Case	How to Handle
startX, startY equals finishX, finishY	Return 0 immediately as no travel is required.
Empty specialRoads array	Calculate and return the Manhattan distance between (startX, startY) and (finishX, finishY) as the only possible cost.
Large specialRoads array and large coordinate values causing potential integer overflow	Use 64-bit integers (long) for calculations involving coordinate differences and cost accumulation to prevent overflow.
specialRoads with cost equal to 0	The Dijkstra algorithm handles this by potentially updating the shortest path with this zero-cost edge.
No path exists using specialRoads to reach finishX, finishY	If the Dijkstra algorithm does not find a path to the finish coordinates, it returns the Manhattan distance as the best possible cost.
specialRoads form a cycle, possibly negative cost cycles.	Dijkstra's algorithm with a priority queue doesn't handle negative cycles, but since the default cost ensures non-negative edge costs, the presence of cycles is not an immediate issue. If costs are negative Dijkstra cannot be used, other algorithms must be used.
startX, startY or finishX, finishY coincides with special road starting or ending coordinates.	The algorithm correctly calculates the cost from the starting coordinates to all the starting points of the special roads, and adds an edge in the graph.
extreme coordinate values	Ensure the memory allocation is sufficient for the vertices and edges, given the size of coordinate values, and use appropriate data structures to represent the graph efficiently.