LeetCode Problem Workspace
Minimum Cost of a Path With Special Roads
Find the minimum cost path between two points, using special roads or direct moves in a 2D space.
4
Topics
5
Code langs
3
Related
Practice Focus
Medium · Array plus Graph
Answer-first summary
Find the minimum cost path between two points, using special roads or direct moves in a 2D space.
Ace coding interviews with Interview AiBoxInterview AiBox guidance for Array plus Graph
In this problem, you need to determine the minimum cost to travel from a starting point to a target in a 2D space. The problem involves moving either directly or by using special roads, each with their own cost. The challenge lies in optimizing the path based on the given special roads and the cost of direct moves.
Problem Statement
You are given a 2D grid where the start and target points are represented by two arrays, start and target. Each array contains two integers, startX, startY for the start position and targetX, targetY for the target position.
The challenge is to find the minimum cost to go from the start position to the target. You can move directly between positions in the grid with a cost calculated by the Manhattan distance formula: |x2 - x1| + |y2 - y1|. Additionally, you are provided with special roads, each defined by a start and end point with a cost to traverse.
Examples
Example 1
Input: start = [1,1], target = [4,5], specialRoads = [[1,2,3,3,2],[3,4,4,5,1]]
Output: 5
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
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
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
Solution Approach
Graph Representation
The grid and special roads can be modeled as a graph where nodes represent positions in the 2D space. Special roads act as directed edges with specific costs. You must consider both the direct distance and the special roads to find the optimal path.
Dijkstra's Algorithm with Priority Queue
Since the problem involves finding the shortest path, you can apply Dijkstra's algorithm with a priority queue to handle the variable costs of moving from one position to another. Each node's cost will be updated based on the direct path or special road with the least cost.
Limitations on Path Selection
It is proven that optimal paths should only involve start, target, or special road endpoints. This drastically reduces the number of possible paths and ensures that we can use Dijkstra’s algorithm efficiently.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity depends on the graph representation and the priority queue operations, typically O((V + E) log V), where V is the number of nodes and E is the number of edges (special roads). The space complexity also depends on the graph structure and is O(V + E).
What Interviewers Usually Probe
- Evaluating how the candidate handles pathfinding in a graph with variable edge costs.
- How the candidate optimizes the graph search based on the start and target points.
- Whether the candidate can apply Dijkstra's algorithm correctly with the given constraints.
Common Pitfalls or Variants
Common pitfalls
- Not reducing the problem space by only considering key points (start, target, special road endpoints).
- Ignoring the optimization of pathfinding using a priority queue, leading to inefficient solutions.
- Failing to handle edge cases such as when no special roads are useful or when a direct path is optimal.
Follow-up variants
- Consider adding more special roads with varied costs to increase the complexity of the problem.
- Modify the problem to allow for multiple target positions and find the shortest path to any of them.
- Allow bidirectional special roads, which would require additional logic to handle the reverse direction.
FAQ
What is the Minimum Cost of a Path With Special Roads problem?
The problem asks you to find the minimum cost path between a start and target point in a 2D grid, considering both direct moves and special roads with defined costs.
How does the special road affect the cost calculation?
Special roads provide a cost-effective way to travel between points in the grid. These roads are factored into the minimum cost path calculation alongside the direct distance.
Why should we only consider start, target, and special road endpoints?
It has been proven that optimal paths only require considering these key points, as any other paths would incur higher costs or be redundant.
What algorithm is most suitable for solving this problem?
Dijkstra’s algorithm is most suitable as it efficiently handles the shortest path in graphs with varying edge weights, especially when combined with a priority queue.
How can GhostInterview assist in solving this problem?
GhostInterview provides step-by-step guidance on applying graph algorithms, optimizing your solution, and avoiding common pitfalls like inefficient path selection.
Solution
Solution 1: Dijkstra
We can find that for each coordinate $(x, y)$ we visit, suppose the minimum cost from the start point to $(x, y)$ is $d$. If we choose to move directly to $(targetX, targetY)$, then the total cost is $d + |x - targetX| + |y - targetY|$. If we choose to go through a special path $(x_1, y_1) \rightarrow (x_2, y_2)$, then we need to spend $|x - x_1| + |y - y_1| + cost$ to move from $(x, y)$ to $(x_2, y_2)$.
class Solution:
def minimumCost(
self, start: List[int], target: List[int], specialRoads: List[List[int]]
) -> int:
def dist(x1: int, y1: int, x2: int, y2: int) -> int:
return abs(x1 - x2) + abs(y1 - y2)
q = [(0, start[0], start[1])]
vis = set()
ans = inf
while q:
d, x, y = heappop(q)
if (x, y) in vis:
continue
vis.add((x, y))
ans = min(ans, d + dist(x, y, *target))
for x1, y1, x2, y2, cost in specialRoads:
heappush(q, (d + dist(x, y, x1, y1) + cost, x2, y2))
return ansContinue Topic
array
Practice more edge cases under the same topic.
arrow_forwardauto_awesomeContinue Pattern
Array plus Graph
Expand the same solving frame across more problems.
arrow_forwardsignal_cellular_altSame Difficulty Track
Medium
Stay on this level to stabilize interview delivery.
arrow_forward