What is the time complexity of Minimum Cost of a Path With Special Roads?

The optimal solution has a time complexity of O(N * M * log(N * M)).

What is the space complexity of Minimum Cost of a Path With Special Roads?

The space complexity is O(N * M).

What data structures are used to solve Minimum Cost of a Path With Special Roads?

This problem uses Array, Graph, Heap (Priority Queue), Shortest Path. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Cost of a Path With Special Roads?

Insight 1: The problem can be modeled as a graph where nodes are the start point, target point, and the start and end points of all special roads. Edges are either Manhattan distances between points or the cost of special roads. Insight 2: Dijkstra's algorithm is suitable for finding the shortest path from the starting point to the target point in this graph, as it guarantees finding the minimum cost path in a weighted graph with non-negative edge weights.

Which companies ask LeetCode 2662. Minimum Cost of a Path With Special Roads in interviews?

This problem is frequently asked at Samsung.

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Minimum Cost of a Path With Special Roads - LeetCode 2662 Solution

Minimum Cost of a Path With Special Roads - Complete Solution Guide

Minimum Cost of a Path With Special Roads is LeetCode problem 2662, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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] = [x1 i , y1 i , x2 i , y2 i , cost i ] in

Detailed Explanation

The problem asks us to find the minimum cost to travel from a starting point `(startX, startY)` to a target point `(targetX, targetY)` in a 2D space. The cost of moving between any two points `(x1, y1)` and `(x2, y2)` is the Manhattan distance `|x2 - x1| + |y2 - y1|`. We are also given a set of special roads, where each special road `[x1_i, y1_i, x2_i, y2_i, cost_i]` allows us to travel from `(x1_i, y1_i)` to `(x2_i, y2_i)` at a cost of `cost_i`. We can use each special road any number of times. The goal is to determine the minimum total cost to reach the target from the starting point.

Solution Approach

The provided solution uses Dijkstra's algorithm to find the minimum cost path. It first creates a graph where the nodes are the start point, the target point, and the beginning and end points of each special road. The edges are the Manhattan distances between all pairs of points, and the cost of the special roads are added as edges between the corresponding points. Dijkstra's algorithm then finds the shortest path from the start node to the target node in this graph using a priority queue to efficiently select the next node to explore. During the algorithm, the direct Manhattan distance to the target is also considered from each node to further improve efficiency.

Step-by-Step Algorithm

Step 1: Create a list of all unique points: the starting point, the target point, and the starting and ending points of all special roads.
Step 2: Construct a graph represented by a cost matrix. The cost between two nodes is the Manhattan distance between them.
Step 3: Add the special road costs to the cost matrix. If a special road connects two points, the cost between those points is updated to the minimum of the Manhattan distance and the special road cost.
Step 4: Initialize distances to all nodes as infinity, except for the starting node, which has a distance of 0.
Step 5: Use a priority queue to store nodes to visit, ordered by their distance from the starting node.
Step 6: While the priority queue is not empty, extract the node with the minimum distance. If this node is the target, return the distance.
Step 7: For each neighbor of the current node, calculate the distance to that neighbor through the current node. If this distance is less than the current distance to the neighbor, update the neighbor's distance and add it to the priority queue.
Step 8: If the target is never reached in the main loop, it means some error has occurred (but given the problem constraints, it's theoretically impossible if graph is constructed correctly) . Return the calculated distance to the target.

Key Insights

Insight 1: The problem can be modeled as a graph where nodes are the start point, target point, and the start and end points of all special roads. Edges are either Manhattan distances between points or the cost of special roads.
Insight 2: Dijkstra's algorithm is suitable for finding the shortest path from the starting point to the target point in this graph, as it guarantees finding the minimum cost path in a weighted graph with non-negative edge weights.
Insight 3: An optimization can be made by considering the direct Manhattan distance to the target from any intermediate node during Dijkstra's search, allowing the algorithm to terminate earlier or find a shorter path.

Complexity Analysis

Time Complexity: O(N * M * log(N * M))

Space Complexity: O(N * M)

Topics

This problem involves: Array, Graph, Heap (Priority Queue), Shortest Path.

Companies

Asked at: Samsung.

Minimum Cost of a Path With Special Roads | LeetCode 2662

Medium

Array Graph Heap (Priority Queue)Shortest Path

Time: O(N * M * log(N * M))Space: O(N * M)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int minimumCost(int[] start, int[] target, int[][] specialRoads) {
        Point start_pt = new Point(start[0], start[1]);
        Point target_pt = new Point(target[0], target[1]);

        Set<Point> points = new HashSet<>();
        points.add(start_pt);
        points.add(target_pt);
        for (int[] road : specialRoads) {
            points.add(new Point(road[0], road[1]));
            points.add(new Point(road[2], road[3]));
        }

        List<Point> points_list = new ArrayList<>(points);
        Map<Point, Integer> point_to_idx = new HashMap<>();
        for (int i = 0; i < points_list.size(); i++) {
            point_to_idx.put(points_list.get(i), i);
        }
        int n = points_list.size();

        int[][] cost_matrix = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cost_matrix[i][j] = Integer.MAX_VALUE;
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                Point p1 = points_list.get(i);
                Point p2 = points_list.get(j);
                int cost = Math.abs(p1.x - p2.x) + Math.abs(p1.y - p2.y);
                cost_matrix[i][j] = cost;
            }
        }

        for (int[] road : specialRoads) {
            int x1 = road[0], y1 = road[1], x2 = road[2], y2 = road[3], cost = road[4];
            int u_idx = point_to_idx.get(new Point(x1, y1));
            int v_idx = point_to_idx.get(new Point(x2, y2));
            cost_matrix[u_idx][v_idx] = Math.min(cost_matrix[u_idx][v_idx], cost);
        }

        int start_idx = point_to_idx.get(start_pt);
        int target_idx = point_to_idx.get(target_pt);

        int[] dist = new int[n];
        Arrays.fill(dist, Integer.MAX_VALUE);
        dist[start_idx] = 0;
        PriorityQueue<Pair> pq = new PriorityQueue<>(Comparator.comparingInt(a -> a.d));
        pq.offer(new Pair(0, start_idx));

        while (!pq.isEmpty()) {
            Pair curr = pq.poll();
            int d = curr.d;
            int u = curr.u;

            if (d > dist[u]) {
                continue;
            }

            if (u == target_idx) {
                return d;
            }

            for (int v = 0; v < n; v++) {
                if (dist[u] != Integer.MAX_VALUE && cost_matrix[u][v] != Integer.MAX_VALUE && dist[u] + cost_matrix[u][v] < dist[v]) {
                    dist[v] = dist[u] + cost_matrix[u][v];
                    pq.offer(new Pair(dist[v], v));
                }
            }
        }

        return dist[target_idx];
    }

    static class Point {
        int x, y;

        public Point(int x, int y) {
            this.x = x;
            this.y = y;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) return true;
            if (o == null || getClass() != o.getClass()) return false;
            Point point = (Point) o;
            return x == point.x && y == point.y;
        }

        @Override
        public int hashCode() {
            return Objects.hash(x, y);
        }
    }

    static class Pair {
        int d, u;

        public Pair(int d, int u) {
            this.d = d;
            this.u = u;
        }
    }
}

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Problem Info

Description

Problem 2662: Minimum Cost of a Path With Special Roads - Detailed explanation and solution approaches.

Topics

Array Graph Heap (Priority Queue)Shortest Path

Companies

Samsung

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2662

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Create a list of all unique points: the starting point, the target point, and the starting and ending points of all special roads.
2
Step 2: Construct a graph represented by a cost matrix. The cost between two nodes is the Manhattan distance between them.
3
Step 3: Add the special road costs to the cost matrix. If a special road connects two points, the cost between those points is updated to the minimum of the Manhattan distance and the special road cost.
4
Step 4: Initialize distances to all nodes as infinity, except for the starting node, which has a distance of 0.
5
Step 5: Use a priority queue to store nodes to visit, ordered by their distance from the starting node.
6
Step 6: While the priority queue is not empty, extract the node with the minimum distance. If this node is the target, return the distance.
7
Step 7: For each neighbor of the current node, calculate the distance to that neighbor through the current node. If this distance is less than the current distance to the neighbor, update the neighbor's distance and add it to the priority queue.
8
Step 8: If the target is never reached in the main loop, it means some error has occurred (but given the problem constraints, it's theoretically impossible if graph is constructed correctly) . Return the calculated distance to the target.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N * M * log(N * M))

💾

Space Complexity

O(N * M)

📝

Why?

The time complexity is dominated by Dijkstra's algorithm. Creating the graph (cost matrix) takes O(V^2) time, where V is the number of points. Dijkstra's algorithm with a priority queue takes O(E log V) time, where E is the number of edges. In this case, E can be at most V^2. The space complexity is O(V^2) because the cost matrix is of size V x V. Therefore, it's a dense graph with E can be O(V^2) and Dijkstra's complexity becomes O(V^2)

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A* Search

A* search is an informed search algorithm that uses a heuristic to estimate the cost from a given node to the target. In this case, the Manhattan distance can be used as a heuristic. A* search may perform better than Dijkstra's algorithm if the heuristic is accurate and consistent.

Time: O(E) in best case, O(V^2) in worst case. Where E is the number of edges and V is the number of vertices.Space: O(V), where V is the number of vertices.

Trade-offs: A* search has the potential to be faster than Dijkstra's algorithm if a good heuristic is available. However, if the heuristic is not well-chosen, A* search may perform worse than Dijkstra's algorithm. Dijkstra's algorithm is guaranteed to find the shortest path, whereas the performance of A* is heavily dependent on the chosen heuristic.

Bellman-Ford Algorithm

Bellman-Ford algorithm can also solve the shortest path problem. However, Dijkstra's algorithm is more efficient when edge weights are non-negative.

Time: O(VE) where V is vertices and E is edges. Due to dense nature of our graph it becomes O(V^3)Space: O(V)

Trade-offs: Dijkstra's algorithm is preferred here due to its better time complexity O(V^2) compared to Bellman-Ford O(V^3).

✨

Key Insights

1Insight 1: The problem can be modeled as a graph where nodes are the start point, target point, and the start and end points of all special roads. Edges are either Manhattan distances between points or the cost of special roads.
2Insight 2: Dijkstra's algorithm is suitable for finding the shortest path from the starting point to the target point in this graph, as it guarantees finding the minimum cost path in a weighted graph with non-negative edge weights.
3Insight 3: An optimization can be made by considering the direct Manhattan distance to the target from any intermediate node during Dijkstra's search, allowing the algorithm to terminate earlier or find a shorter path.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: The direct path (Manhattan distance) from start to target might be the optimal solution. The algorithm considers this as the initial best answer and updates if a better path using special roads is found.
Edge case 2: There might be no special roads. The algorithm correctly calculates the Manhattan distance directly from start to target.
Edge case 3: The cost of the special road might be greater than the Manhattan distance between its start and end points. The algorithm considers the minimum of the two when constructing the graph.
Edge case 4: The start or end point of a special road might coincide with the overall start or target location.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the direct Manhattan distance from any intermediate node to the target. This can lead to suboptimal solutions if a shorter path is possible by simply going directly to the target.

✗Common mistake 2: Not handling duplicate points correctly when constructing the graph (e.g., the start point of a special road coincides with the target point). This can lead to incorrect edge weights in the cost matrix.

✗Common mistake 3: Incorrect implementation of Dijkstra's algorithm, such as not using a priority queue or not updating distances correctly.

✗Common mistake 4: Forgetting to include the initial distance which is the Manhattan distance from start to target.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Route planning in navigation systems, where special roads represent highways with different speed limits or tolls.

🏦Real-world scenario 2: Network routing, where special roads represent direct connections between routers with specific costs or bandwidth limitations. Finding the lowest cost path between two computers on a network.

🎮Real-world scenario 3: Logistics and supply chain optimization, where special roads represent different transportation routes with associated costs. Determining the most efficient way to transport goods from a warehouse to a store.

📱Real-world scenario 4: Robotics path planning in a warehouse environment with obstacles and defined routes. The special roads can define routes where the robot can move freely without obstacles.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost of a Path With Special Roads"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether graph techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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