Reorder Routes to Make All Paths Lead to the City Zero

Problem Statement

In the context of this task, the objective is to manage traffic direction in a network of cities connected by roads. There are n cities, numbered from 0 to n - 1, interconnected by n - 1 roads, forming a tree-like structure, meaning there is exactly one unique path between any two different cities. These roads were initially bidirectional but were later oriented in a specific direction due to their narrowness.

The representation of these directed roads is done using pairs within the connections array, where each pair [ai, bi] signifies a directed road from city ai to city bi. This year, a major event is happening at the capital city, labeled as city 0, and it is required that all other cities should be able to directly or indirectly reach this city.

Your challenge is to determine the minimum number of roads that need to have their directions reversed to ensure that every city can travel to the capital city 0. The guarantee provided is that it's possible for each city to reach the capital after the necessary reversals.

Examples

Example 1

Input:

n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]

Output:

3

Explanation:

Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 2

Input:

n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]

Output:

2

Explanation:

Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3

Input:

n = 3, connections = [[1,0],[2,0]]

Output:

0

Constraints

• 2 <= n <= 5 * 104
• connections.length == n - 1
• connections[i].length == 2
• 0 <= ai, bi <= n - 1
• ai != bi

Approach and Intuition

The task involves reorienting roads in a minimalistic way to ensure all cities can reach the capital city (node 0). Here, we are looking to solve a problem of routing on a directed tree with constraints:

General Considerations

• Think of the problem as an issue of ensuring all nodes (cities) are able to send traffic (or direct a path) towards a root node (city 0) via directed edges (roads).
• Given the tree structure where there is exactly one path between any two cities, if a road does not point towards the capital directly or indirectly, then it potentially needs reversing.

Examples

Example 1

Input:

n = 6, connections = [[0,1],[1,3],[2,3],[4,0],[4,5]]

Output:

3

Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 2

Input:

n = 5, connections = [[1,0],[1,2],[3,2],[3,4]]

Output:

2

Explanation: Change the direction of edges show in red such that each node can reach the node 0 (capital).

Example 3

Input:

n = 3, connections = [[1,0],[2,0]]

Output:

0

Constraints

• 2 <= n <= 5 * 104
• connections.length == n - 1
• connections[i].length == 2
• 0 <= ai, bi <= n - 1
• ai != bi

Steps to Approach

1. Graph Representation: Understand the graphical structure where nodes represent cities and directed edges represent roads. Each node needs to eventually point toward the capital city node.

2. Initial Edge Direction Analysis: Firstly, analyze the number of edges already directed towards the capital. These do not need to change.

3. Reverse Edge Calculation:

   • For each city road not pointing towards the capital, examine if reversing its direction helps in creating a direct or indirect path to the capital.
   • Count these edges as they represent the minimum number of changes needed.

4. Utilize Depth-First Search (DFS) or Breadth-First Search (BFS):

   • Apply a DFS or BFS starting from the node 0, to determine the reachability of all other nodes.
   • Through DFS/BFS, check the outgoing and incoming roads to/from each city, marking those which are incorrectly directed and count them.

5. Optimization Techniques: Consider how to reduce computations, perhaps by marking already visited nodes and using memoization to avoid recalculating paths from the same node multiple times.

Examples Exploration

• In example 1, with n = 6 and connections = [[0,1],[1,3],[2,3],[4,0],[4,5]], observe that by simply reversing the paths from 3 -> 1, 3 ->2, 5 -> 4, each node can then reach node 0. This marks a necessity of changing three connections.

• Example 2 highlights a similar redirection need but with fewer cities and connections. The task is to focus on directly reversing efficient edges minimizing traversal disconnections.

Through such systematic checking and employment of graph traversal algorithms (DFS/BFS), the problem of reorienting roads to centrally point towards the capital becomes feasible, adhering to the minimal change to satisfy the requirement for all cities to reach the capital city 0.

Solutions

class Solution {
public:
    int reorderCount = 0;
    void breadthFirstSearch(int start, int total_nodes, vector<vector<pair<int, int>>>& graph) {
        queue<int> queue;
        vector<bool> visited(total_nodes);
        queue.push(start);
        visited[start] = true;
    
        while (!queue.empty()) {
            int current = queue.front();
            queue.pop();
            for (auto& [next, weight] : graph[current]) {
                if (!visited[next]) {
                    reorderCount += weight;
                    visited[next] = true;
                    queue.push(next);
                }
            }
        }
    }
    
    int minReorder(int total_nodes, vector<vector<int>>& connections) {
        vector<vector<pair<int, int>>> graph(total_nodes);
        for (auto& connection : connections) {
            graph[connection[0]].push_back({connection[1], 1});
            graph[connection[1]].push_back({connection[0], 0});
        }
        breadthFirstSearch(0, total_nodes, graph);
        return reorderCount;
    }
};

class Solution {
    int pathChanges = 0;
    
    public void exploreGraph(int vertex, int verticesCount, Map<Integer, List<List<Integer>>> graphMap) {
        Queue<Integer> queue = new LinkedList<>();
        boolean[] visited = new boolean[verticesCount];
        queue.add(vertex);
        visited[vertex] = true;
    
        while (!queue.isEmpty()) {
            int currentVertex = queue.poll();
            if (!graphMap.containsKey(currentVertex)) {
                continue;
            }
            for (List<Integer> edge : graphMap.get(currentVertex)) {
                int adjacentVertex = edge.get(0);
                int direction = edge.get(1);
                if (!visited[adjacentVertex]) {
                    pathChanges += direction;
                    visited[adjacentVertex] = true;
                    queue.add(adjacentVertex);
                }
            }
        }
    }
    
    public int getMinReorders(int vertexCount, int[][] paths) {
        Map<Integer, List<List<Integer>>> adjacencyList = new HashMap<>();
        for (int[] path : paths) {
            adjacencyList.computeIfAbsent(path[0], k -> new ArrayList<>()).add(Arrays.asList(path[1], 1));
            adjacencyList.computeIfAbsent(path[1], k -> new ArrayList<>()).add(Arrays.asList(path[0], 0));
        }
        exploreGraph(0, vertexCount, adjacencyList);
        return pathChanges;
    }
}