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LeetCode 2368: Reachable Nodes With Restrictions Solution

Master LeetCode problem 2368 (Reachable Nodes With Restrictions), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can perform a Depth First Search (DFS) traversal on the tree starting from node 0 while keeping track of the restricted nodes. We can use a priority queue to prioritize visiting nodes that are not restricted. We will also maintain a set to keep track of visited nodes.

  1. Initialize a graph data structure to store the edges.
  2. Perform a DFS traversal starting from node 0 and keep track of the number of nodes visited that are not restricted.
  3. Use a priority queue to prioritize visiting nodes that are not restricted.
  4. Maintain a set to keep track of visited nodes and restricted nodes.
  5. Keep track of the maximum number of nodes visited without visiting a restricted node.
  6. Return the maximum number of nodes reached from node 0 without visiting a restricted node.

Time Complexity: O(n log n) where n is the number of nodes. Space Complexity: O(n) for storing the graph and sets. :

Solution Code

import java.util.*;

class Solution {
    public int reachableNodes(int n, int[][] edges, int[] restricted) {
        Map<Integer, List<int[]>> graph = new HashMap<>();
        Set<Integer> visited = new HashSet<>();
        Set<Integer> restrictedSet = new HashSet<>();
        for (int r : restricted) {
            restrictedSet.add(r);
        }
        
        for (int[] edge : edges) {
            int u = edge[0], v = edge[1];
            graph.computeIfAbsent(u, k -> new ArrayList<>()).add(new int[]{v, 1});
            graph.computeIfAbsent(v, k -> new ArrayList<>()).add(new int[]{u, 1});
        }
        
        PriorityQueue<int[]> pq = new PriorityQueue<>((a, b) -> b[1] - a[1]);
        pq.offer(new int[]{0, n});
        
        int res = 0;
        
        while (!pq.isEmpty()) {
            int[] curr = pq.poll();
            int node = curr[0];
            int moves = curr[1];
            
            if (visited.contains(node)) {
                continue;
            }
            
            visited.add(node);
            if (!restrictedSet.contains(node)) {
                res++;
            }
            
            if (graph.containsKey(node)) {
                for (int[] next : graph.get(node)) {
                    int nextNode = next[0];
                    int cost = next[1];
                    if (!visited.contains(nextNode)) {
                        int remainingMoves = moves - cost - 1;
                        if (remainingMoves >= 0) {
                            pq.offer(new int[]{nextNode, remainingMoves});
                        }
                    }
                }
            }
        }
        
        return res;
    }
}

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