Reachable Nodes With Restrictions LeetCode Solution

Problem – Reachable Nodes With Restrictions LeetCode Solution

There is an undirected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.

You are given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are also given an integer array restricted which represents restricted nodes.

Return the maximum number of nodes you can reach from node 0 without visiting a restricted node.

Note that node 0 will not be a restricted node.

Example 1:

Input: n = 7, edges = [[0,1],[1,2],[3,1],[4,0],[0,5],[5,6]], restricted = [4,5]
Output: 4
Explanation: The diagram above shows the tree.
We have that [0,1,2,3] are the only nodes that can be reached from node 0 without visiting a restricted node.

Example 2:

Input: n = 7, edges = [[0,1],[0,2],[0,5],[0,4],[3,2],[6,5]], restricted = [4,2,1]
Output: 3
Explanation: The diagram above shows the tree.
We have that [0,5,6] are the only nodes that can be reached from node 0 without visiting a restricted node.

Constraints:

  • 2 <= n <= 105
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= ai, bi < n
  • ai != bi
  • edges represents a valid tree.
  • 1 <= restricted.length < n
  • 1 <= restricted[i] < n
  • All the values of restricted are unique.

Reachable Nodes With Restrictions LeetCode Solution in Java

class Solution {
    public int reachableNodes(int n, int[][] edges, int[] restricted) {
        List<Integer>[] al = new ArrayList[n];
        for(int i=0;i<n;i++) al[i] = new ArrayList<>();
        for(int e[] : edges){
            al[e[0]].add(e[1]); al[e[1]].add(e[0]);
        }
        Set<Integer> set = new HashSet<>();
        for(int i : restricted) set.add(i);
        Queue<Integer> q = new LinkedList<>();
        q.add(0);  set.add(0);
        int ans = 0;
        while(!q.isEmpty()){
            int size = q.size();
            while(size-->0){
                ans++;
                int curr = q.remove();
                for(int next : al[curr]) 
                    if(set.add(next)) q.add(next);
            }
        }
        return ans;
    }
}

Reachable Nodes With Restrictions LeetCode Solution in C++

int reachableNodes(int n, vector<vector<int>>&e , vector<int>& r) {
    unordered_set<int> s(begin(r),end(r));
    vector<vector<int>> graph(n);
    for(auto i:e) graph[i[0]].push_front(i[1]) , graph[i[1]].push_front(i[0]);
    queue<int> q;
    vector<bool> seen(n);
    int cnt=0;
    q.push(0);
    seen[0]=true;
    while(q.size()){
        auto node= q.front(); q.pop();
        if(s.count(node)) continue;
        cnt++;
        for(auto j:graph[node]) //Put Neighbours
            if(!seen[j]) seen[j]=true , q.push(j);
    }
    return cnt;
}

Reachable Nodes With Restrictions LeetCode Solution in Python

class Solution:
    def reachableNodes(self, n: int, edges: List[List[int]], restricted: List[int]) -> int:
        res = set(restricted)
        graph = defaultdict(list)
        vis = set()
        
        for u, v in edges:
            if u in res or v in res: continue
            graph[u].append(v)
            graph[v].append(u)
        
        def dfs(node):
            r = 1
            for nei in graph[node]:
                if nei in vis: continue
                vis.add(nei)
                r += dfs(nei)
            return r
        
        vis.add(0)
        return dfs(0)
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