LeetCode 2368. Reachable Nodes With Restrictions Solution in Java, C++, Python & More | Explanation + Code

Description

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.

Solutions

Solution 1: DFS

First, we construct an adjacency list g based on the given edges, where g[i] represents the list of nodes adjacent to node i. Then we define a hash table vis to record the restricted nodes or nodes that have been visited, and initially add the restricted nodes to vis.

Next, we define a depth-first search function dfs(i), which represents the number of nodes that can be reached starting from node i. In the dfs(i) function, we first add node i to vis, then traverse the nodes j adjacent to node i. If j is not in vis, we recursively call dfs(j) and add the return value to the result.

Finally, we return dfs(0).

The time complexity is O(n), and the space complexity is O(n). Where n is the number of nodes.

PythonJavaC++GoTypeScript

class Solution: def reachableNodes( self, n: int, edges: List[List[int]], restricted: List[int] ) -> int: def dfs(i: int) -> int: vis.add(i) return 1 + sum(j not in vis and dfs(j) for j in g[i]) g = defaultdict(list) for a, b in edges: g[a].append(b) g[b].append(a) vis = set(restricted) return dfs(0)(code-box)

class Solution { private List<Integer>[] g; private boolean[] vis; public int reachableNodes(int n, int[][] edges, int[] restricted) { g = new List[n]; vis = new boolean[n]; Arrays.setAll(g, k -> new ArrayList<>()); for (var e : edges) { int a = e[0], b = e[1]; g[a].add(b); g[b].add(a); } for (int i : restricted) { vis[i] = true; } return dfs(0); } private int dfs(int i) { vis[i] = true; int ans = 1; for (int j : g[i]) { if (!vis[j]) { ans += dfs(j); } } return ans; } }(code-box)

class Solution { public: int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) { vector<int> g[n]; vector<int> vis(n); for (auto& e : edges) { int a = e[0], b = e[1]; g[a].emplace_back(b); g[b].emplace_back(a); } for (int i : restricted) { vis[i] = true; } function<int(int)> dfs = [&](int i) { vis[i] = true; int ans = 1; for (int j : g[i]) { if (!vis[j]) { ans += dfs(j); } } return ans; }; return dfs(0); } };(code-box)

func reachableNodes(n int, edges [][]int, restricted []int) int { g := make([][]int, n) vis := make([]bool, n) for _, e := range edges { a, b := e[0], e[1] g[a] = append(g[a], b) g[b] = append(g[b], a) } for _, i := range restricted { vis[i] = true } var dfs func(int) int dfs = func(i int) (ans int) { vis[i] = true ans = 1 for _, j := range g[i] { if !vis[j] { ans += dfs(j) } } return } return dfs(0) }(code-box)

function reachableNodes(n: number, edges: number[][], restricted: number[]): number { const vis: boolean[] = Array(n).fill(false); const g: number[][] = Array.from({ length: n }, () => []); for (const [a, b] of edges) { g[a].push(b); g[b].push(a); } for (const i of restricted) { vis[i] = true; } const dfs = (i: number): number => { vis[i] = true; let ans = 1; for (const j of g[i]) { if (!vis[j]) { ans += dfs(j); } } return ans; }; return dfs(0); }(code-box)

Solution 2: BFS

Similar to Solution 1, we first construct an adjacency list g based on the given edges, then define a hash table vis to record the restricted nodes or nodes that have been visited, and initially add the restricted nodes to vis.

Next, we use breadth-first search to traverse the entire graph and count the number of reachable nodes. We define a queue q, initially add node 0 to q, and add node 0 to vis. Then we continuously take out node i from q, accumulate the answer, and add the unvisited adjacent nodes of node i to q and vis.

After the traversal, we return the answer.

The time complexity is O(n), and the space complexity is O(n). Where n is the number of nodes.

PythonJavaC++GoTypeScript

class Solution: def reachableNodes( self, n: int, edges: List[List[int]], restricted: List[int] ) -> int: g = defaultdict(list) for a, b in edges: g[a].append(b) g[b].append(a) vis = set(restricted + [0]) q = deque([0]) ans = 0 while q: i = q.popleft() ans += 1 for j in g[i]: if j not in vis: q.append(j) vis.add(j) return ans(code-box)

class Solution { public int reachableNodes(int n, int[][] edges, int[] restricted) { List<Integer>[] g = new List[n]; boolean[] vis = new boolean[n]; Arrays.setAll(g, k -> new ArrayList<>()); for (var e : edges) { int a = e[0], b = e[1]; g[a].add(b); g[b].add(a); } for (int v : restricted) { vis[v] = true; } Deque<Integer> q = new ArrayDeque<>(); q.offer(0); int ans = 0; for (vis[0] = true; !q.isEmpty(); ++ans) { int i = q.pollFirst(); for (int j : g[i]) { if (!vis[j]) { q.offer(j); vis[j] = true; } } } return ans; } }(code-box)

class Solution { public: int reachableNodes(int n, vector<vector<int>>& edges, vector<int>& restricted) { vector<int> g[n]; vector<int> vis(n); for (auto& e : edges) { int a = e[0], b = e[1]; g[a].emplace_back(b); g[b].emplace_back(a); } for (int i : restricted) { vis[i] = true; } queue<int> q{{0}}; int ans = 0; for (vis[0] = true; !q.empty(); ++ans) { int i = q.front(); q.pop(); for (int j : g[i]) { if (!vis[j]) { vis[j] = true; q.push(j); } } } return ans; } };(code-box)

func reachableNodes(n int, edges [][]int, restricted []int) (ans int) { g := make([][]int, n) vis := make([]bool, n) for _, e := range edges { a, b := e[0], e[1] g[a] = append(g[a], b) g[b] = append(g[b], a) } for _, i := range restricted { vis[i] = true } q := []int{0} for vis[0] = true; len(q) > 0; ans++ { i := q[0] q = q[1:] for _, j := range g[i] { if !vis[j] { vis[j] = true q = append(q, j) } } } return }(code-box)

function reachableNodes(n: number, edges: number[][], restricted: number[]): number { const vis: boolean[] = Array(n).fill(false); const g: number[][] = Array.from({ length: n }, () => []); for (const [a, b] of edges) { g[a].push(b); g[b].push(a); } for (const i of restricted) { vis[i] = true; } const q: number[] = [0]; let ans = 0; for (vis[0] = true; q.length; ++ans) { const i = q.pop()!; for (const j of g[i]) { if (!vis[j]) { vis[j] = true; q.push(j); } } } return ans; }(code-box)