LeetCode 2581: Count Number of Possible Root Nodes Solution

Master LeetCode problem 2581 (Count Number of Possible Root Nodes), a hard 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.

2581. Count Number of Possible Root Nodes

Hard

Array Hash Table Dynamic Programming Tree Depth-First Search

Problem Explanation

Explanation

To solve this problem, we can iterate through all possible root nodes and check how many correct guesses each root node leads to. For each root node, we perform a depth-first search (DFS) starting from that node and check if the guesses are consistent with the tree structure. We keep track of the number of correct guesses for each root node and return the count of root nodes that lead to at least k correct guesses.

Initialize a graph data structure from the given edges.
Iterate through all possible root nodes (0 to n-1):
- Perform a DFS starting from the current root node.
- Check if the guesses are consistent with the DFS traversal.
- Count the number of correct guesses.
Return the count of root nodes that lead to at least k correct guesses.

Time complexity: O(n^2) where n is the number of nodes
Space complexity: O(n) for storing the graph and visited nodes

Solution Code

class Solution {
    public int countPossibleRootNodes(int[][] edges, int[][] guesses, int k) {
        int n = edges.length + 1;
        List<Integer>[] graph = new List[n];
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            graph[edge[0]].add(edge[1]);
            graph[edge[1]].add(edge[0]);
        }
        
        int count = 0;
        for (int root = 0; root < n; root++) {
            Set<int[]> correctGuesses = new HashSet<>();
            boolean[] visited = new boolean[n];
            dfs(graph, root, root, visited, correctGuesses, guesses);
            if (correctGuesses.size() >= k) {
                count++;
            }
        }
        return count;
    }
    
    private void dfs(List<Integer>[] graph, int root, int node, boolean[] visited, Set<int[]> correctGuesses, int[][] guesses) {
        visited[node] = true;
        for (int neighbor : graph[node]) {
            if (!visited[neighbor]) {
                correctGuesses.add(new int[]{node, neighbor});
                dfs(graph, root, neighbor, visited, correctGuesses, guesses);
            }
        }
        for (int[] guess : guesses) {
            if (guess[0] == root && guess[1] == node) {
                correctGuesses.add(guess);
            }
        }
    }
}

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