LeetCode 2867: Count Valid Paths in a Tree Solution

Master LeetCode problem 2867 (Count Valid Paths in a Tree), 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.

2867. Count Valid Paths in a Tree

Hard

Math Dynamic Programming Tree Depth-First Search Number Theory

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) on the tree starting from each node. During the DFS, we can keep track of the prime numbers encountered on the current path. If there is exactly one prime number on the path from the current node to any ancestor node, we increment the count of valid paths.

We can maintain an array of prime numbers to check if a number is prime efficiently. Additionally, we can use a hashmap to store the count of prime numbers encountered on the path from each node to its ancestors.

The time complexity of this solution is O(n) as we visit each node once during the DFS, and the space complexity is O(n) to store the prime numbers array and hashmap.

Solution Code

class Solution {
    public int countValidPaths(int n, int[][] edges) {
        List<Integer>[] tree = new ArrayList[n + 1];
        for (int i = 1; i <= n; i++) {
            tree[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            tree[edge[0]].add(edge[1]);
            tree[edge[1]].add(edge[0]);
        }

        int[] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}; // prime numbers
        Map<Integer, Integer>[] countMap = new HashMap[n + 1];
        
        int[] result = new int[1];
        dfs(tree, primes, countMap, 1, 0, result);
        
        return result[0];
    }

    private void dfs(List<Integer>[] tree, int[] primes, Map<Integer, Integer>[] countMap, int node, int parent, int[] result) {
        countMap[node] = new HashMap<>();
        for (int prime : primes) {
            countMap[node].put(prime, countMap.getOrDefault(parent, new HashMap<>()).getOrDefault(prime, 0));
        }
        
        for (int child : tree[node]) {
            if (child == parent) continue;
            
            for (int prime : primes) {
                countMap[node].put(prime, countMap[node].get(prime) + countMap[child].getOrDefault(prime, 0));
            }
        }
        
        for (int prime : primes) {
            if (countMap[node].get(prime) == 1) {
                result[0]++;
                break;
            }
        }
        
        for (int child : tree[node]) {
            if (child == parent) continue;
            dfs(tree, primes, countMap, child, node, result);
        }
    }
}

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