How do you solve Count Valid Paths in a Tree on LeetCode?

Count Valid Paths in a Tree (LeetCode #2867) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Identifying prime numbers efficiently is crucial for this problem.

What is the brute force solution for Count Valid Paths in a Tree?

The brute force approach for Count Valid Paths in a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check all pairs of nodes in the tree and count the valid paths by traversing the tree for each pair. This approach is straightforward but inefficient as it checks every possible path. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Valid Paths in a Tree?

Count Valid Paths in a Tree uses the following concepts: Math, Dynamic Programming, Tree, Depth-First Search, Number Theory. The recommended patterns to study are: Depth-First Search, Sieve of Eratosthenes, Tree Traversal.

What are the interview tips for Count Valid Paths in a Tree?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and the reasoning behind your chosen approach. Practice common tree traversal techniques, as they are frequently used in similar problems.

What are common mistakes when solving Count Valid Paths in a Tree?

Failing to account for the uniqueness of paths (counting (a, b) and (b, a) separately). Not properly managing visited nodes during DFS can lead to incorrect counts.

#2867

Count Valid Paths in a Tree

Hard

MathDynamic ProgrammingTreeDepth-First SearchNumber TheoryDepth-First SearchSieve of EratosthenesTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can use a single DFS traversal to count valid paths by maintaining a count of prime nodes along the way. This approach is efficient as it avoids redundant calculations and leverages the tree structure.

⚙️

Algorithm

3 steps

1Step 1: Use the Sieve of Eratosthenes to identify all prime numbers up to n.
2Step 2: Perform a DFS from each node, keeping track of the count of prime nodes encountered on the path.
3Step 3: For each node, count how many valid paths can be formed with the prime count being exactly one.

solution.py31 lines

1def countValidPaths(n, edges):
2    from collections import defaultdict
3    def sieve(n):
4        is_prime = [True] * (n + 1)
5        is_prime[0], is_prime[1] = False, False
6        for i in range(2, int(n**0.5) + 1):
7            if is_prime[i]:
8                for j in range(i*i, n + 1, i):
9                    is_prime[j] = False
10        return is_prime
11
12    graph = defaultdict(list)
13    for u, v in edges:
14        graph[u].append(v)
15        graph[v].append(u)
16
17    primes = sieve(n)
18    valid_paths = 0
19
20    def dfs(node, parent, prime_count):
21        nonlocal valid_paths
22        if prime_count == 1:
23            valid_paths += 1
24        for neighbor in graph[node]:
25            if neighbor != parent:
26                dfs(neighbor, node, prime_count + (1 if primes[neighbor] else 0))
27
28    for i in range(1, n + 1):
29        dfs(i, -1, 1 if primes[i] else 0)
30
31    return valid_paths

ℹ

Complexity note: The time complexity is O(n) because we traverse each node once in the DFS. The space complexity is O(n) due to the graph representation and the recursion stack.

1Identifying prime numbers efficiently is crucial for this problem.
2Tree traversal techniques like DFS are essential for exploring paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.