What is the time complexity of Count Valid Paths in a Tree?

The optimal solution has a time complexity of O(n + E).

What data structures are used to solve Count Valid Paths in a Tree?

This problem uses Math, Dynamic Programming, Tree, Depth-First Search, Number Theory. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Valid Paths in a Tree?

Insight 1: Identifying prime numbers is crucial. We need an efficient way to determine whether a node's label is a prime number. Insight 2: Paths can be represented as a sequence of nodes. The problem's constraint of 'exactly one prime' simplifies path validation. A path is valid if and only if it contains exactly one prime number node.

Which companies ask LeetCode 2867. Count Valid Paths in a Tree in interviews?

This problem is frequently asked at Sprinklr.

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Count Valid Paths in a Tree - LeetCode 2867 Solution

Count Valid Paths in a Tree - Complete Solution Guide

Count Valid Paths in a Tree is LeetCode problem 2867, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is an undirected tree with n nodes labeled from 1 to n . You are given the integer n and a 2D integer array edges of length n - 1 , where edges[i] = [u i , v i ] indicates that there is an edge between nodes u i and v i in the tree. Return the number of valid paths in the tree . A path (a, b) is valid if there exists exactly one prime number among the node labels in the path from a to b . Note that: The path (a, b) is a sequence of distinct nodes starting with node a and ending with node b

Detailed Explanation

The problem asks us to find the number of 'valid paths' in a given undirected tree. A valid path is defined as a path between two nodes (a, b) in the tree that contains exactly one prime number among the labels of the nodes in that path. We are given the number of nodes (n) and a list of edges representing the tree's structure. The nodes are labeled from 1 to n, and we need to count the number of distinct valid paths.

Solution Approach

The solution uses a combination of prime number identification (Sieve of Eratosthenes) and Disjoint Set Union (DSU) to efficiently count the valid paths. First, it identifies all prime numbers up to 'n'. Then, it uses DSU to merge all non-prime nodes that are connected by edges. Finally, for each prime number, it calculates the number of valid paths that pass through it by considering the sizes of the adjacent non-prime connected components. The sum of paths through all prime nodes gives the final answer.

Step-by-Step Algorithm

Step 1: **Prime Number Identification:** Use the Sieve of Eratosthenes to create a boolean array `is_prime` where `is_prime[i]` indicates whether the number `i` is prime.
Step 2: **Adjacency List Creation:** Create an adjacency list `adj` to represent the tree structure based on the given `edges`. Each `adj[i]` will contain a list of nodes adjacent to node `i`.
Step 3: **Disjoint Set Union (DSU) Initialization:** Initialize DSU data structures: `parent` array (initially `parent[i] = i`) and `sz` array (initially `sz[i] = 1`). The `parent` array represents the parent node in the disjoint set, and the `sz` array represents the size of each set.
Step 4: **DSU Union of Non-Prime Nodes:** Iterate through the `edges`. If both nodes connected by an edge are non-prime, perform a `union` operation using DSU to merge them into the same connected component.
Step 5: **Counting Valid Paths:** Iterate through all nodes from 1 to `n`. If a node `p` is prime, calculate the number of valid paths that pass through it.
Step 6: **Connected Component Size Calculation:** For each prime node `p`, iterate through its neighbors. If a neighbor is not prime, find the root of its connected component using the DSU `find` operation. Maintain a map (`comp_map`) to store the sizes of distinct connected components adjacent to the prime node `p`.
Step 7: **Calculating Paths Through Prime Node:** Let `S` be the sum of sizes of all distinct connected components adjacent to prime node `p`. The number of paths that only connect the prime node to non-prime nodes from its adjacent components is equal to S. Now we need to add the paths that traverse components that are linked through the prime node 'p'. Iterate through component_sizes, calculate paths_for_p += current_sum * s , where `current_sum` stores the sum of all prior component sizes.
Step 8: **Total Path Accumulation:** Add the number of paths found for the current prime node to the `total_paths` counter.
Step 9: **Return Result:** After iterating through all nodes, return the `total_paths` as the final answer.

Key Insights

Insight 1: Identifying prime numbers is crucial. We need an efficient way to determine whether a node's label is a prime number.
Insight 2: Paths can be represented as a sequence of nodes. The problem's constraint of 'exactly one prime' simplifies path validation. A path is valid if and only if it contains exactly one prime number node.
Insight 3: Disjoint Set Union (DSU) can be effectively used to group non-prime nodes into connected components. This helps in efficiently calculating the size of the connected components adjacent to a prime node.
Insight 4: The number of valid paths through each prime node can be calculated independently by considering the sizes of the adjacent non-prime connected components.
Insight 5: Path (a,b) is the same as Path (b,a). This implies undirected graph traversal and avoid counting the same path twice.

Complexity Analysis

Time Complexity: O(n + E)

Space Complexity: O(n)

Topics

This problem involves: Math, Dynamic Programming, Tree, Depth-First Search, Number Theory.

Companies

Asked at: Sprinklr.

Count Valid Paths in a Tree | LeetCode 2867

Hard

Math Dynamic Programming Tree Depth-First Search Number Theory

Time: O(n + E)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public long countPaths(int n, int[][] edges) {
        boolean[] is_prime = new boolean[n + 1];
        Arrays.fill(is_prime, true);
        is_prime[0] = is_prime[1] = false;
        for (int i = 2; i <= Math.sqrt(n); i++) {
            if (is_prime[i]) {
                for (int multiple = i * i; multiple <= n; multiple += i) {
                    is_prime[multiple] = false;
                }
            }
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int[] parent = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            parent[i] = i;
        }
        int[] sz = new int[n + 1];
        Arrays.fill(sz, 1);

        function find = (i) -> {
            if (parent[i] == i) {
                return i;
            }
            parent[i] = find.apply(parent[i]);
            return parent[i];
        };

        BiConsumer<Integer, Integer> union = (i, j) -> {
            int root_i = find.apply(i);
            int root_j = find.apply(j);
            if (root_i != root_j) {
                if (sz[root_i] < sz[root_j]) {
                    int temp = root_i;
                    root_i = root_j;
                    root_j = temp;
                }
                parent[root_j] = root_i;
                sz[root_i] += sz[root_j];
            }
        };

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            if (!is_prime[u] && !is_prime[v]) {
                union.accept(u, v);
            }
        }

        long total_paths = 0;

        for (int p = 1; p <= n; p++) {
            if (is_prime[p]) {
                Map<Integer, Integer> comp_map = new HashMap<>();
                for (int neighbor : adj.get(p)) {
                    if (!is_prime[neighbor]) {
                        int root = find.apply(neighbor);
                        comp_map.put(root, sz[root]);
                    }
                }

                List<Integer> component_sizes = new ArrayList<>(comp_map.values());

                long S = 0;
                for(int s : component_sizes){
                    S += s;
                }
                long paths_for_p = S;

                long current_sum = 0;
                for (int s : component_sizes) {
                    paths_for_p += current_sum * s;
                    current_sum += s;
                }

                total_paths += paths_for_p;
            }
        }

        return total_paths;
    }

    interface function {
        int apply(int i);
    }
}

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Problem Info

Description

Problem 2867: Count Valid Paths in a Tree - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Tree Depth-First Search Number Theory

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2867

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

9 steps

1
Step 1: **Prime Number Identification:** Use the Sieve of Eratosthenes to create a boolean array `is_prime` where `is_prime[i]` indicates whether the number `i` is prime.
2
Step 2: **Adjacency List Creation:** Create an adjacency list `adj` to represent the tree structure based on the given `edges`. Each `adj[i]` will contain a list of nodes adjacent to node `i`.
3
Step 3: **Disjoint Set Union (DSU) Initialization:** Initialize DSU data structures: `parent` array (initially `parent[i] = i`) and `sz` array (initially `sz[i] = 1`). The `parent` array represents the parent node in the disjoint set, and the `sz` array represents the size of each set.
4
Step 4: **DSU Union of Non-Prime Nodes:** Iterate through the `edges`. If both nodes connected by an edge are non-prime, perform a `union` operation using DSU to merge them into the same connected component.
5
Step 5: **Counting Valid Paths:** Iterate through all nodes from 1 to `n`. If a node `p` is prime, calculate the number of valid paths that pass through it.
6
Step 6: **Connected Component Size Calculation:** For each prime node `p`, iterate through its neighbors. If a neighbor is not prime, find the root of its connected component using the DSU `find` operation. Maintain a map (`comp_map`) to store the sizes of distinct connected components adjacent to the prime node `p`.
7
Step 7: **Calculating Paths Through Prime Node:** Let `S` be the sum of sizes of all distinct connected components adjacent to prime node `p`. The number of paths that only connect the prime node to non-prime nodes from its adjacent components is equal to S. Now we need to add the paths that traverse components that are linked through the prime node 'p'. Iterate through component_sizes, calculate paths_for_p += current_sum * s , where `current_sum` stores the sum of all prior component sizes.
8
Step 8: **Total Path Accumulation:** Add the number of paths found for the current prime node to the `total_paths` counter.
9
Step 9: **Return Result:** After iterating through all nodes, return the `total_paths` as the final answer.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n + E)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is dominated by the Sieve of Eratosthenes and the DSU operations. While the DSU 'find' operation can take O(log n) without path compression, path compression amortizes the operations to near O(1). The space complexity is determined by the space required to store the prime number information, the DSU data structures, and the adjacency list.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

Instead of DSU, we could use DFS to traverse the tree and count the number of valid paths. For each node, we can start a DFS and explore all possible paths. In each path, we keep track of the number of prime numbers encountered. A path is valid if it has exactly one prime number.

Time: O(n^2) in the worst case. DFS can explore all possible paths, leading to a quadratic time complexity.Space: O(n) due to the recursion stack and visited set in DFS.

Trade-offs: DFS is simpler to implement but less efficient than the DSU approach for larger trees due to its higher time complexity. The DSU approach is generally preferred for its better performance.

✨

Key Insights

1Insight 1: Identifying prime numbers is crucial. We need an efficient way to determine whether a node's label is a prime number.
2Insight 2: Paths can be represented as a sequence of nodes. The problem's constraint of 'exactly one prime' simplifies path validation. A path is valid if and only if it contains exactly one prime number node.
3Insight 3: Disjoint Set Union (DSU) can be effectively used to group non-prime nodes into connected components. This helps in efficiently calculating the size of the connected components adjacent to a prime node.
4Insight 4: The number of valid paths through each prime node can be calculated independently by considering the sizes of the adjacent non-prime connected components.
5Insight 5: Path (a,b) is the same as Path (b,a). This implies undirected graph traversal and avoid counting the same path twice.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: **n = 1, no edges:** The result should be 0 since there are no paths. The code implicitly handles this case by correctly initializing and iterating through the data structures.
Edge case 2: **n = 2, one edge [1, 2]:** If 1 is prime and 2 is not, or vice-versa, the count is 1. If both are prime or neither are prime, the count is 0. The code correctly identifies the prime numbers and adjacent nodes.
Edge case 3: **All nodes are prime:** The code should correctly compute the number of valid paths, each path should just have one prime number.
Edge case 4: **All nodes are not prime:** The count should be 0, the logic correctly prevents paths formation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: **Incorrect Prime Number Identification:** Using a naive approach for prime number checking, such as iterating through all numbers up to n for each number, would result in a much higher time complexity (O(n^2)).

✗Common mistake 2: **Incorrect DSU Implementation:** Not implementing path compression or union by size can significantly increase the time complexity of the DSU operations.

✗Common mistake 3: **Double Counting Paths:** Failing to account for the fact that path (a, b) is the same as path (b, a) would lead to overcounting the valid paths. This is implicitly handled by the algorithm considering the undirected graph property.

✗Common mistake 4: **Not handling disconnected components correctly:** Failing to correctly identify the connected components around each prime and summing them would result in incorrect path counts.

✗Common mistake 5: **Integer Overflow:** Using int instead of long long for storing the path counts, component sizes, and sums can lead to overflow issues for large trees. In Java/C++, ensure variables storing intermediate and final results are of type `long` or `long long` respectively.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: **Network Routing:** Determining optimal paths in a network where certain nodes have special properties (e.g., high security, specific bandwidth requirements) can be modeled as finding valid paths with constraints on node properties, similar to the prime number constraint in this problem.

🏦Real-world scenario 2: **Bioinformatics:** Analyzing biological networks (e.g., protein-protein interaction networks) where certain proteins (nodes) have specific characteristics (e.g., essential genes, disease markers). Identifying paths with specific combinations of these characteristics can provide insights into biological processes or disease mechanisms.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Valid Paths in a Tree"?

For this hard-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?