What is the time complexity of Maximize the Number of Target Nodes After Connecting Trees II?

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

What is the space complexity of Maximize the Number of Target Nodes After Connecting Trees II?

The space complexity is O(n+m).

What data structures are used to solve Maximize the Number of Target Nodes After Connecting Trees II?

This problem uses Tree, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximize the Number of Target Nodes After Connecting Trees II?

Insight 1: The concept of 'target' node depends on the parity (even/odd) of the path length between two nodes. Nodes at an even distance are 'targets' to each other. Insight 2: Since the problem asks for the *maximum* possible target nodes for each node in the first tree, we only need to find the best connection that maximizes the number of target nodes without worrying about the specific connection node.

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Maximize the Number of Target Nodes After Connecting Trees II - LeetCode 3373 Solution

Q: Which companies ask LeetCode 3373. Maximize the Number of Target Nodes After Connecting Trees II in interviews?

This problem is frequently asked at jio.

Maximize the Number of Target Nodes After Connecting Trees II - Complete Solution Guide

Maximize the Number of Target Nodes After Connecting Trees II is LeetCode problem 3373, 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 exist two undirected trees with n and m nodes, labeled from [0, n - 1] and [0, m - 1] , respectively. You are given two 2D integer arrays edges1 and edges2 of lengths n - 1 and m - 1 , respectively, where edges1[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i in the first tree and edges2[i] = [u i , v i ] indicates that there is an edge between nodes u i and v i in the second tree. Node u is target to node v if the number of edges on the path from u to v is even

Detailed Explanation

The problem presents two undirected trees, Tree 1 with 'n' nodes and Tree 2 with 'm' nodes. The goal is to find, for each node 'i' in Tree 1, the maximum number of nodes that are 'target' to node 'i' after connecting Tree 1 and Tree 2 with a single edge. A node 'u' is 'target' to node 'v' if the path between them has an even number of edges. This implies a node is always a target to itself. The challenge lies in determining the optimal connection between the two trees to maximize target nodes for each starting node 'i' in the first tree. The final output is an array of size 'n' where each element answer[i] stores the maximum number of target nodes achievable for node 'i' in Tree 1.

Solution Approach

The solution involves first determining the parity of the distance of each node from an arbitrarily chosen root node (node 0) in both trees. The counts of nodes with even and odd parity distances are then calculated for each tree. For each node 'i' in the first tree, we determine its parity. If node 'i' has even parity, it's a target for all other even parity nodes in the first tree. If it's odd, it targets all odd parity nodes. To maximize the total number of target nodes for 'i', we connect the first tree to the second tree. In the second tree, either the number of even-parity nodes or odd-parity nodes contributes towards increasing the number of target nodes from the connection, so we take the max. Thus, the total is then adding the targets from the first tree plus the max targets from the second tree.

Step-by-Step Algorithm

Step 1: Create adjacency lists for both trees based on the given edge lists.
Step 2: Implement a Breadth-First Search (BFS) to traverse each tree, starting from node 0. During the traversal, determine the parity (even/odd) of the distance of each node from the starting node and save it.
Step 3: Count the number of nodes with even parity (C_0) and odd parity (C_1) in each tree.
Step 4: For each node 'i' in the first tree, determine its parity from the stored parity array.
Step 5: Calculate the number of target nodes within the first tree. This is either C1_0 (if the parity of 'i' is even) or C1_1 (if the parity of 'i' is odd).
Step 6: Determine the maximum possible number of target nodes from the second tree by taking the maximum of C2_0 and C2_1.
Step 7: Sum the target nodes from the first tree (calculated in step 5) and the maximum target nodes from the second tree (calculated in step 6). This sum represents the maximum possible number of target nodes for node 'i'. Store this value in the answer array.
Step 8: Repeat steps 4-7 for all nodes in the first tree.
Step 9: Return the answer array.

Key Insights

Insight 1: The concept of 'target' node depends on the parity (even/odd) of the path length between two nodes. Nodes at an even distance are 'targets' to each other.
Insight 2: Since the problem asks for the *maximum* possible target nodes for each node in the first tree, we only need to find the best connection that maximizes the number of target nodes without worrying about the specific connection node.
Insight 3: The best connection is to connect to the tree that has the largest number of target nodes from any of its nodes. Target nodes in one tree are determined by the number of nodes with even or odd distances from the starting node. So choose the maximum from tree 2.

Complexity Analysis

Time Complexity: O(n+m)

Space Complexity: O(n+m)

Topics

This problem involves: Tree, Depth-First Search, Breadth-First Search.

Companies

Asked at: jio.

Maximize the Number of Target Nodes After Connecting Trees II | LeetCode 3373

Hard

Tree Depth-First Search Breadth-First Search

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] maxTargetNodes(int[][] edges1, int[][] edges2) {
        int n = edges1.length + 1;
        int m = edges2.length + 1;

        class ParityResult {
            int[] parities;
            int count0;
            int count1;

            public ParityResult(int[] parities, int count0, int count1) {
                this.parities = parities;
                this.count0 = count0;
                this.count1 = count1;
            }
        }

        class GraphData {
            Map<Integer, List<Integer>> adj;
            int[] parities;
            int count0;
            int count1;

            public GraphData(Map<Integer, List<Integer>> adj, int[] parities, int count0, int count1) {
                this.adj = adj;
                this.parities = parities;
                this.count0 = count0;
                this.count1 = count1;
            }
        }


        ParityResult parityResult1 = getParityCounts(n, edges1);
        int[] p1 = parityResult1.parities;
        int C1_0 = parityResult1.count0;
        int C1_1 = parityResult1.count1;

        ParityResult parityResult2 = getParityCounts(m, edges2);
        int C2_0 = parityResult2.count0;
        int C2_1 = parityResult2.count1;


        int max_targets_from_tree2 = Math.max(C2_0, C2_1);

        int[] answer = new int[n];
        for (int i = 0; i < n; i++) {
            int targets_from_tree1;
            if (p1[i] == 0) {
                targets_from_tree1 = C1_0;
            } else {
                targets_from_tree1 = C1_1;
            }

            answer[i] = targets_from_tree1 + max_targets_from_tree2;
        }

        return answer;
    }

    private ParityResult getParityCounts(int num_nodes, int[][] edges) {
        Map<Integer, List<Integer>> adj = new HashMap<>();
        for (int i = 0; i < num_nodes; i++) {
            adj.put(i, 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[] parities = new int[num_nodes];
        Arrays.fill(parities, -1);
        Queue<Pair<Integer, Integer>> q = new LinkedList<>();
        q.offer(new Pair<>(0, 0)); // (node, parity)
        parities[0] = 0;

        int[] counts = new int[2];
        counts[0] = 1;

        while (!q.isEmpty()) {
            Pair<Integer, Integer> pair = q.poll();
            int u = pair.getKey();
            int p = pair.getValue();

            for (int v : adj.get(u)) {
                if (parities[v] == -1) {
                    int new_p = 1 - p;
                    parities[v] = new_p;
                    counts[new_p]++;
                    q.offer(new Pair<>(v, new_p));
                }
            }
        }
        return new ParityResult(parities, counts[0], counts[1]);
    }

    static class Pair<K, V> {
        private final K key;
        private final V value;

        public Pair(K key, V value) {
            this.key = key;
            this.value = value;
        }

        public K getKey() {
            return key;
        }

        public V getValue() {
            return value;
        }
    }
}

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Problem Info

Description

Problem 3373: Maximize the Number of Target Nodes After Connecting Trees II - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search

Companies

jio

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3373

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: Create adjacency lists for both trees based on the given edge lists.
2
Step 2: Implement a Breadth-First Search (BFS) to traverse each tree, starting from node 0. During the traversal, determine the parity (even/odd) of the distance of each node from the starting node and save it.
3
Step 3: Count the number of nodes with even parity (C_0) and odd parity (C_1) in each tree.
4
Step 4: For each node 'i' in the first tree, determine its parity from the stored parity array.
5
Step 5: Calculate the number of target nodes within the first tree. This is either C1_0 (if the parity of 'i' is even) or C1_1 (if the parity of 'i' is odd).
6
Step 6: Determine the maximum possible number of target nodes from the second tree by taking the maximum of C2_0 and C2_1.
7
Step 7: Sum the target nodes from the first tree (calculated in step 5) and the maximum target nodes from the second tree (calculated in step 6). This sum represents the maximum possible number of target nodes for node 'i'. Store this value in the answer array.
8
Step 8: Repeat steps 4-7 for all nodes in the first tree.
9
Step 9: Return the answer array.

💡

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+m)

💾

Space Complexity

O(n+m)

📝

Why?

The time complexity is derived from the BFS traversals on both trees, which visit each node and edge once. The space complexity is mainly due to storing the adjacency lists and parity arrays for both trees.

📖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 BFS, Depth-First Search (DFS) can be used to traverse the trees and determine the parity of each node's distance from the starting node. This also traverses all nodes and edges.

Time: O(n + m)Space: O(n + m) in worst case (skewed tree), but potentially less in average case compared to BFS due to call stack usage. If tree is balanced, it's O(log n) for space complexity.

Trade-offs: DFS might be slightly more space-efficient in some cases due to smaller stack usage, but the worst-case space complexity is the same.

✨

Key Insights

1Insight 1: The concept of 'target' node depends on the parity (even/odd) of the path length between two nodes. Nodes at an even distance are 'targets' to each other.
2Insight 2: Since the problem asks for the *maximum* possible target nodes for each node in the first tree, we only need to find the best connection that maximizes the number of target nodes without worrying about the specific connection node.
3Insight 3: The best connection is to connect to the tree that has the largest number of target nodes from any of its nodes. Target nodes in one tree are determined by the number of nodes with even or odd distances from the starting node. So choose the maximum from tree 2.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Single node tree: The code handles single node tree correctly. When n = 1 or m = 1, the loop through the tree is either not executed or executed once, which results in the correct result of maximum number of target nodes.
Edge case 2: Disconnected graph: The code assumes the input is a tree, i.e. connected graph. If the graph isn't connected, the BFS only calculates the number of even and odd nodes in the connected component containing node 0, and other disconnected component will lead to wrong answer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating parity: Ensure that the parity is calculated based on the distance (number of edges) from the starting node, not just the neighbor node's parity.

✗Common mistake 2: Not considering all possible connections: The optimal connection is not to connect the closest node between 2 trees, but should be considered globally. Since the trees will be connected, it's enough to get the maximum number of target nodes in tree 2, by either the number of nodes with even or odd parity from the root node.

✅

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: Social Network Analysis: Determining influence or connectivity patterns in a social network, where the even/odd distance between individuals may indicate certain types of relationships or communities.

🏦Real-world scenario 2: Network Routing: Optimizing network routes to minimize latency or maximize throughput based on path lengths. The concept of even/odd path lengths could be used to determine preferred routes or nodes for data transmission.

💡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

Most tree problems can be solved recursively. The base case is usually a null node. Think about what information needs to flow up vs down the tree.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Not handling null nodes properly. Always check if the current node exists before accessing its children.

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: October 12, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

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 "Maximize the Number of Target Nodes After Connecting Trees II"?

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

🤖 Need help with this problem?