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

The optimal solution has a time complexity of O(n^3 + m^2).

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

The space complexity is O(n^2 + m^2).

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

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

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

Insight 1: We can precompute all pairwise distances within each tree using Breadth-First Search (BFS). This avoids redundant distance calculations for each connection attempt. Insight 2: For each node in the first tree, we need to efficiently find the best node to connect to in the second tree. We can precompute the maximum number of 'target' nodes reachable from any node in the second tree within a given distance (up to k).

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Maximize the Number of Target Nodes After Connecting Trees I - LeetCode 3372 Solution

Q: Which companies ask LeetCode 3372. Maximize the Number of Target Nodes After Connecting Trees I in interviews?

This problem is frequently asked at jio.

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

Maximize the Number of Target Nodes After Connecting Trees I is LeetCode problem 3372, a Medium 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, with distinct labels in ranges [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. You are also given an integer k . Node u is target to node v if

Detailed Explanation

The problem describes two undirected trees with distinct node labels. The goal is to find, for each node in the first tree, the maximum number of nodes that are 'target' nodes to it after connecting one node from the first tree to one node from the second tree. A node `u` is a 'target' node to `v` if the distance (number of edges) between them is less than or equal to `k`. The critical part is that we need to independently evaluate each node in the first tree by trying all possible connections between the two trees and selecting the connection that yields the maximum number of target nodes.

Solution Approach

The solution involves precomputing all pairwise distances in both trees using BFS. Then, for each node `i` in the first tree, we iterate through each other node `u` in the first tree. The distance `d_iu` is precalculated. If we consider that we connect node `i` of the first tree to *any* node in the second tree, we now need to determine which node in tree 2 provides the maximum 'target' benefit. A crucial optimization is precomputing the maximum possible reachable nodes with a certain distance in tree 2. `count1[i]` holds the number of targets from node i in tree1. `max_added` tracks the maximum targets we can gain from the other tree (tree2). The final answer then becomes adding existing targets `count1[i]` to the maximum potential gain `max_added`.

Step-by-Step Algorithm

Step 1: Build adjacency lists for both trees (edges1 and edges2).
Step 2: Calculate all-pairs shortest path distances in tree 1 using BFS. Store the distances in the `dist1` matrix. `dist1[i][j]` is the distance between node i and node j in the first tree.
Step 3: Calculate `count1`, where `count1[i]` represents the number of target nodes reachable from node `i` in the first tree without any connection to the second tree. This is based on the precomputed distances `dist1` and the threshold `k`.
Step 4: Calculate the maximum number of reachable nodes within a given distance in the second tree. `count2_within_dist[v][d]` represents the number of nodes within distance `d` from node `v` in tree 2. The helper array `max_count2[d]` represents the maximum number of nodes reachable within distance `d` from *any* node in tree 2.
Step 5: For each node `i` in the first tree, iterate through all nodes `u` in the first tree. Calculate the distance `d_iu` between `i` and `u`. If `d_iu` is small enough that the connecting node `i` to node `u` benefits from the connection we established, calculate the remaining allowed distance `k_rem = k - 1 - d_iu`. Connect the trees in a single edge which accounts for a distance of 1. Determine the number of additional target nodes reachable in tree 2 based on `k_rem` using precomputed `max_count2` table and update maximum reachable.
Step 6: Add the number of original targets from the first tree (count1[i]) to the maximum added targets from the other tree (max_added) and return.

Key Insights

Insight 1: We can precompute all pairwise distances within each tree using Breadth-First Search (BFS). This avoids redundant distance calculations for each connection attempt.
Insight 2: For each node in the first tree, we need to efficiently find the best node to connect to in the second tree. We can precompute the maximum number of 'target' nodes reachable from any node in the second tree within a given distance (up to k).
Insight 3: The core idea is to precompute distance matrices for both trees using BFS. Then, for each node in the first tree, we iterate through all possible connection points in the second tree. When connecting at specific nodes, we calculate how many additional target nodes are reached on the other tree and select the connection that maximizes target count.

Complexity Analysis

Time Complexity: O(n^3 + m^2)

Space Complexity: O(n^2 + m^2)

Topics

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

Companies

Asked at: jio.

Maximize the Number of Target Nodes After Connecting Trees I | LeetCode 3372

Medium

Tree Depth-First Search Breadth-First Search

Time: O(n^3 + m^2)Space: O(n^2 + m^2)

Solution Code

Available in:

import java.util.*;

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

        List<List<Integer>> adj1 = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj1.add(new ArrayList<>());
        }
        for (int[] edge : edges1) {
            int u = edge[0];
            int v = edge[1];
            adj1.get(u).add(v);
            adj1.get(v).add(u);
        }

        List<List<Integer>> adj2 = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            adj2.add(new ArrayList<>());
        }
        for (int[] edge : edges2) {
            int u = edge[0];
            int v = edge[1];
            adj2.get(u).add(v);
            adj2.get(v).add(u);
        }

        class BFSResult {
            int[] dist;

            BFSResult(int[] dist) {
                this.dist = dist;
            }
        }

        class Helper {
            BFSResult bfs(int start_node, int num_nodes, List<List<Integer>> adj) {
                int[] dist = new int[num_nodes];
                Arrays.fill(dist, -1);
                dist[start_node] = 0;
                Queue<Integer> q = new LinkedList<>();
                q.offer(start_node);
                while (!q.isEmpty()) {
                    int u_node = q.poll();
                    for (int v_node : adj.get(u_node)) {
                        if (dist[v_node] == -1) {
                            dist[v_node] = dist[u_node] + 1;
                            q.offer(v_node);
                        }
                    }
                }
                return new BFSResult(dist);
            }
        }

        Helper helper = new Helper();

        List<BFSResult> dist1 = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            dist1.add(helper.bfs(i, n, adj1));
        }

        int[] count1 = new int[n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (dist1.get(i).dist[j] <= k) {
                    count1[i]++;
                }
            }
        }

        int[] max_count2 = new int[m];

        int[][] count2_within_dist = new int[m][m];
        for (int v = 0; v < m; v++) {
            BFSResult bfsResult = helper.bfs(v, m, adj2);
            int[] distances_from_v = bfsResult.dist;
            int[] counts_at_dist = new int[m];
            for (int dist_val : distances_from_v) {
                if (dist_val < m && dist_val != -1) {
                    counts_at_dist[dist_val]++;
                }
            }

            int current_sum = 0;
            for (int d = 0; d < m; d++) {
                current_sum += counts_at_dist[d];
                count2_within_dist[v][d] = current_sum;
            }
        }

        for (int d = 0; d < m; d++) {
            int max_val = 0;
            for (int v = 0; v < m; v++) {
                max_val = Math.max(max_val, count2_within_dist[v][d]);
            }
            max_count2[d] = max_val;
        }

        int[] answer = new int[n];
        for (int i = 0; i < n; i++) {
            int max_added = 0;
            for (int u = 0; u < n; u++) {
                int d_iu = dist1.get(i).dist[u];
                int k_rem = k - 1 - d_iu;

                if (k_rem >= 0) {
                    int effective_k_rem = Math.min(k_rem, m - 1);
                    int added = max_count2[effective_k_rem];
                    max_added = Math.max(max_added, added);
                }
            }

            answer[i] = count1[i] + max_added;
        }

        return answer;
    }
}

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Problem Info

Description

Problem 3372: Maximize the Number of Target Nodes After Connecting Trees I - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search

Companies

jio

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3372

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

6 steps

1
Step 1: Build adjacency lists for both trees (edges1 and edges2).
2
Step 2: Calculate all-pairs shortest path distances in tree 1 using BFS. Store the distances in the `dist1` matrix. `dist1[i][j]` is the distance between node i and node j in the first tree.
3
Step 3: Calculate `count1`, where `count1[i]` represents the number of target nodes reachable from node `i` in the first tree without any connection to the second tree. This is based on the precomputed distances `dist1` and the threshold `k`.
4
Step 4: Calculate the maximum number of reachable nodes within a given distance in the second tree. `count2_within_dist[v][d]` represents the number of nodes within distance `d` from node `v` in tree 2. The helper array `max_count2[d]` represents the maximum number of nodes reachable within distance `d` from *any* node in tree 2.
5
Step 5: For each node `i` in the first tree, iterate through all nodes `u` in the first tree. Calculate the distance `d_iu` between `i` and `u`. If `d_iu` is small enough that the connecting node `i` to node `u` benefits from the connection we established, calculate the remaining allowed distance `k_rem = k - 1 - d_iu`. Connect the trees in a single edge which accounts for a distance of 1. Determine the number of additional target nodes reachable in tree 2 based on `k_rem` using precomputed `max_count2` table and update maximum reachable.
6
Step 6: Add the number of original targets from the first tree (count1[i]) to the maximum added targets from the other tree (max_added) and return.

💡

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^3 + m^2)

💾

Space Complexity

O(n^2 + m^2)

📝

Why?

The `dist1` matrix has size n x n, and the `count2_within_dist` matrix has size m x m. These are the dominant factors in space complexity. Other arrays such as adjaceny list are secondary to the square matrices.

📖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

Floyd-Warshall Algorithm

Floyd-Warshall can compute all-pairs shortest paths, but it is generally less efficient than repeated BFS for sparse graphs like trees. It has a time complexity of O(V^3) where V is number of nodes. Applying this separately to both graphs gives O(n^3 + m^3).

Time: O(n^3 + m^3)Space: O(n^2 + m^2)

Trade-offs: Floyd-Warshall could be considered for denser graphs. However, for trees, repeated BFS is typically faster due to its ability to terminate early and the sparser nature of the graph.

✨

Key Insights

1Insight 1: We can precompute all pairwise distances within each tree using Breadth-First Search (BFS). This avoids redundant distance calculations for each connection attempt.
2Insight 2: For each node in the first tree, we need to efficiently find the best node to connect to in the second tree. We can precompute the maximum number of 'target' nodes reachable from any node in the second tree within a given distance (up to k).
3Insight 3: The core idea is to precompute distance matrices for both trees using BFS. Then, for each node in the first tree, we iterate through all possible connection points in the second tree. When connecting at specific nodes, we calculate how many additional target nodes are reached on the other tree and select the connection that maximizes target count.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: `k = 0`. In this case, a node is only target to itself. The algorithm correctly handles this because the BFS correctly calculates the distance from a node to itself as 0 and updates the dist matrix accordingly.
Edge case 2: Disconnected graph. The prompt states input are valid trees, and no disconnected subgraphs should exist.
Edge case 3: Large k value. If `k` is larger than or equal to the diameter of the second tree, connecting to any node in the second tree will allow reaching all nodes. The `effective_k_rem = Math.min(k_rem, m - 1)` line in the Java code handles this.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not precomputing distances. Calculating distances repeatedly within the main loop for each node `i` leads to a TLE (Time Limit Exceeded) error. The correct approach is to precompute the distance matrices.

✗Common mistake 2: Incorrectly handling the 'target' node condition. Forgetting that a node is always target to itself (distance 0) or miscalculating the remaining allowed distance (`k_rem`) after connecting to the other tree.

✅

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 the influence or reach of a user in a network, where 'target' nodes are those within a certain degree of separation (akin to `k`). Connecting two disjoint social networks and assessing changes in reach.

🏦Real-world scenario 2: Network routing. Finding optimal routes in a communication network where the 'distance' is latency or cost, and 'target' nodes are servers or resources that must be accessed within a certain limit.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 11, 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 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I choose between recursive and iterative tree traversal?

Use recursive traversal for cleaner code when stack depth isn't a concern (tree height < ~1000). Use iterative with explicit stack for very deep trees to avoid stack overflow. Iterative is also preferred when you need fine-grained control over the traversal process.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximize the Number of Target Nodes After Connecting Trees I"?

For this medium-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.

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