What is the time complexity of Find Minimum Diameter After Merging Two Trees?

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

What data structures are used to solve Find Minimum Diameter After Merging Two Trees?

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

What are the key insights for solving Find Minimum Diameter After Merging Two Trees?

Insight 1: The diameter of the merged tree depends on the diameters of the individual trees and the path created by connecting them. We need to consider how these three values interact. Insight 2: The radius of a tree (minimum eccentricity among all nodes) is relevant because connecting at the center of each tree minimizes the added path length.

Which companies ask LeetCode 3203. Find Minimum Diameter After Merging Two Trees in interviews?

This problem is frequently asked at ServiceNow.

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Find Minimum Diameter After Merging Two Trees - LeetCode 3203 Solution

Find Minimum Diameter After Merging Two Trees - Complete Solution Guide

Find Minimum Diameter After Merging Two Trees is LeetCode problem 3203, 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, numbered from 0 to n - 1 and from 0 to 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 must connect one node from the first tree with another node from the se

Detailed Explanation

The problem asks us to find the minimum possible diameter of a tree formed by merging two given undirected trees. We are given two sets of edges representing the two trees. The task is to connect one node from the first tree to one node from the second tree by adding a single edge, and then find the diameter of the resulting merged tree. The diameter of a tree is the longest path between any two nodes in the tree. We want to find the connection that results in the smallest possible diameter for the combined tree.

Solution Approach

The solution calculates the diameter and radius of each tree independently. Then, it considers the diameter of the merged tree as the maximum of the individual tree diameters and the diameter of the path formed by connecting the 'centers' of the two trees. The 'center' is indirectly obtained by using the radius. The radius is defined as the minimum eccentricity, where eccentricity of a node is the maximum distance to any other node in the tree from the given node. Eccentricity is indirectly determined by finding the distance to all the nodes farthest from both ends of the diameter.

Step-by-Step Algorithm

Step 1: Define a function `get_tree_properties(num_nodes, edges)` to calculate the diameter and radius of a given tree.
Step 2: In `get_tree_properties`, construct the adjacency list representation of the tree from the given edges.
Step 3: Use Breadth-First Search (BFS) to find the farthest node from an arbitrary starting node (node 0). Let this node be 'u'.
Step 4: Use BFS again to find the farthest node from 'u'. Let this node be 'v'. The distance between 'u' and 'v' is the diameter of the tree.
Step 5: Use BFS from u and v again to calculate the eccentricity of each node by taking max of the distance from u and v to the node.
Step 6: Find the radius by taking the minimum eccentricity calculated in step 5.
Step 7: Calculate the diameter and radius for both input trees using `get_tree_properties`.
Step 8: Calculate the diameter of the merged tree by connecting the two trees using the path created as radius1 + radius2 + 1.
Step 9: Return the maximum of the diameters of the two original trees and the calculated merged tree diameter.

Key Insights

Insight 1: The diameter of the merged tree depends on the diameters of the individual trees and the path created by connecting them. We need to consider how these three values interact.
Insight 2: The radius of a tree (minimum eccentricity among all nodes) is relevant because connecting at the center of each tree minimizes the added path length.
Insight 3: We don't need to try all possible connections. Connecting at the approximate 'centers' (nodes with minimum eccentricity) of the original trees is sufficient to find the minimum diameter of the combined tree.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

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

Companies

Asked at: ServiceNow.

Find Minimum Diameter After Merging Two Trees | LeetCode 3203

Hard

Tree Depth-First Search Breadth-First Search Graph

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.*;

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

        int[] tree1Properties = getTreeProperties(n, edges1);
        int diam1 = tree1Properties[0];
        int radius1 = tree1Properties[1];

        int[] tree2Properties = getTreeProperties(m, edges2);
        int diam2 = tree2Properties[0];
        int radius2 = tree2Properties[1];

        int pathThroughBridge = radius1 + radius2 + 1;

        return Math.max(Math.max(diam1, diam2), pathThroughBridge);
    }

    private int[] getTreeProperties(int numNodes, int[][] edges) {
        if (numNodes <= 1) {
            return new int[] {0, 0};
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < numNodes; 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[] bfsResult1 = bfs(0, numNodes, adj);
        int u = bfsResult1[1];

        int[] bfsResult2 = bfs(u, numNodes, adj);
        int[] distsFromU = bfsResult2[0];
        int v = bfsResult2[1];
        int diameter = distsFromU[v];

        int[] bfsResult3 = bfs(v, numNodes, adj);
        int[] distsFromV = bfsResult3[0];

        int radius = Integer.MAX_VALUE;
        for (int i = 0; i < numNodes; i++) {
            int eccentricity = Math.max(distsFromU[i], distsFromV[i]);
            radius = Math.min(radius, eccentricity);
        }

        return new int[] {diameter, radius};
    }

    private int[] bfs(int startNode, int numNodes, List<List<Integer>> adj) {
        int[] distances = new int[numNodes];
        Arrays.fill(distances, -1);
        distances[startNode] = 0;
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(startNode);

        while (!queue.isEmpty()) {
            int u = queue.poll();
            for (int vNeighbor : adj.get(u)) {
                if (distances[vNeighbor] == -1) {
                    distances[vNeighbor] = distances[u] + 1;
                    queue.offer(vNeighbor);
                }
            }
        }

        int maxDist = -1;
        int farthestNode = -1;
        for (int i = 0; i < numNodes; i++) {
            if (distances[i] > maxDist) {
                maxDist = distances[i];
                farthestNode = i;
            }
        }

        return new int[] {distances, farthestNode};
    }
}

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Problem Info

Description

Problem 3203: Find Minimum Diameter After Merging Two Trees - Detailed explanation and solution approaches.

Topics

Tree Depth-First Search Breadth-First Search Graph

Companies

ServiceNow

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3203

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: Define a function `get_tree_properties(num_nodes, edges)` to calculate the diameter and radius of a given tree.
2
Step 2: In `get_tree_properties`, construct the adjacency list representation of the tree from the given edges.
3
Step 3: Use Breadth-First Search (BFS) to find the farthest node from an arbitrary starting node (node 0). Let this node be 'u'.
4
Step 4: Use BFS again to find the farthest node from 'u'. Let this node be 'v'. The distance between 'u' and 'v' is the diameter of the tree.
5
Step 5: Use BFS from u and v again to calculate the eccentricity of each node by taking max of the distance from u and v to the node.
6
Step 6: Find the radius by taking the minimum eccentricity calculated in step 5.
7
Step 7: Calculate the diameter and radius for both input trees using `get_tree_properties`.
8
Step 8: Calculate the diameter of the merged tree by connecting the two trees using the path created as radius1 + radius2 + 1.
9
Step 9: Return the maximum of the diameters of the two original trees and the calculated merged tree diameter.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The BFS traversal visits each node and edge at most once. The adjacency list stores each edge twice (once for each endpoint). Therefore, the time and space complexity are directly proportional to the number of nodes and edges, resulting in O(n+m).

📖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

Calculate all-pairs shortest paths using dynamic programming. The maximum shortest path gives the diameter.

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

Trade-offs: This is not efficient for sparse graphs like trees where BFS is much faster. Useful for dense graphs but has higher time and space complexity.

Using Diameter Endpoints to Compute Eccentricity

Instead of BFS from the endpoints of the diameter, we can do a single BFS from each node and save the shortest distances for each node. After this the Eccentricity can be computed and Radius can be calculated.

Time: O(n^2)Space: O(n)

Trade-offs: This may be helpful where the adjacency structure is given as an input, instead of the edges as an input. This is useful when the edges have to be constructed every time.

✨

Key Insights

1Insight 1: The diameter of the merged tree depends on the diameters of the individual trees and the path created by connecting them. We need to consider how these three values interact.
2Insight 2: The radius of a tree (minimum eccentricity among all nodes) is relevant because connecting at the center of each tree minimizes the added path length.
3Insight 3: We don't need to try all possible connections. Connecting at the approximate 'centers' (nodes with minimum eccentricity) of the original trees is sufficient to find the minimum diameter of the combined tree.

📚

Key Terms Explained

For Students

Tree

Hierarchical structure with nodes. Like a family tree!

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: A tree with only one node. The diameter and radius should be 0. The code handles this case by checking if `num_nodes <= 1` and returning 0.
Edge case 2: When the input trees are just single nodes. This situation is correctly handled by the initial check and BFS implementation as the return values are 0 for both diameter and radius.
Edge case 3: Disconnected graph as input. The BFS might not reach all nodes, causing incorrect diameters. Problem description guarantees valid trees as input so this is not handled. If the assumption is invalid, error handling is necessary

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to consider the case when one or both trees have only one node. This requires a special check to avoid errors.

✗Common mistake 2: Incorrectly implementing BFS, leading to infinite loops or incorrect distance calculations. Careful attention to the queue management and visited node tracking is essential.

✗Common mistake 3: Not realizing the impact of connecting the trees at points other than their approximate 'centers'. Connecting at arbitrary points can lead to a larger diameter.

✗Common mistake 4: Not properly handling edge cases in the C implementation like memory allocation and deallocation for the queue and the adjacency list. This can lead to memory leaks or segmentation faults.

✅

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 design: Finding the minimum delay path in a network formed by connecting two subnetworks. Each subnetwork represents a tree, and the goal is to minimize the maximum latency.

🏦Real-world scenario 2: Transportation planning: Optimizing the route between two regions by connecting their road networks. The goal is to minimize the longest travel distance between any two points in the combined network.

🎮Real-world scenario 3: Social Networks: Suppose you have two different social networks. You want to connect them. Using the same approach, this problem will find you the best way to connect these two networks so that the distance between any two people on the network is minimised.

💡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

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

⚠️ 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 22, 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 "Find Minimum Diameter After Merging Two Trees"?

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?