Google Technical Interview: Social Network Connectivity Time

Social network graph problems are frequently used in technical interviews to test a candidate's algorithm design and graph theory knowledge. Today, we use a real Google interview problem to demonstrate how to efficiently solve graph connectivity problems and showcase the value of oavoservice's interview assistance.

📋 Problem Description

Given a series of friendship logs, each containing a timestamp and a relationship change between two users. You need to find the earliest timestamp at which all users in the network become connected (i.e., everyone can reach everyone else via friends).

Input:

• users: List of users
• logs: Friendship logs, each entry [timestamp, user1, user2]

Output:

• Return the earliest timestamp when the network is fully connected.
• If it never becomes fully connected, return -1.

Constraints:

• Friendships are undirected.
• Logs are sorted by timestamp.
• Duplicate friendships can be ignored.

🎯 Core Concepts Analysis

This problem mainly tests:

1. Union-Find Data Structure: The best solution for dynamic connectivity problems.
2. Graph Theory Thinking: Understanding connected components.
3. Time Complexity Optimization: Path compression and union by rank.
4. Edge Case Handling: Handling cases that never connect.

💡 Solution Strategy (oavoservice Guidance)

Method: Union-Find

Why Union-Find?

oavoservice suggestion: Union-Find is perfect for this because it can:

• Efficiently determine if two users are in the same component.
• Support fast union operations.
• Achieve near O(1) time complexity with path compression and union by rank.

Implementation Steps

class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [0] * n
        self.components = n
    
    def find(self, x):
        # Path compression
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    
    def union(self, x, y):
        root_x, root_y = self.find(x), self.find(y)
        if root_x == root_y:
            return False
        
        # Union by rank
        if self.rank[root_x] < self.rank[root_y]:
            self.parent[root_x] = root_y
        elif self.rank[root_x] > self.rank[root_y]:
            self.parent[root_y] = root_x
        else:
            self.parent[root_y] = root_x
            self.rank[root_x] += 1
        
        self.components -= 1
        return True
    
    def is_connected(self):
        return self.components == 1

def earliest_connection_time(users, logs):
    n = len(users)
    user_index = {user: i for i, user in enumerate(users)}
    uf = UnionFind(n)
    
    for timestamp, user1, user2 in logs:
        idx1, idx2 = user_index[user1], user_index[user2]
        uf.union(idx1, idx2)
        
        # Check if fully connected
        if uf.is_connected():
            return timestamp
    
    return -1

📊 Complexity Analysis

Time Complexity: O(n × α(n))

• n is the number of log entries
• α is the Inverse Ackermann function, effectively constant
• Overall near O(n)

Space Complexity: O(u)

• u is the total number of users
• Stores parent and rank for each user

🚀 Optimization Strategies (oavoservice Advanced Guidance)

1. Batch Processing Optimization

For massive logs (millions), you can:

• Process logs in batches to reduce check frequency.
• Check connectivity only when necessary.

2. Early Termination

# Return immediately if connected
if uf.is_connected():
    return timestamp

3. Preprocessing

• Filter duplicate relationships beforehand.
• Index users for faster lookup.

🎤 Key Interview Questions

Q1: What if the logs are huge?

oavoservice Suggestion:

• Use batch processing.
• Consider distributed processing.
• Use sampling to estimate connection time.

Q2: How to handle dynamic friend removal?

oavoservice Suggestion:

• Union-Find doesn't support efficient removal.
• Consider dynamic graph data structures.
• Or rebuild Union-Find.

Q3: Can BFS/DFS solve this?

oavoservice Suggestion:

• Yes, but less efficient.
• Requires traversing the graph after each addition.
• Time complexity reaches O(n × (V + E)).

💼 How oavoservice Helps You Succeed

In real interviews, oavoservice provides:

Real-time Logic Guidance - Help you choose the optimal algorithm quickly. Code Implementation Assistance - Ensure clear and correct logic. Follow-up Strategy - Prepare for deep dive questions. Complexity Analysis - Accurate time and space analysis. Optimization Discussion - Demonstrate engineering skills.

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Tags: #Google #GraphTheory #UnionFind #SocialNetwork #InterviewPrep #VOHelp #AlgorithmInterview #1point3acres

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