What is the time complexity of Maximum Employees to Be Invited to a Meeting?

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

What data structures are used to solve Maximum Employees to Be Invited to a Meeting?

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

What are the key insights for solving Maximum Employees to Be Invited to a Meeting?

Insight 1: The problem can be modeled as a directed graph where an edge `i -> favorite[i]` means employee `i` likes employee `favorite[i]`. Insight 2: The optimal solution involves finding the largest cycle in the graph (cycles of length greater than 2) and combining it with the longest chains attached to 2-cycles (cycles of length 2).

Which companies ask LeetCode 2127. Maximum Employees to Be Invited to a Meeting in interviews?

This problem is frequently asked at Barclays, SAP.

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Maximum Employees to Be Invited to a Meeting - LeetCode 2127 Solution

Maximum Employees to Be Invited to a Meeting - Complete Solution Guide

Maximum Employees to Be Invited to a Meeting is LeetCode problem 2127, 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

A company is organizing a meeting and has a list of n employees, waiting to be invited. They have arranged for a large circular table, capable of seating any number of employees. The employees are numbered from 0 to n - 1 . Each employee has a favorite person and they will attend the meeting only if they can sit next to their favorite person at the table. The favorite person of an employee is not themself. Given a 0-indexed integer array favorite , where favorite[i] denotes the favorite person o

Detailed Explanation

The problem asks us to find the maximum number of employees that can be invited to a meeting given a list of each employee's favorite person. The employees will be seated at a circular table, and an employee will only attend if they can sit next to their favorite person. The input is an array `favorite` where `favorite[i]` is the favorite person of employee `i`. The output is the maximum number of employees that can be invited.

Solution Approach

The solution identifies two types of possible arrangements: 1. A cycle of length greater than 2. Any cycle means all the employees in that cycle can sit together. We iterate through the graph and find the longest cycle. 2. Chains connected to a 2-cycle (a pair of employees that like each other). For each 2-cycle `(a, b)`, we can include `a` and `b` plus the longest chains (paths) that lead into `a` and `b` but are not part of a cycle. The algorithm calculates these chains using topological sort.

Step-by-Step Algorithm

Step 1: Calculate the in-degrees of each node in the graph. This helps identify nodes that have no incoming edges, which is useful for topological sorting.
Step 2: Calculate the length of the longest chain ending at each node using a modified topological sort approach. Starting from nodes with an in-degree of 0, iteratively remove them and update the length of the longest chain ending at their favorite person.
Step 3: Identify cycles in the graph. Iterate through the nodes, and for each unvisited node with a positive in-degree (part of a cycle), traverse the cycle to calculate its length.
Step 4: If a cycle of length 2 is found (a mutual liking relationship), sum the longest chain ending at each of the two nodes in the cycle. These chains represents trees connected to this 'flower'.
Step 5: If a cycle with length greater than 2 is found, update the maximum cycle length found so far.
Step 6: Return the maximum of the largest cycle length (cycles of length > 2) and the sum of the lengths of all 2-cycle 'flower' chains.

Key Insights

Insight 1: The problem can be modeled as a directed graph where an edge `i -> favorite[i]` means employee `i` likes employee `favorite[i]`.
Insight 2: The optimal solution involves finding the largest cycle in the graph (cycles of length greater than 2) and combining it with the longest chains attached to 2-cycles (cycles of length 2).
Insight 3: Topological sort is helpful to process the graph by removing nodes with no incoming edges, which allows us to calculate the longest chain lengths leading to cycles.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Depth-First Search, Graph, Topological Sort.

Companies

Asked at: Barclays, SAP.

Maximum Employees to Be Invited to a Meeting | LeetCode 2127

Hard

Depth-First Search Graph Topological Sort

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Arrays;
import java.util.Queue;

class Solution {
    public int maximumInvitations(int[] favorite) {
        int n = favorite.length;

        // Calculate in-degrees for all nodes
        int[] inDegree = new int[n];
        for (int favPerson : favorite) {
            inDegree[favPerson]++;
        }

        // longest_chain[i] will store the number of nodes in the longest chain ending at node i.
        // A chain is a directed path of nodes that are not part of a cycle, leading into a cycle.
        // We initialize each node as a chain of length 1 (itself).
        int[] longestChain = new int[n];
        Arrays.fill(longestChain, 1);

        // Queue for topological sort-like processing
        // Start with nodes that are not favorited by anyone (these are the starting points of chains)
        Queue<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) {
                q.offer(i);
            }
        }

        // Process nodes in topological order to find the longest chain ending at each node.
        // This processes all the tree structures pointing towards cycles.
        while (!q.isEmpty()) {
            int u = q.poll();
            int v = favorite[u];

            // The chain ending at u can be extended to a chain ending at v.
            // We take the max because multiple chains can point to v.
            longestChain[v] = Math.max(longestChain[v], longestChain[u] + 1);

            inDegree[v]--;
            if (inDegree[v] == 0) {
                q.offer(v);
            }
        }

        // After the above loop, all non-cycle nodes are processed.
        // The remaining nodes (with in_degree > 0) are all part of cycles.

        // Candidate 1: The largest cycle of length k > 2.
        int maxKCycleLen = 0;
        // Candidate 2: The sum of sizes of all 2-cycle "flower" structures.
        int sumOf2CyclesChains = 0;

        boolean[] visited = new boolean[n];

        for (int i = 0; i < n; i++) {
            // If in_degree is still > 0, this node is part of a cycle and hasn't been visited yet.
            if (inDegree[i] > 0 && !visited[i]) {

                // Traverse the cycle to find its length.
                int cycleLen = 0;
                int curr = i;
                while (!visited[curr]) {
                    visited[curr] = true;
                    cycleLen++;
                    curr = favorite[curr];
                }

                if (cycleLen == 2) {
                    // This is a 2-cycle. The two nodes are i and favorite[i].
                    // The total size of this "flower" is the sum of the longest chains
                    // ending at each of the two cycle nodes.
                    sumOf2CyclesChains += longestChain[i] + longestChain[favorite[i]];
                } else {
                    // This is a cycle of length k > 2.
                    // It's a candidate for the maximum number of invitees.
                    maxKCycleLen = Math.max(maxKCycleLen, cycleLen);
                }
            }
        }

        return Math.max(maxKCycleLen, sumOf2CyclesChains);
    }
}

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Problem Info

Description

Problem 2127: Maximum Employees to Be Invited to a Meeting - Detailed explanation and solution approaches.

Topics

Depth-First Search Graph Topological Sort

Companies

BarclaysSAP

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2127

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: Calculate the in-degrees of each node in the graph. This helps identify nodes that have no incoming edges, which is useful for topological sorting.
2
Step 2: Calculate the length of the longest chain ending at each node using a modified topological sort approach. Starting from nodes with an in-degree of 0, iteratively remove them and update the length of the longest chain ending at their favorite person.
3
Step 3: Identify cycles in the graph. Iterate through the nodes, and for each unvisited node with a positive in-degree (part of a cycle), traverse the cycle to calculate its length.
4
Step 4: If a cycle of length 2 is found (a mutual liking relationship), sum the longest chain ending at each of the two nodes in the cycle. These chains represents trees connected to this 'flower'.
5
Step 5: If a cycle with length greater than 2 is found, update the maximum cycle length found so far.
6
Step 6: Return the maximum of the largest cycle length (cycles of length > 2) and the sum of the lengths of all 2-cycle 'flower' chains.

💡

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 time complexity is linear because we iterate through the input array `favorite` a constant number of times. The space complexity is linear because we create arrays of size `n` to store in-degrees, longest chain lengths, and visited status.

📖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 topological sort, DFS can be used to detect cycles and find the longest path leading to a cycle. However, this can be less efficient than topological sort in certain scenarios.

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

Trade-offs: DFS might be more intuitive for some to implement for cycle detection, but can be less efficient due to the overhead of recursive calls in cases with many cycles.

✨

Key Insights

1Insight 1: The problem can be modeled as a directed graph where an edge `i -> favorite[i]` means employee `i` likes employee `favorite[i]`.
2Insight 2: The optimal solution involves finding the largest cycle in the graph (cycles of length greater than 2) and combining it with the longest chains attached to 2-cycles (cycles of length 2).
3Insight 3: Topological sort is helpful to process the graph by removing nodes with no incoming edges, which allows us to calculate the longest chain lengths leading to cycles.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: When the graph is acyclic. The algorithm handles this by the initial topological sort, which effectively determines the longest path in the graph.
Edge case 2: When the graph contains only 2-cycles. The algorithm correctly sums the lengths of the chains attached to these cycles to provide the correct result.
Edge case 3: When there is a single large cycle involving all the nodes. The algorithm correctly identifies this cycle and outputs its length.
Edge case 4: When all employees favorite each other. Then algorithm would calculate one large cycle correctly.
Edge case 5: Input where every node has in_degree > 0 (everything is part of a cycle). Then topological sort does nothing, and the algorithm will still correctly find the longest cycle.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling 2-cycles (mutual liking relationships) separately from longer cycles. The chains attached to 2-cycles can be combined, whereas employees in cycles of length > 2 must be invited as a whole cycle.

✗Common mistake 2: Incorrectly calculating the longest chains leading into the cycles. Topological sort needs to be done carefully to ensure that the correct chain lengths are computed.

✗Common mistake 3: Not considering the case where there are no cycles at all.

✅

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. Identifying the largest group of mutually connected friends or followers.

🏦Real-world scenario 2: Task scheduling. Determining the longest sequence of dependent tasks in a project where tasks must be completed in a specific order.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: October 26, 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

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

@TechLeadJun 30

❓

Frequently Asked Questions

When should I use BFS vs DFS for graph problems?

Use BFS for shortest path in unweighted graphs, level-order processing, or when exploring neighbors systematically. Use DFS for exhaustive search, detecting cycles, topological sort, or when you need to explore complete paths. DFS uses less memory for very wide graphs.

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 "Maximum Employees to Be Invited to a Meeting"?

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

🤖 Need help with this problem?