LeetCode #2127 — HARD

Maximum Employees to Be Invited to a Meeting

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 of the i^th employee, return the maximum number of employees that can be invited to the meeting.

Example 1:

Input: favorite = [2,2,1,2]
Output: 3
Explanation:
The above figure shows how the company can invite employees 0, 1, and 2, and seat them at the round table.
All employees cannot be invited because employee 2 cannot sit beside employees 0, 1, and 3, simultaneously.
Note that the company can also invite employees 1, 2, and 3, and give them their desired seats.
The maximum number of employees that can be invited to the meeting is 3.

Example 2:

Input: favorite = [1,2,0]
Output: 3
Explanation: 
Each employee is the favorite person of at least one other employee, and the only way the company can invite them is if they invite every employee.
The seating arrangement will be the same as that in the figure given in example 1:
- Employee 0 will sit between employees 2 and 1.
- Employee 1 will sit between employees 0 and 2.
- Employee 2 will sit between employees 1 and 0.
The maximum number of employees that can be invited to the meeting is 3.

Example 3:

Input: favorite = [3,0,1,4,1]
Output: 4
Explanation:
The above figure shows how the company will invite employees 0, 1, 3, and 4, and seat them at the round table.
Employee 2 cannot be invited because the two spots next to their favorite employee 1 are taken.
So the company leaves them out of the meeting.
The maximum number of employees that can be invited to the meeting is 4.

Constraints:

n == favorite.length
2 <= n <= 10⁵
0 <= favorite[i] <= n - 1
favorite[i] != i

Step 01

Brute Force Baseline

Problem summary: 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 of the ith employee, return the maximum number of employees that can be invited to the meeting.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Topological Sort

Example 1

[2,2,1,2]

Example 2

[1,2,0]

Example 3

[3,0,1,4,1]

Core Insight

What unlocks the optimal approach

From the given array favorite, create a graph where for every index i, there is a directed edge from favorite[i] to i. The graph will be a combination of cycles and chains of acyclic edges. Now, what are the ways in which we can choose employees to sit at the table?
The first way by which we can choose employees is by selecting a cycle of the graph. It can be proven that in this case, the employees that do not lie in the cycle can never be seated at the table (unless the cycle has a length of 2).
The second way is by combining acyclic chains. At most two chains can be combined by a cycle of length 2, where each chain ends on one of the employees in the cycle.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2127: Maximum Employees to Be Invited to a Meeting
class Solution {
    public int maximumInvitations(int[] favorite) {
        return Math.max(maxCycle(favorite), topologicalSort(favorite));
    }

    private int maxCycle(int[] fa) {
        int n = fa.length;
        boolean[] vis = new boolean[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (vis[i]) {
                continue;
            }
            List<Integer> cycle = new ArrayList<>();
            int j = i;
            while (!vis[j]) {
                cycle.add(j);
                vis[j] = true;
                j = fa[j];
            }
            for (int k = 0; k < cycle.size(); ++k) {
                if (cycle.get(k) == j) {
                    ans = Math.max(ans, cycle.size() - k);
                }
            }
        }
        return ans;
    }

    private int topologicalSort(int[] fa) {
        int n = fa.length;
        int[] indeg = new int[n];
        int[] dist = new int[n];
        Arrays.fill(dist, 1);
        for (int v : fa) {
            indeg[v]++;
        }
        Deque<Integer> q = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            if (indeg[i] == 0) {
                q.offer(i);
            }
        }
        int ans = 0;
        while (!q.isEmpty()) {
            int i = q.pollFirst();
            dist[fa[i]] = Math.max(dist[fa[i]], dist[i] + 1);
            if (--indeg[fa[i]] == 0) {
                q.offer(fa[i]);
            }
        }
        for (int i = 0; i < n; ++i) {
            if (i == fa[fa[i]]) {
                ans += dist[i];
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2127: Maximum Employees to Be Invited to a Meeting
func maximumInvitations(favorite []int) int {
	a, b := maxCycle(favorite), topologicalSort(favorite)
	return max(a, b)
}

func maxCycle(fa []int) int {
	n := len(fa)
	vis := make([]bool, n)
	ans := 0
	for i := range fa {
		if vis[i] {
			continue
		}
		j := i
		cycle := []int{}
		for !vis[j] {
			cycle = append(cycle, j)
			vis[j] = true
			j = fa[j]
		}
		for k, v := range cycle {
			if v == j {
				ans = max(ans, len(cycle)-k)
				break
			}
		}
	}
	return ans
}

func topologicalSort(fa []int) int {
	n := len(fa)
	indeg := make([]int, n)
	dist := make([]int, n)
	for i := range fa {
		dist[i] = 1
	}
	for _, v := range fa {
		indeg[v]++
	}
	q := []int{}
	for i, v := range indeg {
		if v == 0 {
			q = append(q, i)
		}
	}
	for len(q) > 0 {
		i := q[0]
		q = q[1:]
		dist[fa[i]] = max(dist[fa[i]], dist[i]+1)
		indeg[fa[i]]--
		if indeg[fa[i]] == 0 {
			q = append(q, fa[i])
		}
	}
	ans := 0
	for i := range fa {
		if i == fa[fa[i]] {
			ans += dist[i]
		}
	}
	return ans
}

# Accepted solution for LeetCode #2127: Maximum Employees to Be Invited to a Meeting
class Solution:
    def maximumInvitations(self, favorite: List[int]) -> int:
        def max_cycle(fa: List[int]) -> int:
            n = len(fa)
            vis = [False] * n
            ans = 0
            for i in range(n):
                if vis[i]:
                    continue
                cycle = []
                j = i
                while not vis[j]:
                    cycle.append(j)
                    vis[j] = True
                    j = fa[j]
                for k, v in enumerate(cycle):
                    if v == j:
                        ans = max(ans, len(cycle) - k)
                        break
            return ans

        def topological_sort(fa: List[int]) -> int:
            n = len(fa)
            indeg = [0] * n
            dist = [1] * n
            for v in fa:
                indeg[v] += 1
            q = deque(i for i, v in enumerate(indeg) if v == 0)
            while q:
                i = q.popleft()
                dist[fa[i]] = max(dist[fa[i]], dist[i] + 1)
                indeg[fa[i]] -= 1
                if indeg[fa[i]] == 0:
                    q.append(fa[i])
            return sum(dist[i] for i, v in enumerate(fa) if i == fa[fa[i]])

        return max(max_cycle(favorite), topological_sort(favorite))

// Accepted solution for LeetCode #2127: Maximum Employees to Be Invited to a Meeting
/**
 * [2127] Maximum Employees to Be Invited to a Meeting
 *
 * 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 of the i^th employee, return the maximum number of employees that can be invited to the meeting.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/12/14/ex1.png" style="width: 236px; height: 195px;" />
 * Input: favorite = [2,2,1,2]
 * Output: 3
 * Explanation:
 * The above figure shows how the company can invite employees 0, 1, and 2, and seat them at the round table.
 * All employees cannot be invited because employee 2 cannot sit beside employees 0, 1, and 3, simultaneously.
 * Note that the company can also invite employees 1, 2, and 3, and give them their desired seats.
 * The maximum number of employees that can be invited to the meeting is 3.
 *
 * Example 2:
 *
 * Input: favorite = [1,2,0]
 * Output: 3
 * Explanation:
 * Each employee is the favorite person of at least one other employee, and the only way the company can invite them is if they invite every employee.
 * The seating arrangement will be the same as that in the figure given in example 1:
 * - Employee 0 will sit between employees 2 and 1.
 * - Employee 1 will sit between employees 0 and 2.
 * - Employee 2 will sit between employees 1 and 0.
 * The maximum number of employees that can be invited to the meeting is 3.
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/12/14/ex2.png" style="width: 219px; height: 220px;" />
 * Input: favorite = [3,0,1,4,1]
 * Output: 4
 * Explanation:
 * The above figure shows how the company will invite employees 0, 1, 3, and 4, and seat them at the round table.
 * Employee 2 cannot be invited because the two spots next to their favorite employee 1 are taken.
 * So the company leaves them out of the meeting.
 * The maximum number of employees that can be invited to the meeting is 4.
 *
 *  
 * Constraints:
 *
 * 	n == favorite.length
 * 	2 <= n <= 10^5
 * 	0 <= favorite[i] <= n - 1
 * 	favorite[i] != i
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximum-employees-to-be-invited-to-a-meeting/
// discuss: https://leetcode.com/problems/maximum-employees-to-be-invited-to-a-meeting/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn maximum_invitations(favorite: Vec<i32>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2127_example_1() {
        let favorite = vec![2, 2, 1, 2];

        let result = 3;

        assert_eq!(Solution::maximum_invitations(favorite), result);
    }

    #[test]
    #[ignore]
    fn test_2127_example_2() {
        let favorite = vec![1, 2, 0];

        let result = 3;

        assert_eq!(Solution::maximum_invitations(favorite), result);
    }

    #[test]
    #[ignore]
    fn test_2127_example_3() {
        let favorite = vec![3, 0, 1, 4, 1];

        let result = 4;

        assert_eq!(Solution::maximum_invitations(favorite), result);
    }
}

// Accepted solution for LeetCode #2127: Maximum Employees to Be Invited to a Meeting
function maximumInvitations(favorite: number[]): number {
    return Math.max(maxCycle(favorite), topologicalSort(favorite));
}

function maxCycle(fa: number[]): number {
    const n = fa.length;
    const vis: boolean[] = Array(n).fill(false);
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (vis[i]) {
            continue;
        }
        const cycle: number[] = [];
        let j = i;
        for (; !vis[j]; j = fa[j]) {
            cycle.push(j);
            vis[j] = true;
        }
        for (let k = 0; k < cycle.length; ++k) {
            if (cycle[k] === j) {
                ans = Math.max(ans, cycle.length - k);
            }
        }
    }
    return ans;
}

function topologicalSort(fa: number[]): number {
    const n = fa.length;
    const indeg: number[] = Array(n).fill(0);
    const dist: number[] = Array(n).fill(1);
    for (const v of fa) {
        ++indeg[v];
    }
    const q: number[] = [];
    for (let i = 0; i < n; ++i) {
        if (indeg[i] === 0) {
            q.push(i);
        }
    }
    let ans = 0;
    while (q.length) {
        const i = q.pop()!;
        dist[fa[i]] = Math.max(dist[fa[i]], dist[i] + 1);
        if (--indeg[fa[i]] === 0) {
            q.push(fa[i]);
        }
    }
    for (let i = 0; i < n; ++i) {
        if (i === fa[fa[i]]) {
            ans += dist[i];
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

REPEATED DFS

O(V × E) time

O(V) space

Repeatedly find a vertex with no incoming edges, remove it and its outgoing edges, and repeat. Finding the zero-in-degree vertex scans all V vertices, and we do this V times. Removing edges touches E edges total. Without an in-degree array, this gives O(V × E).

TOPOLOGICAL SORT

O(V + E) time

O(V + E) space

Build an adjacency list (O(V + E)), then either do Kahn's BFS (process each vertex once + each edge once) or DFS (visit each vertex once + each edge once). Both are O(V + E). Space includes the adjacency list (O(V + E)) plus the in-degree array or visited set (O(V)).

Shortcut: Process each vertex once + each edge once → O(V + E). Same as BFS/DFS on a graph.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.