<p>A company is organizing a meeting and has a list of <code>n</code> employees, waiting to be invited. They have arranged for a large <strong>circular</strong> table, capable of seating <strong>any number</strong> of employees.</p>

<p>The employees are numbered from <code>0</code> to <code>n - 1</code>. Each employee has a <strong>favorite</strong> person and they will attend the meeting <strong>only if</strong> they can sit next to their favorite person at the table. The favorite person of an employee is <strong>not</strong> themself.</p>

<p>Given a <strong>0-indexed</strong> integer array <code>favorite</code>, where <code>favorite[i]</code> denotes the favorite person of the <code>i<sup>th</sup></code> employee, return <em>the <strong>maximum number of employees</strong> that can be invited to the meeting</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/14/ex1.png" style="width: 236px; height: 195px;" />
<pre>
<strong>Input:</strong> favorite = [2,2,1,2]
<strong>Output:</strong> 3
<strong>Explanation:</strong>
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. 
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> favorite = [1,2,0]
<strong>Output:</strong> 3
<strong>Explanation:</strong> 
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.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/14/ex2.png" style="width: 219px; height: 220px;" />
<pre>
<strong>Input:</strong> favorite = [3,0,1,4,1]
<strong>Output:</strong> 4
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == favorite.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= favorite[i] &lt;=&nbsp;n - 1</code></li>
	<li><code>favorite[i] != i</code></li>
</ul>

from collections import deque
from typing import List

class Solution:
    def maximumInvitations(self, favorite: List[int]) -> int:
        n = len(favorite)
        graph = [[] for _ in range(n)]
        
        # Build the graph where each employee points to their favorite
        for i, x in enumerate(favorite):
            graph[x].append(i)
        
        def bfs(x, seen):
            """Return the longest arm of x."""
            ans = 0
            queue = deque([x])
            while queue:
                for _ in range(len(queue)):
                    u = queue.popleft()
                    for v in graph[u]:
                        if v not in seen:
                            seen.add(v)
                            queue.append(v)
                ans += 1
            return ans
        
        ans = 0
        seen = [False] * n
        
        # Handle pairs of mutual favorites
        for i, x in enumerate(favorite):
            if favorite[x] == i and not seen[i]:
                seen[i] = seen[x] = True
                # Calculate the longest arms for both employees in the pair
                ans += bfs(i, {i, x}) + bfs(x, {i, x})
        
        dp = [0] * n
        
        # Detect cycles and calculate their lengths
        for i, x in enumerate(favorite):
            if dp[i] == 0:
                ii, val = i, 0
                memo = {}
                while ii not in memo:
                    if dp[ii]:
                        cycle = dp[ii]
                        break
                    memo[ii] = val
                    val += 1
                    ii = favorite[ii]
                else:
                    cycle = val - memo[ii]
                for k in memo:
                    dp[k] = cycle
        
        # Return the maximum of the longest arms and the longest cycle
        return max(ans, max(dp))
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