<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>

class UnionFind:
    def __init__(self, n):
        self.parents = list(range(n))
        self.size = [1] * n
    
    def find(self, x):
        if x != self.parents[x]:
            self.parents[x] = self.find(self.parents[x])
        return self.parents[x]
    
    def union(self, x, y):
        xr, yr = self.find(x), self.find(y)
        if xr != yr:
            if self.size[xr] < self.size[yr]:
                x, y = y, x
            self.parents[yr] = xr
            self.size[xr] += self.size[yr]

class Solution:
    def maximumInvitations(self, favorite: List[int]) -> int:
        n = len(favorite)
        tail_length = [1] * n
        inDegree = [0] * n
        # Building Topological Sequences
        for curr in range(n):
            inDegree[favorite[curr]] += 1
        # Calculating the length of tail of each popular person
        queue = deque([c for c in range(n) if inDegree[c] == 0])
        while queue:
            curr = queue.popleft()
            fav = favorite[curr]
            # beacuse `fav` also have his own preference, as a result, if it previously has a tail
            # in other word, ans[fav] > 1, then we want to pick the largest one between the previous
            # tail and current tail
            tail_length[fav] = max(tail_length[fav], tail_length[curr] + 1)
            inDegree[fav] -= 1
            if inDegree[fav] == 0:
                queue.append(fav)
        # Building the Union Find of popular person
        uf = UnionFind(n)
        for i in range(n):
            if inDegree[i]:
                uf.union(i, favorite[i])
        # Two Situation:
        # Either we pick two popular person who like each other and join them together (Union Find size = 2)
        # or we we can select the size of this popluar person group (Union Find size > 2)
        join = 0
        res = 0
        for i in range(n):
            if not inDegree[i]:
                continue
            curr = uf.size[uf.find(i)]
            if curr == 2:
                fav = favorite[i]
                inDegree[fav] -= 1
                curr = tail_length[i] + tail_length[fav]
                join += curr
            else:
                res = max(res, curr)
        return max(res, join)
32400