<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 Solution:
    def maximumInvitations(self, A: List[int]) -> int:
		# First, we find the largest circle.
        n, maxc = len(A), 0
        seen = [0] * n
        for idx in range(n):
		
			# If a people hasn't been visited:
            if seen[idx] == 0:
				# start is for locating the first visited people, cur_people stands 
				#for the current people we are visiting, we use curset to store all 
				# the visited people in this iteration.
                start = idx
                cur_people = idx
                curset = set()
				
				# As long as we are visiting new people, we keep finding his/her favorite.
                while seen[cur_people] == 0:
                    seen[cur_people] = 1
                    curset.add(cur_people)
                    cur_people = A[cur_people]
					
				# Until we find the first visited people. Depends on if this 
				# visited people has been visited in eariler iteration or just this iteration.
                if cur_people in curset:       # if current people is in current set, meaning we have found a new circle
                    cursum = len(curset)
					
					# use 'start' to find the distance from the first visited people in this iteration 
					# to this current people.
                    while start != cur_people:
                        cursum -= 1
                        start = A[start]
                    maxc = max(maxc, cursum)
                                       
		# Then we try to find the sum of largest arms. Firstly, find all mutal-favorite peoples.
        pair = []
        visited = [0] * n
        for i in range(n):
		
			# If a is b's favorite and vise versa, we put them in 'pair'.
            if A[A[i]] == i and visited[i] == 0:
                pair.append([i, A[i]])
                visited[i] = 1
                visited[A[i]] = 1
		
		# for every people I, find out all the people whos favorite is I.
        res = 0
        child = collections.defaultdict(list)
        for i in range(n):
            child[A[i]].append(i)
        
        for a, b in pair:
            # max arm length start from first people a
            maxa = 0
            dq = collections.deque()
            for cand in child[a]:
                if cand != b:
                    dq.append([cand, 1])
            while dq:
                cur, n = dq.popleft()
                maxa = max(maxa, n)
                for nxt in child[cur]:
                    dq.append([nxt, n + 1])
                    
            # max arm length start from first people b
            maxb = 0
            dq = collections.deque()
            for cand in child[b]:
                if cand != a:
                    dq.append([cand, 1])
            while dq:
                cur, n = dq.popleft()
                maxb = max(maxb, n)
                for nxt in child[cur]:
                    dq.append([nxt, n + 1])
            
			# Thus the total length is the two longest arm plus 2 (a and b themselves)
            res += 2 + maxa + maxb
			
		# select the larger one as the answer.
        return max(maxc, res)     
51400