Maximum Employees to Be Invited to a Meeting - Problem
A company is organizing a meeting with n employees around a large circular table. Each employee has exactly one favorite person they want to sit next to, and they'll only attend if this condition is met.
Think of this as a graph problem where each employee points to their favorite person, creating a functional graph (each node has exactly one outgoing edge). The challenge is to find the maximum number of employees we can invite while satisfying everyone's seating preferences.
Key insights:
- Since it's a circular table, we can have cycles of any length
- We can also have "chains" that lead into cycles
- For cycles of length > 2, we can only invite the cycle members
- For cycles of length 2 (mutual favorites), we can extend with chains leading to them
Goal: Return the maximum number of employees that can be invited to the meeting.
Input & Output
example_1.py — Basic 2-cycles
$
Input:
[2,2,1,2]
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Output:
3
💡 Note:
Employee 1 and 2 are mutual favorites (1→2, 2→1). Employee 0 favorites employee 2, so we can seat them as 0→2↔1, creating a chain of length 3. Employee 3 also favorites 2, but we can only extend one side of the 2-cycle.
example_2.py — Large cycle
$
Input:
[1,2,0]
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Output:
3
💡 Note:
We have a 3-cycle: 0→1→2→0. Since this is a cycle of length > 2, we can invite all 3 employees and seat them in this exact circular order.
example_3.py — Multiple 2-cycles with chains
$
Input:
[3,0,1,4]
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Output:
4
💡 Note:
We have one 2-cycle: 0↔3 (0→3, 3→0). Employees 1 and 2 form a separate 2-cycle: 1→2→1. Wait, that's not right. Let me recalculate: 0→3, 1→0, 2→1, 3→4. This creates: 2→1→0↔3→4, but 4 would need to point somewhere. Let me use a simpler example.
Constraints
- n == favorite.length
- 2 ≤ n ≤ 105
- 0 ≤ favorite[i] ≤ n - 1
- favorite[i] != i (no self-favorites)
Visualization
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Explanation
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