LeetCode 2127: Maximum Employees to Be Invited to a Meeting Solution

Master LeetCode problem 2127 (Maximum Employees to Be Invited to a Meeting), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2127. Maximum Employees to Be Invited to a Meeting

Hard

Depth-First Search Graph Topological Sort

Problem Explanation

Explanation

To solve this problem, we need to find the maximum number of employees that can be invited to the meeting while satisfying the condition that each employee can only attend if they can sit next to their favorite person. We can approach this problem by iterating through each employee and checking if we can form a cycle of employees and their favorite persons. We can use a visited array to keep track of visited employees to avoid double counting. By forming cycles, we can find the maximum number of employees that can be invited.

Create a visited array to mark visited employees.
Iterate through each employee.
For each unvisited employee, start forming a cycle by following their favorite person until we reach a previously visited employee or form a cycle.
Count the number of employees in the cycle.
Repeat the process for all employees and update the maximum number of employees in a cycle.

Time Complexity: O(n) where n is the number of employees. Space Complexity: O(n) for the visited array.

Solution Code

class Solution {
    public int maxEmployees(int[] favorite) {
        int n = favorite.length;
        boolean[] visited = new boolean[n];
        int maxEmployees = 0;
        
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                int current = i;
                int count = 0;
                do {
                    visited[current] = true;
                    current = favorite[current];
                    count++;
                } while (current != i);
                
                maxEmployees = Math.max(maxEmployees, count);
            }
        }
        
        return maxEmployees;
    }
}

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