LeetCode 2988: Manager of the Largest Department Solution

Master LeetCode problem 2988 (Manager of the Largest Department), a medium 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.

2988. Manager of the Largest Department

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we need to find the manager of the largest department in a company. We are given a list of employees where each employee has an id and a report_to field indicating the id of their manager. We can represent this information using a graph where each employee is a node and the report_to field indicates an edge from an employee to their manager.

We can perform a Depth First Search (DFS) starting from each employee to find the size of their department. While performing the DFS, we can keep track of the size of each department and store the manager of the largest department. Solution:

Solution Code

class Solution {
    Map<Integer, Integer> departmentSizes = new HashMap<>();
    int largestDepartmentManager = -1;
    int largestDepartmentSize = 0;

    public int manageDepartment(List<Employee> employees) {
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for (Employee employee : employees) {
            int employeeId = employee.getId();
            int managerId = employee.getReportTo();
            graph.putIfAbsent(managerId, new ArrayList<>());
            graph.get(managerId).add(employeeId);
        }

        for (Employee employee : employees) {
            int employeeId = employee.getId();
            int departmentSize = dfs(graph, employeeId);
            if (departmentSize > largestDepartmentSize) {
                largestDepartmentSize = departmentSize;
                largestDepartmentManager = employeeId;
            }
        }

        return largestDepartmentManager;
    }

    private int dfs(Map<Integer, List<Integer>> graph, int start) {
        if (!graph.containsKey(start)) {
            return 1; // employee has no subordinates
        }

        int size = 1;
        for (int subordinate : graph.get(start)) {
            size += dfs(graph, subordinate);
        }

        departmentSizes.put(start, size);
        return size;
    }
}

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