690 Employee Importance

690. Employee Importance

1. Question

You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.

For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.

Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.

Example 1:

Input:
 [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1

Output:
 11

Explanation:

Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3. So the total importance value of employee 1 is 5 + 3 + 3 = 11.

Note:

1. One employee has at most one direct leader and may have several subordinates.

2. The maximum number of employees won't exceed 2000.

2. Implementation

(1) BFS

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<>();

        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }

        Queue<Employee> queue = new LinkedList<>();
        queue.add(map.get(id));

        int res = 0;

        while (!queue.isEmpty()) {
            Employee curEmployee = queue.remove();
            res += curEmployee.importance;

            for (Integer employeeId : curEmployee.subordinates) {
                queue.add(map.get(employeeId));
            }
        }
        return res;
    }
}

(2) DFS

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<>();

        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }

        int[] res = new int[1];
        getImportanceByDFS(map, id, res);
        return res[0];

    }

    public void getImportanceByDFS(Map<Integer, Employee> map, int id, int[] res) {
        res[0] += map.get(id).importance;

        for (Integer subId : map.get(id).subordinates) {
            getImportanceByDFS(map, subId, res);
        }
    }
}

3. Time & Space Complexity

BFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap

DFS: 时间复杂度O(n), 空间复杂度O(n),因为用了HashMap