Employee Importance

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

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.

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

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

Note

DFS through the HashMap of ID-Employee

Code

// Employee info
class Employee {
    // It's the unique id of each node;
    // unique id of this employee
    public int id;
    // the importance value of this employee
    public int importance;
    // the id of direct subordinates
    public List<Integer> subordinates;
}

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);
        }

        return dfs(id, map);
    }

    private int dfs(int id,  Map<Integer, Employee> map) {
        int ans = map.get(id).importance;
        for (Integer i : map.get(id).subordinates) {
            ans += dfs(i, map);
        }

        return ans;
    }
}