EmployeeImportance.java

package graphSolutions;
import java.util.*;
//LeetCode Problem

/*
 * 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:
One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won't exceed 2000.
 */

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;
};

public class EmployeeImportance 
{
	int value = 0;
    public int getImportance(List<Employee> employees, int id) 
    {
        for(Employee e : employees)
        {
            if (e.id==id)
            {
                dfs(e,employees);
                value+= e.importance;
            }
        }
        System.out.println("edited");
        return value;
    }
    
    public void dfs(Employee e, List<Employee> employees)
    {
        for(Integer s : e.subordinates)
        {
            for(Employee e1 : employees)
            {
                if (e1.id==s)
                {
                    dfs(e1,employees);
                    value+= e1.importance;
                }
            }
        }
    }
}