LeetCode 690 - Employee Importance

https://leetcode.com/problems/employee-importance/description/

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

  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;

  };

  Map<Integer, Employee> emap;

  public int getImportance(List<Employee> employees, int queryid) {

    emap = new HashMap();

    for (Employee e : employees)

      emap.put(e.id, e);

    return dfs(queryid);

  }

  public int dfs(int eid) {

    Employee employee = emap.get(eid);

    int ans = employee.importance;

    for (Integer subid : employee.subordinates)

      ans += dfs(subid);

    return ans;

  }

public int getImportance(List<Employee> employees, int id) {

  Map<Integer, Employee> map = employees.stream().collect(Collectors.toMap(e -> e.id, Function.identity()));

  return getImportanceImpl(map, id);

}

public int getImportanceImpl(Map<Integer, Employee> employees, int id) {

  Employee employee = employees.get(id);

  int result = employee.importance;

  if (employee.subordinates.isEmpty()) {

    return result;

  }

  for (Integer subId : employee.subordinates) {

    result += getImportanceImpl(employees, subId);

  }

  return result;

}

    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }
        return getImportanceHelper(map, id);
    }
    
    private int getImportanceHelper(Map<Integer, Employee> map, int rootId) {
        Employee root = map.get(rootId);
        int total = root.importance;
        for (int subordinate : root.subordinates) {
            total += getImportanceHelper(map, subordinate);
        }
        return total;
    }




X. BFS
https://leetcode.com/problems/employee-importance/discuss/112587/Java-HashMap-bfs-dfs

    public int getImportance(List<Employee> employees, int id) {
        int total = 0;
        Map<Integer, Employee> map = new HashMap<>();
        for (Employee employee : employees) {
            map.put(employee.id, employee);
        }
        Queue<Employee> queue = new LinkedList<>();
        queue.offer(map.get(id));
        while (!queue.isEmpty()) {
            Employee current = queue.poll();
            total += current.importance;
            for (int subordinate : current.subordinates) {
                queue.offer(map.get(subordinate));
            }
        }
        return total;
    }