# 690. Employee Importance

## Question 690

[https://leetcode.com/problems/employee-importance/description/](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 **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.

### Answer  <a href="#answer" id="answer"></a>

### **v1. Recursive**

### v**2. Dict + queue**

1. 將 employees 存成 dictionary，以 id 當作 key
2. 建立一個 list 用 queue 的概念去想它
3. 有 subordinate 時就將其加入 list 中
4. 一直相加 list 中的員工 important 值，相加後 pop 該員工出去
5. 最後 list 會是空的，結束。

{% tabs %}
{% tab title="v1" %}
```python
'''
class Employee(object): 
    def init(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
 '''
  
class Solution(object):
    def getImportance(self, employees, id): 
    for employee in employees:
        if employee.id == id:
            if employee.subordinates != []:
                for subordinate in employee.subordinates:
                    employee.importance += self.getImportance(employees,subordinate)
                return employee.importance
            return employee.importance
    return 0
```
{% endtab %}

{% tab title="v2" %}
```python
'''
class Employee(object): 
    def init(self, id, importance, subordinates):
        # It's the unique id of each node.
        # unique id of this employee
        self.id = id
        # the importance value of this employee
        self.importance = importance
        # the id of direct subordinates
        self.subordinates = subordinates
 '''
  
class Solution(object):
    def getImportance(self, employees, id): 
        DictE = { employee.id:employee for employee in employees }
        SelectL = [id]
        Total = 0
        while SelectL:
            select = SelectL.pop()
            Total += DictE[select].importance
            for sub in DictE[select].subordinates:
                SelectL.append(sub)
        return Total
```
{% endtab %}
{% endtabs %}

