# Employee Importance in Python

Suppose we have a data structure of employee information, there are employee's unique id, his importance value and his direct subordinates' id. As an example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. And suppose their importance values are 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, []].

So, if we have the employee information of a company, and an employee id, we have to find the total importance value of this employee and all his subordinates.

So, if the input is like [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1, then the output will be 11, as Emp1 has importance value 5, and there are two direct subordinates of Emp1, they are − Emp2 and Emp3. Now both have importance value 3. So, the total importance value of Emp1 is 5 + 3 + 3 = 11.

To solve this, we will follow these steps −

• weight := a new map, leader := a new map
• for each e in employees, do
• weight[e[0]] := e[1]
• res := 0
• res := res + weight[id]
• while queue is non-zero, do
• new_queue := a new list
• node := delete last element from queue
• res := res + weight[node]
• if leader[node] is non-zero, then
• queue := queue + new_queue
• return res

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution(object):
def getImportance(self, employees, id):
weight = {}
for e in employees:
weight[e[0]] = e[1]
res = 0
res += weight[id]
while queue:
new_queue = []
node = queue.pop()
res += weight[node]
queue += new_queue
return res
ob = Solution()
print(ob.getImportance([[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1))

## Input

[[1, 5, [2, 3]], [2, 3, []], [3, 3, []]], 1

## Output

11