Employee Importance in Python

PythonServer Side ProgrammingProgramming

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]
    • leader[e[0]] := e[2]
  • res := 0
  • res := res + weight[id]
  • queue := leader[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
      • new_queue := new_queue + leader[size of leader]
    • 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 = {}
      leader = {}
      for e in employees:
         weight[e[0]] = e[1]
         leader[e[0]] = e[2]
      res = 0
      res += weight[id]
      queue = leader[id]
      while queue:
         new_queue = []
         node = queue.pop()
         res += weight[node]
         if leader[node]:
            new_queue += leader[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
raja
Published on 04-Jul-2020 09:37:01
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