LeetCode 690. Employee Importance

Description

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

You have a data structure of employee information, which includes the employee’s unique id, their importance value, and their direct subordinates’ id.

You are given an array of employees employees where:

  • employees[i].id is the ID of the ith employee.
  • employees[i].importance is the importance value of the ith employee.
  • employees[i].subordinates is a list of the IDs of the subordinates of the ith employee.

Given an integer id that represents the ID of an employee, return the total importance value of this employee and all their subordinates.

Example 1:

Input: employees = [[1,5,[2,3]],[2,3,[]],[3,3,[]]], id = 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.

Example 2:

Input: employees = [[1,2,[5]],[5,-3,[]]], id = 5
Output: -3

Constraints:

  • 1 <= employees.length <= 2000
  • 1 <= employees[i].id <= 2000
  • All employees[i].id are unique.
  • -100 <= employees[i].importance <= 100
  • One employee has at most one direct leader and may have several subordinates.
  • id is guaranteed to be a valid employee id.

Explanation

Conduct Breadth-first search to accumulate all the importances from the employee and the subordinates.

Python Solution

"""
# Definition for Employee.
class Employee:
    def __init__(self, id: int, importance: int, subordinates: List[int]):
        self.id = id
        self.importance = importance
        self.subordinates = subordinates
"""

class Solution:
    def getImportance(self, employees: List['Employee'], id: int) -> int:
        
        employees_dict = {}

        result = 0
        queue = []
                
        for employee in employees:                        
            if employee.id == id:        
                queue.append(employee)
            
            employees_dict[employee.id] = employee
        
        while queue:
            employee = queue.pop(0)
                        
            result += employee.importance
                        
            for subordinate_id in employee.subordinates:                    
                queue.append(employees_dict[subordinate_id])
        

        return result
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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