Leetcode: Employee Importance

Employee Importance



Similar Problems:


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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/employee-importance
## Basic Ideas: BFS + hasmap
##         hasmap: find subordinate by id
##         visited_set: visited nodes
## Complexity: Time O(n), Space O(n)
"""
# Employee info
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):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        m, visited_set, queue = {}, set([]), []
        for employee in employees: m[employee.id] = employee

        res = m[id].importance
        queue.append(id)
        visited_set.add(id)
        while len(queue) != 0:
            for i in xrange(len(queue)):
                employee_id = queue[0]
                del queue[0]
                for subordinate_id in m[employee_id].subordinates:
                    if subordinate_id not in visited_set:
                        res += m[subordinate_id].importance
                        queue.append(subordinate_id)
                        visited_set.add(subordinate_id)
        return res
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