employee-importance.py

# Time:  O(n)
# Space: O(h)

# 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:
# One employee has at most one direct leader and may have several subordinates.
# The maximum number of employees won't exceed 2000.

import collections


"""
# 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
        """
        if employees[id-1] is None:
            return 0
        result = employees[id-1].importance
        for id in employees[id-1].subordinates:
            result += self.getImportance(employees, id)
        return result


# Time:  O(n)
# Space: O(w), w is the max number of nodes in the levels of the tree
class Solution2(object):
    def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        result, q = 0, collections.deque([id])
        while q:
            curr = q.popleft()
            employee = employees[curr-1]
            result += employee.importance;
            for id in employee.subordinates:
                q.append(id)
        return result