"""
690. Employee Importance

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.
"""

emap1 = {}
emap2 = {}


def sumImportance(id):
    if len(emap2[id]) == 0:
        return emap1[id]
    else:
        total = emap1[id]
        for i in emap2[id]:
            total += sumImportance(i)
        return total


def getImportance(employees, id):
    """
    :type employees: Employee
    :type id: int
    :rtype: int
    """
    for ecnt in employees:
        emap1[ecnt[0]] = ecnt[1]
        emap2[ecnt[0]] = ecnt[2]

    if id not in emap1:
        return 0

    return sumImportance(id)


e = [[1, 5, [2, 3]], [2, 3, []], [3, 3, []]]
i = 1
print (getImportance(e, i))
