Leetcode 690. Employee Importance (Python)

Related Topic

Depth-First-Search.

Description

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.

Sample I/O

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.

Methodology

Find the current employee id and employee’s importance value and recursively find the employee’s subordinate’s id and subordinate’s importance value and add important values together.

Code(DFS)

def getImportance(self, employees, id):
        """
        :type employees: Employee
        :type id: int
        :rtype: int
        """
        dict = {employee.id: employee for employee in employees}
        def dfs(id):
            total_importance = dict[id].importance
            for sub_id in dict[id].subordinates:
                total_importance+=dfs(sub_id)
            return total_importance
        return dfs(id)

BigO

We traversal all elements of the list once so time complexity is O(n)