Employee Importance

Similar Problems:

- Tag: graph

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.

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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