Employee Importance

• LeetCode: Employee Importance
• Problem:

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.

• 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.

• Example:

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.

• Analysis:

Because we should calculate total importance for all suborindates. 

Use DFS or BFS to find all direct subordinates.

• Code-Java:

/*
// Employee info
class Employee {
    // It's the unique id of each node;
    // unique id of this employee
    public int id;
    // the importance value of this employee
    public int importance;
    // the id of direct subordinates
    public List<Integer> subordinates;
};
*/
class Solution {
    public int getImportance(List<Employee> employees, int id) {
        int result = 0;
        Map<Integer, Employee> employeesMap = new HashMap<Integer, Employee>();
        for (Employee employee: employees) {
            employeesMap.put(employee.id, employee);
        }
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.add(id);
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int val = queue.poll();
                Employee employee = employeesMap.get(val);
                if (employee == null) {
                    continue;
                }
                queue.addAll(employee.subordinates);
                result += employee.importance;
                employeesMap.remove(val);
            }
        }
        return result;
    }
}

• Code - C++:

/*
// Employee info
class Employee {
public:
    // It's the unique ID of each node.
    // unique id of this employee
    int id;
    // the importance value of this employee
    int importance;
    // the id of direct subordinates
    vector<int> subordinates;
};
*/
class Solution {
public:
    int getImportance(vector<Employee*> employees, int id) {
        int result = 0;
        map<int, Employee*> employeesMap;
        int size = employees.size();
        for (int i = 0; i < size; i++) {
            employeesMap[employees[i]->id] = employees[i];
        }
        queue<int> q;
        q.push(id);
        while (q.size() != 0) {
            size = q.size();
            for (int i = 0; i < size; i++) {
                int val = q.front();
                Employee *employee = employeesMap[val];
                if (employee == NULL) {
                    continue;
                }
                for (int i = 0; i < employee->subordinates.size(); i++) {
                    q.push(employee->subordinates[i]);
                }
                result += employee->importance;
                employeesMap.erase(val);
                q.pop();
            }
        }
        return result;
    }
};

• Code - Python:

"""
# 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
        """
        result = 0
        employeesMap = {}
        for employee in employees:
            employeesMap[employee.id] = employee
        q = []
        q.append(id)
        while (len(q) != 0):
            size = len(q)
            for i in range(size):
                val = q[0]
                employee = employeesMap[val]
                if employee == None:
                    continue
                for sub in employee.subordinates:
                    q.append(sub)
                result += employee.importance
                del employeesMap[val]
                del q[0]
        return result

• Time Complexity: O(N), N is the number of employees
• Space Complexity: O(N), N is the nubmer of employees