690. Employee Importance

package Leetcode_690

import java.util.*
import kotlin.collections.HashMap

/**
 * 690. Employee Importance
 * https://leetcode.com/problems/employee-importance/
 *
 * You have a data structure of employee information, including the employee's unique ID, importance value, and direct subordinates' IDs.
You are given an array of employees employees where:
employees[i].id is the ID of the ith employee.
employees[i].importance is the importance value of the ith employee.
employees[i].subordinates is a list of the IDs of the direct subordinates of the ith employee.
Given an integer id that represents an employee's ID, return the total importance value of this employee and all their direct and indirect subordinates.

Example1:
Input: employees = [[1,5,[2,3]],[2,3,[]],[3,3,[]]], id = 1
Output: 11
Explanation: Employee 1 has an importance value of 5 and has two direct subordinates: employee 2 and employee 3.
They both have an importance value of 3.
Thus, the total importance value of employee 1 is 5 + 3 + 3 = 11.

Constraints:
1 <= employees.length <= 2000
1 <= employees[i].id <= 2000
All employees[i].id are unique.
-100 <= employees[i].importance <= 100
One employee has at most one direct leader and may have several subordinates.
The IDs in employees[i].subordinates are valid IDs.
 * */

// Definition for Employee.
class Employee {
    var id: Int = 0
    var importance: Int = 0
    var subordinates: List<Int> = listOf()
}

class Solution {
    /**
     * Solution: Mapping + BFS, Time complexity:O(n), Space complexity:O(n)
     * */
    fun getImportance(employees: List<Employee?>, id: Int): Int {
        var result = 0
        val map = HashMap<Int, Employee?>()
        for (employee in employees) {
            if (employee == null) {
                continue
            }
            map.put(employee.id, employee)
        }
        val queue = LinkedList<Employee>()
        val first = map.get(id)
        queue.offer(first)
        while (queue.isNotEmpty()) {
            val top = queue.pop()
            result += top.importance
            for (item in top.subordinates) {
                queue.offer(map.get(item))
            }
        }
        return result
    }
}