690. Employee Importance

Problem

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**.

  Example 1:

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.

Example 2:

Input: employees = [[1,2,[5]],[5,-3,[]]], id = 5
Output: -3
Explanation: Employee 5 has an importance value of -3 and has no direct subordinates.
Thus, the total importance value of employee 5 is -3.

  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.

Solution (Java)

/*
// Definition for Employee.
class Employee {
    public int id;
    public int importance;
    public List<Integer> subordinates;
};
*/

class Solution {
    public int getImportance(List<Employee> employees, int id) {
        Map<Integer, Employee> map = new HashMap<>();
        for (Employee emp : employees) {
            map.put(emp.id, emp);
        }
        return calculateImportance(id, map);
    }

    private int calculateImportance(int id, Map<Integer, Employee> map) {
        Employee employee = map.get(id);
        int sum = employee.importance;
        for (int sub : employee.subordinates) {
            sum += calculateImportance(sub, map);
        }
        return sum;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).