690. Employee Importance

You are given a data structure of employee information, which includes the employee’s unique id, their importance value and their 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 their subordinates.

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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.

Solution

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/*
// Definition for Employee.
class Employee {
    public int id;
    public int importance;
    public List<Integer> subordinates;
};
*/
class Solution {
    int sum = 0;
    boolean[] visited = new boolean[2001];
    public int getImportance(List<Employee> employees, int id) {
        for (var e: employees) {
            if (e.id == id) {
                dfs(employees, id);
            }
        }
        return sum;
    }
    
    void dfs(List<Employee> employees, int id) {
        visited[id] = true;
        for (var e: employees) {
            if (e.id == id) {
               sum += e.importance;
               for (Integer sub: e.subordinates) {
                   if (!visited[sub]) {
                       dfs(employees, sub);
                   }
               }
            }
        }
    }
}

Solution#

Solution