[LeetCode] Problem 690 - Employee Importance

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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

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

  1. One employee has at most one direct leader and may have several subordinates.
  2. The maximum number of employees won’t exceed 2000.

Code

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public class Employee {
    public int id;
    public int importance;
    public List<Integer> subordinates;
}
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public int getImportance(List<Employee> employees, int id) {
    int result = 0;
    Map<Integer, Employee> map = new HashMap<>();
    Queue<Integer> queue = new LinkedList<>();
    queue.offer(id);

    for (Employee employee : employees)
        map.put(employee.id, employee);

    while (!queue.isEmpty()) {
        Integer employeeId = queue.poll();
        Employee employee = map.get(employeeId);
        result += employee.importance;

        for (Integer subordinate : employee.subordinates)
            queue.offer(subordinate);
    }

    return result;
}