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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0601_0700.s0690_employee_importance;

// #Medium #Hash_Table #Depth_First_Search #Breadth_First_Search
// #2022_03_22_Time_7_ms_(77.84%)_Space_58.8_MB_(21.60%)

import com_github_leetcode.Employee;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/*
// Definition for Employee.
class Employee {
    public int id;
    public int importance;
    public List subordinates;
};
*/
/**
 * 690 - Employee Importance\.
 *
 * Medium
 *
 * 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 i^th employee.
 * *   `employees[i].importance` is the importance value of the i^th employee.
 * *   `employees[i].subordinates` is a list of the IDs of the direct subordinates of the i^th 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/31/emp1-tree.jpg)
 *
 * **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:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/05/31/emp2-tree.jpg)
 *
 * **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.
**/
public class Solution {
    public int getImportance(List employees, int id) {
        Map map = new HashMap<>();
        for (Employee emp : employees) {
            map.put(emp.id, emp);
        }
        return calculateImportance(id, map);
    }

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