690. Employee Importance LeetCode Solution

In this guide, you will get 690. Employee Importance LeetCode Solution with the best time and space complexity. The solution to Employee Importance problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Employee Importance solution in C++
4. Employee Importance solution in Java
5. Employee Importance solution in Python
6. Additional Resources

Problem Statement of 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.

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.

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(n)

690. Employee Importance LeetCode Solution in C++

class Solution {
 public:
  int getImportance(vector<Employee*> employees, int id) {
    unordered_map<int, Employee*> idToEmployee;

    for (Employee* employee : employees)
      idToEmployee[employee->id] = employee;

    return dfs(id, idToEmployee);
  }

 private:
  int dfs(int id, const unordered_map<int, Employee*>& idToEmployee) {
    int values = 0;

    for (const int subId : idToEmployee.at(id)->subordinates)
      values += dfs(subId, idToEmployee);

    return idToEmployee.at(id)->importance + values;
  }
};
/* code provided by PROGIEZ */

690. Employee Importance LeetCode Solution in Java

class Solution {
  public int getImportance(List<Employee> employees, int id) {
    Map<Integer, Employee> idToEmployee = new HashMap<>();

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

    return dfs(id, idToEmployee);
  }

  private int dfs(int id, Map<Integer, Employee> idToEmployee) {
    int values = 0;

    for (final int subId : idToEmployee.get(id).subordinates)
      values += dfs(subId, idToEmployee);

    return idToEmployee.get(id).importance + values;
  }
}
// code provided by PROGIEZ

690. Employee Importance LeetCode Solution in Python

class Solution:
  def getImportance(self, employees: list['Employee'], id: int) -> int:
    idToEmployee = {employee.id: employee for employee in employees}

    def dfs(id: int) -> int:
      values = idToEmployee[id].importance
      for subId in idToEmployee[id].subordinates:
        values += dfs(subId)
      return values

    return dfs(id)
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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