LeetCode: Employee-importance BFS

Question

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 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.

Please comment about performance.

using System.Collections.Generic;
using Microsoft.VisualStudio.TestTools.UnitTesting;
namespace GraphsQuestions
{
 /// <summary>
 /// https://leetcode.com/problems/employee-importance/
 /// </summary>
 [TestClass]
 public class EmployeeImportanceBFS
 {
 [TestMethod]
 public void ChildrenSumTo30Test()
 {
 List<Employee> employees = new List<Employee>();
 Employee one = new Employee { Id = 1, Importance = 15 };
 Employee two = new Employee { Id = 2, Importance = 10 };
 one.Subordinates.Add(two.Id);
 Employee three = new Employee { Id = 3, Importance = 5 };
 two.Subordinates.Add(three.Id);
 employees.Add(one);
 employees.Add(two);
 employees.Add(three);
 Assert.AreEqual(30, GetImportance(employees, 1));
 }
 [TestMethod]
 public void ChildrenSumTo11Test()
 {
 List<Employee> employees = new List<Employee>();
 Employee one = new Employee { Id = 1, Importance = 5 };
 Employee two = new Employee { Id = 2, Importance = 3 };
 Employee three = new Employee { Id = 3, Importance = 3 };
 one.Subordinates.Add(three.Id);
 one.Subordinates.Add(two.Id);
 employees.Add(one);
 employees.Add(two);
 employees.Add(three);
 Assert.AreEqual(11, GetImportance(employees, 1));
 }
 int GetImportance(IList<Employee> employees, int id)
 {
 Dictionary<int, Employee> idToEmployee = new Dictionary<int, Employee>();
 foreach (var employee in employees)
 {
 idToEmployee.Add(employee.Id, employee);
 }
 int result = 0;
 if (employees == null || employees.Count == 0)
 {
 return result;
 }
 Queue<Employee> Q = new Queue<Employee>();
 Q.Enqueue(idToEmployee[id]);
 while (Q.Count > 0)
 {
 var current = Q.Dequeue();
 result += current.Importance;
 foreach (var childIdSubordinate in current.Subordinates)
 {
 if (idToEmployee.ContainsKey(childIdSubordinate))
 {
 Q.Enqueue(idToEmployee[childIdSubordinate]);
 }
 }
 }
 return result;
 }
 }
 // Employee info
 class Employee
 {
 public Employee()
 {
 Subordinates = new List<int>();
 }
 // It's the unique ID of each node.
 // unique id of this employee
 public int Id { get; set; }
 // the importance value of this employee
 public int Importance { get; set; }
 // the id of direct subordinates
 public List<int> Subordinates { get; set; }
 }
}

Answer 1

You'll have to have this check before the initialization of the dictionary:

 if (employees == null || employees.Count == 0)
 {
 return 0;
 }

or else the initialization of the dictionary may throw if employees == null

---

You can initialize the dictionary this way:

Dictionary<int, Employee> idToEmployee = employees.ToDictionary(e => e.Id);

---

Instead of:

if (idToEmployee.ContainsKey(childIdSubordinate))
 {
 Q.Enqueue(idToEmployee[childIdSubordinate]);
 }

You can do:

 if (idToEmployee.TryGetValue(childIdSubordinate, out Employee subordinate))
 {
 Q.Enqueue(subordinate);
 }

• Rick Davin: I think the TryGetValue needs explanation that it's not just an alternative technique but is more efficient since it queries the dictionary once instead of twice.
• user73941: @RickDavin: One should think so, but according to the reference it actually does the same as when using ContainsKey etc. - it just combines them into one method.
• VisualMelon: @HenrikHansen it only calls FindEntry once, which is the expensive bit. (As opposed to calling ContainsKey and this[TKey] which both call it).
• user73941: @RickDavin: You're right - I didn't check this[]