Employee Importance: Cache + DFS

Here is a problem (marked as easy) on LC: Employee Importance - LeetCode

690. Employee Importance

Easy

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.

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:

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

A simple approach would be to:

1/ Create a cache ID --> Employee for quick access

2/ From there, perform a simple DFS to calculate the importance

That's it. Code is below, cheers, ACC.

public int GetImportance(IList employees, int id)
{
    //Build a hash map for quick access
    Hashtable cache = new Hashtable();
    foreach (Employee e in employees)
        if (!cache.ContainsKey(e.id)) cache.Add(e.id, e);

    int retVal = 0;
    GetImportance(id, cache, ref retVal);
    return retVal;
}

public void GetImportance(int id, Hashtable cache, ref int importance)
{
    if (!cache.ContainsKey(id)) return;

    Employee e = (Employee)cache[id];
    importance += e.importance;
    foreach (int index in e.subordinates) GetImportance(index, cache, ref importance);
}