Employee Importance - Hash Table - Easy - LeetCode
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Employee Importance - Hash Table - Easy - LeetCode

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  • 1The article discusses a data structure representing employee information, including unique id, importance value, and direct subordinates.
  • 2To calculate total importance, the algorithm sums the importance values of an employee and their direct and indirect subordinates.
  • 3The solution utilizes a dictionary for efficient lookups, achieving a time complexity of O(n) and space complexity of O(n).

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"The article discusses a data structure representing employee information, including unique id, importance value, and direct subordinates."

Employee Importance - Hash Table - Easy - LeetCode

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:

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

/*
// Definition for Employee.
class Employee {
 public int id;
 public int importance;
 public IList<int> subordinates;
}
*/

class Solution {
 public int GetImportance(IList<Employee> employees, int id) {
 var map = new Dictionary<int,Employee>();
 
 foreach(var emp in employees){ 
 map.Add(emp.id,emp); 
 }
 
 var sum = map[id].importance + Helper(map,id);
 return sum;
 }
 
 private int Helper(Dictionary<int,Employee> map, int id){ 
 int sum = 0;
 foreach(var subordinate in map[id].subordinates){
 sum+= map[subordinate].importance;
 sum+= Helper(map, subordinate);
 }
 return sum;
 }
}

Time Complexity: O(n)

Space Complexity: O(n)

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sumitc91

Published on 29 September 2020 · 1 min read · 277 words

Part of AskGif Blog · coding

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Employee Importance - Hash Table - Easy - LeetCode | AskGif Blog