LeetCode 0690. Employee Importance Solution in Java, C++, Python & More | Explanation + Code

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0690. Employee Importance

Description

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.

Solutions

Solution 1: Hash Table + DFS

We use a hash table d to store all employee information, where the key is the employee's ID, and the value is the employee object. Then we start a depth-first search from the given employee ID. Each time we traverse to an employee, we add the employee's importance to the answer, and recursively traverse all the subordinates of the employee, adding the importance of the subordinates to the answer as well.

The time complexity is O(n), and the space complexity is O(n). Where n is the number of employees.

PythonJavaC++GoTypeScriptJavaScript
""" # Definition for Employee. class Employee: def __init__(self, id: int, importance: int, subordinates: List[int]): self.id = id self.importance = importance self.subordinates = subordinates """ class Solution: def getImportance(self, employees: List["Employee"], id: int) -> int: def dfs(i: int) -> int: return d[i].importance + sum(dfs(j) for j in d[i].subordinates) d = {e.id: e for e in employees} return dfs(id)(code-box)

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