Leetcode #690: Employee Importance
In this guide, we solve Leetcode #690 Employee Importance in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Tree, Depth-First Search, Breadth-First Search, Array, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
"""
# 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)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.