How do you solve Employee Importance on LeetCode?

Employee Importance (LeetCode #690) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows for faster lookups of employee data, which is crucial for optimizing the solution.

What is the brute force solution for Employee Importance?

The brute force approach for Employee Importance is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we can start by finding the employee with the given ID and then recursively calculate the importance of all their subordinates. This method is straightforward but can be inefficient due to repeated calculations. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Employee Importance?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as employees with no subordinates or negative importance values. Practice explaining your thought process as you code, as this demonstrates your understanding to the interviewer.

What are common mistakes when solving Employee Importance?

Not using a hash map for quick access to employee data, leading to inefficient searches. Failing to consider the base case in recursive functions, which can lead to infinite recursion.

#690

Employee Importance

Medium

ArrayHash TableTreeDepth-First SearchBreadth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) approach with a hash map to store employee data. This allows us to efficiently access each employee's information and calculate the total importance without redundant searches.

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Algorithm

3 steps

1Step 1: Create a hash map to store employees by their ID for O(1) access.
2Step 2: Define a recursive function that takes an employee ID and calculates their total importance by summing their importance and the importance of their subordinates.
3Step 3: Call the recursive function with the given employee ID and return the result.

solution.py15 lines

1class Employee:
2    def __init__(self, id, importance, subordinates):
3        self.id = id
4        self.importance = importance
5        self.subordinates = subordinates
6
7def getImportance(employees, id):
8    employee_map = {e.id: e for e in employees}
9    def dfs(emp_id):
10        emp = employee_map[emp_id]
11        total = emp.importance
12        for sub_id in emp.subordinates:
13            total += dfs(sub_id)
14        return total
15    return dfs(id)

ℹ

Complexity note: The time complexity is O(n) because we visit each employee exactly once. The space complexity is also O(n) due to the hash map storing employee data.

1Using a hash map allows for faster lookups of employee data, which is crucial for optimizing the solution.
2Recursive approaches can simplify the logic of traversing hierarchical data structures like employee subordinates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.