How do you solve Employees Earning More Than Their Managers on LeetCode?

Employees Earning More Than Their Managers (LeetCode #181) 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: Self-joins can simplify comparisons between related records in the same table.

What is the brute force solution for Employees Earning More Than Their Managers?

The brute force approach for Employees Earning More Than Their Managers is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will compare each employee's salary with their manager's salary. If an employee earns more than their manager, we will include them in the result. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Employees Earning More Than Their Managers?

Employees Earning More Than Their Managers uses the following concepts: Database. The recommended patterns to study are: Self-Join, Filtering with Conditions.

What are the interview tips for Employees Earning More Than Their Managers?

Always think about how to optimize your queries, especially with joins. Be cautious with NULL values and how they affect comparisons. Practice writing SQL queries to become comfortable with syntax and logic.

What are common mistakes when solving Employees Earning More Than Their Managers?

Using nested subqueries unnecessarily, leading to performance issues. Not considering NULL values for managers, which can lead to incorrect results.

#181

Employees Earning More Than Their Managers

Easy

DatabaseSelf-JoinFiltering with Conditions

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)

In this approach, we will use a self-join to directly compare employees with their managers in one query. This is more efficient because we avoid the nested subquery.

⚙️

Algorithm

3 steps

1Step 1: Perform a self-join on the Employee table to link employees with their managers.
2Step 2: Filter the results where the employee's salary is greater than their manager's salary.
3Step 3: Select the employee names from the filtered results.

solution.py4 lines

1SELECT e1.name AS Employee
2FROM Employee e1
3JOIN Employee e2 ON e1.managerId = e2.id
4WHERE e1.salary > e2.salary;

ℹ

Complexity note: This complexity is O(n) because we are scanning through the Employee table once to perform the join and comparison, making it much more efficient than the brute force approach.

1Self-joins can simplify comparisons between related records in the same table.
2Understanding how to leverage SQL joins can significantly improve query performance.

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