How do you solve Find Consistently Improving Employees on LeetCode?

Find Consistently Improving Employees (LeetCode #3580) 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: Employees need at least 3 reviews to qualify.

What is the time complexity of Find Consistently Improving Employees?

The optimal solution for Find Consistently Improving Employees runs in O(n) time and uses O(n) space. Single pass through reviews with efficient filtering leads to linear time complexity.

What is the brute force solution for Find Consistently Improving Employees?

The brute force approach for Find Consistently Improving Employees is Brute Force, with O(n²) time complexity and O(1) space complexity. Check each employee's reviews one by one to see if their last three ratings are strictly increasing. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Consistently Improving Employees?

Find Consistently Improving Employees uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Consistently Improving Employees?

Clarify the requirements before coding. Think about edge cases like equal ratings. Discuss your thought process while coding.

What are common mistakes when solving Find Consistently Improving Employees?

Forgetting to check for at least 3 reviews. Not ordering reviews correctly before checking ratings.

#3580

Find Consistently Improving Employees

Medium

DatabaseHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Use a single query to efficiently find employees with at least three reviews and check the last three ratings for strict improvement.

⚙️

Algorithm

3 steps

1Step 1: Join employees with performance_reviews and filter those with at least 3 reviews.
2Step 2: Order reviews by date and select the last three for each employee.
3Step 3: Check if the last three ratings are strictly increasing.

solution.py1 lines

1SELECT e.name FROM employees e JOIN (SELECT employee_id, rating FROM performance_reviews WHERE employee_id IN (SELECT employee_id FROM performance_reviews GROUP BY employee_id HAVING COUNT(*) >= 3) ORDER BY review_date DESC LIMIT 3) r ON e.employee_id = r.employee_id GROUP BY e.employee_id HAVING COUNT(r.rating) = 3 AND MAX(r.rating) > MIN(r.rating) AND MAX(r.rating) > (SELECT rating FROM performance_reviews WHERE employee_id = e.employee_id ORDER BY review_date DESC LIMIT 1 OFFSET 1) AND (SELECT rating FROM performance_reviews WHERE employee_id = e.employee_id ORDER BY review_date DESC LIMIT 1 OFFSET 2) > (SELECT rating FROM performance_reviews WHERE employee_id = e.employee_id ORDER BY review_date DESC LIMIT 1 OFFSET 1);

ℹ

Complexity note: Single pass through reviews with efficient filtering leads to linear time complexity.

1Employees need at least 3 reviews to qualify.
2Strictly increasing ratings imply no equal ratings.

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