How do you solve Patients With a Condition on LeetCode?

Patients With a Condition (LeetCode #1527) 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 SQL's LIKE operator allows for efficient string matching.

What is the brute force solution for Patients With a Condition?

The brute force approach for Patients With a Condition is Brute Force, with O(n²) time complexity and O(1) space complexity. In this approach, we will check each patient's conditions to see if any of them start with 'DIAB1'. This is straightforward but inefficient, as it requires examining each condition for every patient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Patients With a Condition?

Patients With a Condition uses the following concepts: Database. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Patients With a Condition?

Always think about how to filter data efficiently. Practice writing SQL queries to get comfortable with syntax. Understand the data structure and types before writing queries.

What are common mistakes when solving Patients With a Condition?

Not using the correct LIKE syntax for string matching. Overcomplicating the query with unnecessary joins or conditions.

#1527

Patients With a Condition

Easy

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)

We can optimize the search by using the SQL 'LIKE' operator directly to filter patients who have conditions starting with 'DIAB1'. This avoids unnecessary checks and is efficient.

⚙️

Algorithm

3 steps

1Step 1: Use a SQL query to select patient_id, patient_name, and conditions from the Patients table.
2Step 2: Apply a WHERE clause to filter conditions that start with 'DIAB1'.
3Step 3: Return the results.

solution.py1 lines

1SELECT patient_id, patient_name, conditions FROM Patients WHERE conditions LIKE 'DIAB1%';

ℹ

Complexity note: The time complexity is O(n) because we are scanning through the list of patients once. The space complexity is O(n) for storing the results.

1Using SQL's LIKE operator allows for efficient string matching.
2Filtering directly in the query reduces unnecessary computations.

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