Table: Patients
+--------------+---------+ | Column Name | Type | +--------------+---------+ | patient_id | int | | patient_name | varchar | | conditions | varchar | +--------------+---------+ patient_id is the primary key (column with unique values) for this table. 'conditions' contains 0 or more code separated by spaces. This table contains information of the patients in the hospital.
Write a solution to find the patient_id, patient_name, and conditions of the patients who have Type I Diabetes. Type I Diabetes always starts with DIAB1
prefix.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Patients table: +------------+--------------+--------------+ | patient_id | patient_name | conditions | +------------+--------------+--------------+ | 1 | Daniel | YFEV COUGH | | 2 | Alice | | | 3 | Bob | DIAB100 MYOP | | 4 | George | ACNE DIAB100 | | 5 | Alain | DIAB201 | +------------+--------------+--------------+ Output: +------------+--------------+--------------+ | patient_id | patient_name | conditions | +------------+--------------+--------------+ | 3 | Bob | DIAB100 MYOP | | 4 | George | ACNE DIAB100 | +------------+--------------+--------------+ Explanation: Bob and George both have a condition that starts with DIAB1.
This problem requires querying a database table (Patients
) to find patients with a specific condition. The condition is identified by a string prefix "DIAB1" within the conditions
column. The solution uses SQL to efficiently perform this task.
The core idea is to utilize SQL's LIKE
operator with wildcard characters (%
) to search for the "DIAB1" prefix within the conditions
column. Since the "DIAB1" prefix might appear at the beginning, in the middle, or at the end of the conditions
string (separated by spaces), we need two LIKE
conditions in the WHERE
clause to cover all possibilities.
SELECT
patient_id,
patient_name,
conditions
FROM patients
WHERE conditions LIKE 'DIAB1%' OR conditions LIKE '% DIAB1%';
Explanation:
SELECT patient_id, patient_name, conditions
: This selects the desired columns from the Patients
table.FROM patients
: This specifies the table to query.WHERE conditions LIKE 'DIAB1%' OR conditions LIKE '% DIAB1%'
: This is the crucial part.
conditions LIKE 'DIAB1%'
: This condition finds rows where the conditions
column starts with "DIAB1" followed by any characters.conditions LIKE '% DIAB1%'
: This condition finds rows where the conditions
column contains "DIAB1" anywhere within the string, preceded and followed by any characters (including spaces). The space before DIAB1
is crucial to avoid matching cases like "DIAB100" when searching for "DIAB1".OR
: The OR
operator combines the two LIKE
conditions, ensuring that rows matching either condition are included in the result.The time complexity of this SQL query is primarily determined by the database's query processing engine. In general, a LIKE
query with a prefix search can be optimized using indexes if an index exists on the conditions
column. With an index, the time complexity would be close to O(log n) for searching (where n is the number of rows in the Patients
table) due to the efficiency of index lookups. However, without an index, a full table scan would be required, leading to a time complexity of O(n).
The OR
condition does not significantly alter the time complexity because the database optimizer can efficiently process both LIKE
conditions.
The space complexity is O(k), where k is the size of the output (the number of rows satisfying the condition multiplied by the size of each row). The space needed to store the output depends on the number of patients who have "DIAB1" in their conditions. The space used for processing the query by the database system is generally considered constant relative to the input size.