LEETCODE 1527 - Patients With a Condition - Solution in MySQL

LEETCODE 1527 - Patients With a Condition - Solution in MySQL | LEETCODE 1527 Database SQL Solution

LEETCODE | 1527

 Patients With a Condition

LEETCODE 1527 - Patients With a Condition - Solution in MySQL | LEETCODE 1527 Database SQL Solution :

Table: Patients

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| patient_id   | int     |
| patient_name | varchar |
| conditions   | varchar |
+--------------+---------+
patient_id is the primary key for this table.
'conditions' contains 0 or more code separated by spaces. 
This table contains information of the patients in the hospital.

 

Write an SQL query to report the patient_id, patient_name all conditions of patients who have Type I Diabetes. Type I Diabetes always starts with DIAB1 prefix

Return the result table in any order.

The query 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.

Solution of LEETCODE 1527 - Patients With a Condition in MySQL:

SELECT *

FROM Patients

WHERE conditions LIKE '% DIAB1%' OR conditions LIKE 'DIAB1%'

Another Solution :

SELECT * FROM patients WHERE conditions REGEXP '\\bDIAB1'

The expression conditions REGEXP '\\bDIAB1' is actually the same as conditions LIKE '% DIAB1%' OR conditions LIKE 'DIAB1%';, but it is obviously shorter. 😉

The reason they are the same is that \b matches either a non-word character (in our case, a space) or the position before the first character in the string. Also, you need to escape a backslash with another backslash, like so: \\b. Otherwise, the regular expression won't evaluate.

P.S. \b also matches the position after the last character, but it doesn't matter in the context of this problem.

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