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#1527
Easy Database

Patients with a condition

Database
38.8% acceptance
Feb 27, 2026
864
649
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.

Solution

SQL
LeetCode
solution.sql
#
# 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.
#

# Write your MySQL query statement below

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