diff --git a/LeetCode SQL 50 Solution/1527. Patients With a Condition.sql b/LeetCode SQL 50 Solution/1527. Patients With a Condition.sql
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+1527. Patients With a Condition
+Solved
+Easy
+Topics
+Companies
+SQL Schema
+Pandas Schema
+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%'
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