MySQL Solution for Patients With A Condition Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the solution is to use the LIKE operator in the SQL query to match
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the solution is to use the LIKE operator in the SQL query to match the conditions that indicate Type I Diabetes. The query searches for conditions starting with "DIAB1" (\'DIAB1%\') or conditions that have "DIAB1" preceded by a space (\'% DIAB1%\').\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nThe approach is to retrieve the patient_id, patient_name, and conditions from the Patients table where the conditions column matches the specified patterns. By using the LIKE operator with wildcard characters (%), we can perform pattern matching and retrieve the desired results.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution depends on the size of the Patients table and the length of the conditions column. In the worst case, the query needs to scan through each row and evaluate the LIKE conditions, resulting in a linear time complexity O(N), where N is the number of rows in the table.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity of this solution is determined by the result set returned by the query. It will require memory to store the selected patient_id, patient_name, and conditions for the matching rows. The space complexity is typically O(M), where M is the number of rows that satisfy the conditions.\n\n# Code\n```\n# Write your MySQL query statement below\n# SELECT patient_id, patient_name, conditions\n# FROM Patients\n# WHERE conditions LIKE \'DIAB1%\'\n\nSELECT patient_id, patient_name, conditions\nFROM Patients\nWHERE conditions LIKE \'DIAB1%\' OR conditions LIKE \'% DIAB1%\'\n\n\n```
['MySQL']