USING LOCATE() FUNCTION | EFFICIENT
Intuition\nThe goal is to retrieve records from the Patients table where the condition \'DIAB1\' is either the first entry in the conditions string or is preced
# Intuition\nThe goal is to retrieve records from the `Patients` table where the condition \'DIAB1\' is either the first entry in the `conditions` string or is preceded by a space. This indicates that \'DIAB1\' is either the only condition listed or part of a space-separated list of conditions. My initial thought was to use the `LOCATE` function to identify the presence and position of \'DIAB1\' within the string.\n\n# Approach\nTo solve this problem, we will use the `LOCATE` function in MySQL, which returns the position of a substring within a string. The approach involves the following steps:\n\n1. Check if \'DIAB1\' is at the beginning of the `conditions` string by using `LOCATE(\'DIAB1\', conditions) = 1`.\n2. Check if \'DIAB1\' is found elsewhere in the string, ensuring it is preceded by a space by using `LOCATE(\' DIAB1\', conditions) != 0`.\n3. Combine both conditions using the `OR` operator to retrieve the relevant records.\n\n# Complexity\n- **Time complexity**:\n  - The time complexity of this query is approximately **O(n)**, where `n` is the number of rows in the `Patients` table. This is because each row\'s `conditions` string must be evaluated to determine the presence and position of \'DIAB1\'.\n\n- **Space complexity**:\n  - The space complexity is **O(1)**, as we are not utilizing any additional data structures that grow with the input size; we are only returning results based on the existing data.\n\n# Code\n```mysql\nSELECT *\nFROM Patients\nWHERE LOCATE(\'DIAB1\', conditions) = 1\n   OR LOCATE(\' DIAB1\', conditions) != 0;\n
['MySQL']