SQL data engineering interview problem. Difficulty: beginner. Pattern: String Functions. About 8 minutes. Free to practice.
Find employees whose name starts with 'A'. Columns: employee_id, name. Order by name.
Input: employees employee_id | name 1 | Alice Chen 2 | Bob Kumar 5 | Omar Ali Output: employee_id | name 1 | Alice Chen Why this passes: LIKE 'A%' matches the first character only. Bob and Omar are excluded.
Topics: lakebench, sql, like, strings.
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Interview-style drill: LIKE filter for employee names starting with A.
Find employees whose `name` starts with 'A'. Columns: `employee_id`, `name`. Order by `name`.