Classic question but the null-return requirement is where people slip up.
Use a subquery with the DISTINCT keyword to get unique salaries, order them in descending order, and use LIMIT/OFFSET to pick the second one. Wrap the query so that if no second salary exists, it returns NULL. Alternatively, use a window function like DENSE_RANK() to rank distinct salaries and filter for rank = 2.
Pro tip: Always clarify the SQL dialect (e.g., MySQL, PostgreSQL) because LIMIT/OFFSET syntax varies, and mention that you'd test edge cases like all employees having the same salary or fewer than two distinct salaries.
Confirm that 'second highest distinct salary' means the second unique salary value, and that if it doesn't exist, the result should be NULL. Ask about the SQL dialect if not specified.
Decide between using a subquery with LIMIT/OFFSET or a window function like DENSE_RANK(). The subquery approach is simpler but may not be portable; the window function is more standard and handles ties.
For the subquery approach: SELECT (SELECT DISTINCT salary FROM Employee ORDER BY salary DESC LIMIT 1 OFFSET 1) AS SecondHighestSalary. For the window function approach: SELECT MAX(salary) FROM (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) as rnk FROM Employee) t WHERE rnk = 2.
Ensure the query returns NULL when there is no second highest salary. The subquery approach naturally returns NULL if no rows; the window function approach may need a COALESCE or a check.
Mentally test with: empty table, one distinct salary, multiple employees with the same salary, and multiple distinct salaries. Verify the query returns the correct result in each case.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.