2nd highest without MAX: SELECT salary FROM employees e1 WHERE 1 = (SELECT COUNT(DISTINCT salary) FROM employees e2 WHERE e2.salary > e1.salary). Or: SELECT MIN(salary) FROM (SELECT salary FROM employees ORDER BY salary DESC LIMIT 2) t. Or DENSE_RANK: WITH r AS (SELECT salary, DENSE_RANK() OVER (ORDER BY salary DESC) rn FROM employees) SELECT salary FROM r WHERE rn=2. **Why it matters**: Design choices compound at scale—wrong approach can cause 100× overhead....
The complete answer continues with detailed implementation patterns, architectural trade-offs, and production-grade considerations. It covers performance optimization strategies, common pitfalls to avoid, and real-world examples from companies like Persistent Systems. The answer also includes follow-up discussion points that interviewers commonly explore.
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