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[SQL문제풀기] Second Highest Salary

Nov 21, 2025

[SQL문제풀기] Second Highest Salary

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문제

Second Highest Salary - LeetCode

Can you solve this real interview question? Second Highest Salary - Table: Employee +-------------+------+ | Column Name | Type | +-------------+------+ | id | int | | salary | int | +-------------+------+ id is the primary key (column with unique values) for this table. Each row of this table contains information about the salary of an employee. Write a solution to find the second highest distinct salary from the Employee table. If there is no second highest salary, return null (return None in Pandas). The result format is in the following example. Example 1: Input: Employee table: +----+--------+ | id | salary | +----+--------+ | 1 | 100 | | 2 | 200 | | 3 | 300 | +----+--------+ Output: +---------------------+ | SecondHighestSalary | +---------------------+ | 200 | +---------------------+ Example 2: Input: Employee table: +----+--------+ | id | salary | +----+--------+ | 1 | 100 | +----+--------+ Output: +---------------------+ | SecondHighestSalary | +---------------------+ | null | +---------------------+

Second Highest Salary - LeetCode

https://leetcode.com/problems/second-highest-salary/?envType=study-plan-v2&envId=top-sql-50

Second Highest Salary - LeetCode

MySQL

내가 작성한 오답


-- 1
select salary SecondHighestSalary
from (select distinct salary
        from Employee
        order by salary
        limit 1 offset 1) a;

notion image


-- 2
-- DENSE_RANK() 때문에 s = 2인 행이 여러 개 생겨서, 서브쿼리가 값 1개가 아니라 여러 행을 리턴
select (select salary
        from(select dense_rank() over(order by salary desc) s, salary
                from Employee) a
        where s = 2) SecondHighestSalary;

내가 작성한 정답


-- 1
select (select distinct salary
        from Employee
        order by salary desc
        limit 1 offset 1) SecondHighestSalary


-- 2 
select (select distinct salary
        from(select dense_rank() over(order by salary desc) s, salary
                from Employee) a
        where s = 2) SecondHighestSalary;


-- 3
select max(salary) SecondHighestSalary
from Employee
where salary < (select max(salary) from Employee);

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