SQL Top 10 Interview QnA By Rishabh Mishra in Hindi.pdf

TOP 10 SQL
INTERVIEW QUERIES
By Rishabh Mishra

2
Rishabh Mishra
EmpID EmpName Gender Salary City
1 Arjun M 75000 Pune
2 Ekadanta M 125000 Bangalore
3 Lalita F 150000 Mathura
4 Madhav M 250000 Delhi
5 Visakha F 120000 Mathura
Practice Dataset
Employee Table
EmployeeDetail Table
EmpID Project EmpPosition DOJ
1 P1 Executive 26-01-2019
2 P2 Executive 04-05-2020
3 P1 Lead 21-10-2021
4 P3 Manager 29-11-2018
5 P2 Manager 01-08-2020

CREATE TABLE Employee (
EmpID int NOT NULL,
EmpName Varchar,
Gender Char,
Salary int,
City Char(20) )
--- first run the above code then below code
INSERT INTO Employee
VALUES (1, 'Arjun', 'M', 75000, 'Pune'),
(2, 'Ekadanta', 'M', 125000, 'Bangalore'),
(3, 'Lalita', 'F', 150000 , 'Mathura'),
(4, 'Madhav', 'M', 250000 , 'Delhi'),
(5, 'Visakha', 'F', 120000 , 'Mathura')
3
Rishabh Mishra
CREATE TABLE EmployeeDetail (
EmpID int NOT NULL,
Project Varchar,
EmpPosition Char(20),
DOJ date )
--- first run the above code then below code
INSERT INTO EmployeeDetail
VALUES (1, 'P1', 'Executive', '26-01-2019'),
(2, 'P2', 'Executive', '04-05-2020'),
(3, 'P1', 'Lead', '21-10-2021'),
(4, 'P3', 'Manager', '29-11-2019'),
(5, 'P2', 'Manager', '01-08-2020')
Create Tables: Employee and EmployeeDetail

Q1(b): Write a query to retrieve the list of employees from the same city.
4
Rishabh Mishra
SELECT E1.EmpID, E1.EmpName, E1.City
FROM Employee E1, Employee E2
WHERE E1.City = E2.City AND E1.EmpID != E2.EmpID
Q1(a): Find the list of employees whose salary ranges between 2L to 3L.
SELECT EmpName, Salary FROM Employee
WHERE Salary > 200000 AND Salary < 300000
--- OR –--
SELECT EmpName, Salary FROM Employee
WHERE Salary BETWEEN 200000 AND 300000
SELECT * FROM Employee
WHERE EmpID IS NULL
Q1(c): Query to find the null values in the Employee table.

5
Rishabh Mishra
SELECT
(COUNT(*) FILTER (WHERE Gender = 'M') * 100.0 / COUNT(*)) AS MalePct,
(COUNT(*) FILTER (WHERE Gender = 'F') * 100.0 / COUNT(*)) AS FemalePct
FROM Employee;
Q2(b): What’s the male and female employees ratio.
Q2(a): Query to find the cumulative sum of employee’s salary.
SELECT EmpID, Salary, SUM(Salary) OVER (ORDER BY EmpID) AS CumulativeSum
FROM Employee
Q2(c): Write a query to fetch 50% records from the Employee table.
WHERE EmpID <= (SELECT COUNT(EmpID)/2 from Employee)
If EmpID is not auto-increment field or numeric, then we can use Row NUMBER function

6
Rishabh Mishra
Q3: Query to fetch the employee’s salary but replace the LAST 2 digits with ‘XX’
i.e 12345 will be 123XX
SELECT Salary,
CONCAT(LEFT(Salary, LEN(Salary)-2), 'XX') as masked_salary
FROM Employee
SELECT Salary,
CONCAT(SUBSTRING(Salary::text, 1, LENGTH(Salary::text)-2), 'XX') as masked_number
FROM Employee
--- OR –--
SELECT Salary, CONCAT(LEFT(CAST(Salary AS text), LENGTH(CAST(Salary AS text))-2), 'XX')
AS masked_number
FROM Employee
MySQL

Q4: Write a query to fetch even and odd rows from Employee table.
7
Rishabh Mishra
If you have an auto-increment field
like EmpID then we can use the
MOD() function:
---Fetch even rows
WHERE MOD(EmpID,2)=0;
---Fetch odd rows
WHERE MOD(EmpID,2)=1;
---Fetch even rows
SELECT * FROM
(SELECT *, ROW_NUMBER() OVER(ORDER BY EmpId) AS
RowNumber
FROM Employee) AS Emp
WHERE Emp.RowNumber % 2 = 0
---Fetch odd rows
SELECT * FROM
(SELECT *, ROW_NUMBER() OVER(ORDER BY EmpId) AS
RowNumber
FROM Employee) AS Emp
WHERE Emp.RowNumber % 2 = 1
Alternative Solution
General Solution using ROW_NUMBER()

8
Rishabh Mishra
Q5(a): Write a query to find all the Employee names whose name:
• Begin with ‘A’
• Contains ‘A’ alphabet at second place
• Contains ‘Y’ alphabet at second last place
• Ends with ‘L’ and contains 4 alphabets
• Begins with ‘V’ and ends with ‘A’
SELECT * FROM Employee WHERE EmpName LIKE 'A%';
SELECT * FROM Employee WHERE EmpName LIKE '_a%';
SELECT * FROM Employee WHERE EmpName LIKE '%y_';
SELECT * FROM Employee WHERE EmpName LIKE '____l';
SELECT * FROM Employee WHERE EmpName LIKE 'V%a'

9
Rishabh Mishra
Q5(b): Write a query to find the list of Employee names which is:
• starting with vowels (a, e, i, o, or u), without duplicates
• ending with vowels (a, e, i, o, or u), without duplicates
• starting & ending with vowels (a, e, i, o, or u), without duplicates
SELECT DISTINCT EmpName
FROM Employee
WHERE LOWER(EmpName) SIMILAR TO '[aeiou]%'
FROM Employee
WHERE LOWER(EmpName) SIMILAR TO '%[aeiou]'
FROM Employee
WHERE LOWER(EmpName) SIMILAR TO '[aeiou]%[aeiou]'
FROM Employee
WHERE LOWER(EmpName) REGEXP '^[aeiou]'
FROM Employee
WHERE LOWER(EmpName) REGEXP '[aeiou]$'
FROM Employee
WHERE LOWER(EmpName) REGEXP
'^[aeiou].*[aeiou]$'
MySQL Solution: REGEXP

Q6: Find Nth highest salary from employee table with and without using the
TOP/LIMIT keywords.
10
Rishabh Mishra
SELECT Salary FROM Employee
ORDER BY Salary DESC
LIMIT 1 OFFSET N-1
SELECT Salary FROM Employee E1
WHERE N-1 = (
SELECT COUNT( DISTINCT ( E2.Salary ) )
FROM Employee E2
WHERE E2.Salary > E1.Salary );
--- OR ---
SELECT Salary FROM Employee E1
WHERE N = (
SELECT COUNT( DISTINCT ( E2.Salary ) )
FROM Employee E2
WHERE E2.Salary >= E1.Salary );
Using LIMIT
General Solution without using TOP/LIMIT
SELECT TOP 1 Salary
FROM Employee
WHERE Salary < (
SELECT MAX(Salary) FROM Employee)
AND Salary NOT IN (
SELECT TOP 2 Salary
FROM Employee
ORDER BY Salary DESC)
ORDER BY Salary DESC;
Using TOP

11
Rishabh Mishra
Q7(a): Write a query to find and remove duplicate records from a table.
SELECT EmpID, EmpName, gender, Salary, city,
COUNT(*) AS duplicate_count
FROM Employee
GROUP BY EmpID, EmpName, gender, Salary, city
HAVING COUNT(*) > 1;
Q7(b): Query to retrieve the list of employees working in same project.
WITH CTE AS
(SELECT e.EmpID, e.EmpName, ed.Project
FROM Employee AS e
INNER JOIN EmployeeDetail AS ed
ON e.EmpID = ed.EmpID)
SELECT c1.EmpName, c2.EmpName, c1.project
FROM CTE c1, CTE c2
WHERE c1.Project = c2.Project AND c1.EmpID != c2.EmpID AND c1.EmpID < c2.EmpID
DELETE FROM Employee
WHERE EmpID IN
(SELECT EmpID FROM Employee
GROUP BY EmpID
HAVING COUNT(*) > 1);

12
Rishabh Mishra
Q8: Show the employee with the highest salary for each project
SELECT ed.Project, MAX(e.Salary) AS ProjectSal
FROM Employee AS e
ON e.EmpID = ed.EmpID
GROUP BY Project
ORDER BY ProjectSal DESC;
WITH CTE AS
(SELECT project, EmpName, salary,
ROW_NUMBER() OVER (PARTITION BY project ORDER BY salary DESC) AS row_rank
FROM Employee AS e
ON e.EmpID = ed.EmpID)
SELECT project, EmpName, salary
FROM CTE
WHERE row_rank = 1;
Alternative, more dynamic solution: here you can fetch EmpName, 2nd/3rd highest value, etc
Similarly we can find Total Salary for each
project, just use SUM() instead of MAX()

13
Rishabh Mishra
Q9: Query to find the total count of employees joined each year
SELECT EXTRACT('year' FROM doj) AS JoinYear, COUNT(*) AS EmpCount
FROM Employee AS e
INNER JOIN EmployeeDetail AS ed ON e.EmpID = ed.EmpID
GROUP BY JoinYear
ORDER BY JoinYear ASC
Q10: Create 3 groups based on salary col, salary less than 1L is low, between 1 -
2L is medium and above 2L is High
SELECT EmpName, Salary,
CASE
WHEN Salary > 200000 THEN 'High'
WHEN Salary >= 100000 AND Salary <= 200000 THEN 'Medium'
ELSE 'Low'
END AS SalaryStatus
FROM Employee

14
Rishabh Mishra
BONUS: Query to pivot the data in the Employee table and retrieve the total
salary for each city.
The result should display the EmpID, EmpName, and separate columns for each city
(Mathura, Pune, Delhi), containing the corresponding total salary.
SELECT
EmpID,
EmpName,
SUM(CASE WHEN City = 'Mathura' THEN Salary END) AS "Mathura",
SUM(CASE WHEN City = 'Pune' THEN Salary END) AS "Pune",
SUM(CASE WHEN City = 'Delhi' THEN Salary END) AS "Delhi"
FROM Employee
GROUP BY EmpID, EmpName;

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SQL Top 10 Interview QnA By Rishabh Mishra in Hindi.pdf

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