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shreyashjishreyashji
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adding SQL PROBLEMS
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‎58. AK-finding missing quater.sql

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CREATE TABLE STORES (
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Store varchar(10),
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Quarter varchar(10),
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Amount int);
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INSERT INTO STORES (Store, Quarter, Amount)
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VALUES ('S1', 'Q1', 200),
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('S1', 'Q2', 300),
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('S1', 'Q4', 400),
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('S2', 'Q1', 500),
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('S2', 'Q3', 600),
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('S2', 'Q4', 700),
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('S3', 'Q1', 800),
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('S3', 'Q2', 750),
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('S3', 'Q3', 900);
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select * from STORES;
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#Method 1 Aggregation : 1+2+3+4=10 ,whatevver be missing 10 - summationn of other quater
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select store,10-sum((right(quarter,1))) as q_no from STORES
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group by store;
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select store,'Q'+ cast(10-sum(cast(right(quarter,1) as int)) as char(2)) as q_no from STORES
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group by store;
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#Method 2 Recursive CTE
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WITH ctq as(
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select distinct store, 1 as q_no from stores
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union all
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select store, q_no+1 as q_no from ctq
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where q_no < 4
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)
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select * from ctq
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order by stores;
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WITH ctq as(
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select distinct store, 1 as q_no from stores
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union all
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select store, q_no+1 as q_no from ctq
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where q_no < 4
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)
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, q as (select store,'Q'+ cast(q_no as char(1)) as q_no from ctq)
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select q.* from q left join stores s
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on q.store = s.store and q.q_no = s.quarter
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where s.store is null;
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#Method 3 Cross Join
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WITH ctq as(
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select distinct s1.store,s2.quarter
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from stores s1, stores s2
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)
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select q.* from cte q left join stores s
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on q.store = s.store and q.quarter = s.quarter
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where s.store is null;

‎59. AK-find students with same marks in physics & chemistry Deadly Combination of Group By and Having Clause in SQL.sql

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create table exams (student_id int, subject varchar(20), marks int);
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#delete from exams;
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insert into exams values (1,'Chemistry',91),(1,'Physics',91)
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,(2,'Chemistry',80),(2,'Physics',90)
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,(3,'Chemistry',80)
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,(4,'Chemistry',71),(4,'Physics',54);
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#find students with same marks in physics & chemistry
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SELECT * FROM exams
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where subject in("Physics","Chemistry")
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group by student_id
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Having count(distinct subject)=2 and count(distinct marks)=1; #if marks are same distinct count will be 1

‎59.Sort by one country always at top and others in ascending order Custom sorting.sql

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#Sort by one country always at top and others in ascending order | Custom sorting
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create table counntry (
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country_id int,
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countrynames varchar(20)
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);
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#drop table country;
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#insert into counntry values (1,"Afganistan"),(2,"Australia"),(3,"China"),(1,"France"),(1,"Germany"),(1,"India"),(1,"Italy");
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select * from counntry
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ORDER BY
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(CASE WHEN countrynames="India" then 0
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WHEN countrynames="China" then 1 else 2 end),
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countrynames;

‎60. SQL Queries for experienced.sql

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select * from empt;
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select * from dept;
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#find duplicate records in a table
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SELECT emp_name,salary, COUNT(*) AS CNT
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FROM empt
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GROUP BY emp_name,salary
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HAVING COUNT(*)>1 ;
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#-- delete all the duplicate records in a table
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WITH cte AS (
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SELECT
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emp_name,salary,
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ROW_NUMBER() OVER (
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PARTITION BY
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emp_name,salary
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ORDER BY
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emp_name,salary
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) row_num
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FROM
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empt
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)
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DELETE FROM cte WHERE row_num > 1;
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# find the manager name for the employee
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# --where empid and managerid are on the same table
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SELECT e.emp_id, e.emp_name, m.emp_name
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FROM empt e
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LEFT JOIN empt m
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on e.Manager_Id = m.Emp_Id;
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# find the second highest salary
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Select max(Salary) as Salary
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FROM empt
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WHERE Salary <(Select max(Salary) from empt) ;
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-- 1. Inner Query - Get the highest salary
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-- 2. Outer Query - Get the highest salary excluding the highest salary
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-- gives the second highest salary
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-- find the employee with the second highest salary
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SELECT * FROM empt where Salary in
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(SELECT max(Salary) as Salary
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FROM empt
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WHERE Salary < (Select max(Salary) FROM empt) );
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-- 3rd and Nth highest salary
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SELECT MIN(Salary) FROM -- OUTER QUERY
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( SELECT DISTINCT TOP 3 Salary -- INNER QUERY
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FROM empt
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ORDER BY Salary DESC
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) AS O
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-- Here 3 can be changed to N ; can be applied for any number.
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-- 1. Inner Query - Get the highest 3 salaries
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-- 2. Outer Query - Get the minimum salary from those salaries
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-- query to find maximum salary from each department
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SELECT Department_Id, MAX(Salary) as Salary
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FROM empt
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GROUP BY Department_Id
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#--alternative for TOP clause in SQL
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SELECT TOP 3 * FROM empt
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#Alternative
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SET ROWCOUNT 3
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Select * from empt
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SET ROWCOUNT 0
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-- showing single row from a table twice in the results
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SELECT dep_name FROM dept d WHERE d.dep_name='IT'
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UNION ALL
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SELECT dep_name FROM dept d WHERE d.dep_name='IT'
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-- find departments that have less than 3 employees
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SELECT e.department_id,d.dep_name
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FROM empt e
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JOIN dept d on e.department_id = d.dep_id
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GROUP BY e.department_id,d.dep_name HAVING COUNT(Emp_Id) < 3

‎61. User has to display all these fields along with maximum TranAmt for each CustID and ratio of TranAmt and maximum TranAmt for each transaction..sql

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#CREATE TABLE Transaction_Tbl_1
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(
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CustID int ,
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TranID int ,
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TranAmt float ,
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TranDate date
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) ;
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#drop table Transaction_Tbl_1 ;
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#INSERT into Transaction_Tbl_1 VALUES (1001, 20001, 10000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1001, 20002, 15000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1001, 20003, 80000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1001, 20004, 20000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1002, 30001, 7000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1002, 30002, 15000, CAST('2020-04-25' AS Date));
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#INSERT into Transaction_Tbl_1 VALUES (1002, 30003, 22000, CAST('2020-04-25' AS Date));
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select * from Transaction_Tbl_1;
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#Method 1: windows function
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select
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*,
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TranAmt/maximum_TranAmt
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from(
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select
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*,
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max(TranAmt) over(partition by CustID ) as maximum_TranAmt
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from
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Transaction_Tbl_1
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) t;
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#Method 2: Subquery
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select t2.*,
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t1.max_amt,
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(t2.TranAmt/t1.max_amt) as ratio
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from
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Transaction_Tbl_1 t2
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inner join
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(select CustID,Max(TranAmt) as max_amt
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from
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Transaction_Tbl_1
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group by CustID ) t1
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on t1.CustID=t2.CustID;
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#method 3
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with t1 as
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(
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select * from Transaction_Tbl_1
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),
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t2 as
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(
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Select CustID,max(TranAmt) As max_amt from Transaction_Tbl_1 group by CustID
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)
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select t1.*,
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t2.max_amt,
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(t1.TranAmt/t2.max_amt) as ratio
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from t1
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join t2
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on t1.CustID=t2.CustID;
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‎62.1 Amazon Complex SQL Interview Questions Group BY JOINS OR MULTIPLE QUERIES.sql

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#create table Contests ( contest_id INT, hacker_id INT, name VARCHAR(200) );
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#insert into Contests (contest_id, hacker_id, name) values (66406, 17973, 'Rose'), (66556, 79153, 'Angela'), (94828, 80275, 'Frank');
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#create table Colleges( college_id INT, contest_id INT );
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#insert into Colleges (college_id, contest_id) values (11219, 66406), (32473, 66556), (56685, 94828);
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#create table Challenges ( challenge_id INT, college_id INT );
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#insert into Challenges (challenge_id, college_id) values (18765, 11219), (47127, 11219), (60292, 32473), (72974, 56685);
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#create table View_Stats ( challenge_id INT, total_views INT, total_unique_views INT );
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#insert into View_Stats (challenge_id, total_views, total_unique_views) values (47127, 26, 19), (47127, 15, 14), (18765, 43, 10), (18765, 72, 13), (75516, 35, 17), (60292, 11, 10), (72974, 41, 15), (75516, 75, 11);
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#create table Submission_Stats ( challenge_id INT, total_submissions INT, total_accepted_submissions INT );
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#insert into Submission_Stats (challenge_id, total_submissions, total_accepted_submissions) values (75516, 34, 12), (47127, 27, 10), (47127, 56, 18), (75516, 74, 12), (75516, 83, 8), (72974, 68, 24), (72974, 82, 14), (47127, 28, 11);
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/* Question*
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John interviews many candidates from different colleges using coding challenges and contests.
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Write a query which displays the contest_id, hacker_id, name, sums of total_submissions,
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total_accepted_submissions, total_views, and total_unique_views for each contest sorted by contest_id.
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Exclude the contest from the result if all four sums are zero .
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*/
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select * from Contests;
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select * from Colleges;
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select * from Challenges;
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select * from View_Stats;
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select * from Submission_Stats;
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select con.contest_id,
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con.hacker_id,
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con.name,
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sum(total_submissions), #stats table
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sum(total_accepted_submissions), #stats table
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sum(total_views), #from view_stats
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sum(total_unique_views)
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from contests con
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join colleges col on con.contest_id = col.contest_id
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join challenges cha on col.college_id = cha.college_id
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left join
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(select challenge_id, sum(total_views) as total_views, sum(total_unique_views) as total_unique_views
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from view_stats group by challenge_id) vs on cha.challenge_id = vs.challenge_id
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left join
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(select challenge_id, sum(total_submissions) as total_submissions, sum(total_accepted_submissions) as total_accepted_submissions
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from submission_stats group by challenge_id) ss on cha.challenge_id = ss.challenge_id
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group by con.contest_id, con.hacker_id, con.name;
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/*
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having sum(total_submissions)!=0 or
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sum(total_accepted_submissions)!=0 or
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sum(total_views)!=0 or
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sum(total_unique_views)!=0
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order by contest_id;
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*/

‎63. AK - find cities where covid cases increasing continuously.sql

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create table covid(city varchar(50),days date,cases int);
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/*
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#delete from covid;
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insert into covid values('DELHI','2022-01-01',100);
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insert into covid values('DELHI','2022-01-02',200);
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insert into covid values('DELHI','2022-01-03',300);
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insert into covid values('MUMBAI','2022-01-01',100);
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insert into covid values('MUMBAI','2022-01-02',100);
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insert into covid values('MUMBAI','2022-01-03',300);
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insert into covid values('CHENNAI','2022-01-01',100);
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insert into covid values('CHENNAI','2022-01-02',200);
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insert into covid values('CHENNAI','2022-01-03',150);
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insert into covid values('BANGALORE','2022-01-01',100);
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insert into covid values('BANGALORE','2022-01-02',300);
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insert into covid values('BANGALORE','2022-01-03',200);
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insert into covid values('BANGALORE','2022-01-04',400);
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*/
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select * from covid;
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#find cities where covid cases increasing continuously
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select * ,
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rank() over(partition by city order by days asc) as rn_days
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,rank() over(partition by city order by cases asc) as rn_cases
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from covid
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order by city,cases;
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with abcd as(
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select *
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,rank() over(partition by city order by days asc) - rank() over(partition by city order by cases asc) as diff
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from covid )
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select city from abcd
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group by city
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having count(distinct diff) =1 and max(diff)=0 ;

‎64.AK- Google SQL Interview Question for Data Analyst Position.sql

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#create table company_users
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(
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company_id int,
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user_id int,
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language varchar(20)
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);
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#insert into company_users values (1,1,'English')
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,(1,1,'German')
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,(1,2,'English')
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,(1,3,'German')
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,(1,3,'English')
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,(1,4,'English')
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,(2,5,'English')
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,(2,5,'German')
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,(2,5,'Spanish')
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,(2,6,'German')
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,(2,6,'Spanish')
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,(2,7,'English');
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#find companines who have atleast 2 users who speaks both the language english & german
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select company_id,count(1) as num_of_users from
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(
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select company_id,user_id
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from company_users
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where language in('English','German')
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group by company_id,user_id
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having count(1)=2 ) a
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group by company_id
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having count(1)>=2
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;
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with temp as (
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select *, row_number() over (partition by user_id) as rn
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from company_users
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where language in('English','German')
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)
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select company_id, count(user_id) as num_of_users
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from temp
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where rn > 1
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group by company_id
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having count(user_id) > 1;

‎65-AK Sorting Happiness Index Data with India on Top.sql

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select * from happiness_index;
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select * from
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happiness_index
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order by
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case when country='India' then 3
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when country='Pakistan' then 2
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when country='Sri Lanka' then 1
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else 0 end desc,Happiness_2021 desc

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