SQL Puzzle | The COUNT Puzzle – A SINGLE SELECT | SQL Interview Question

In this puzzle you have to find out how many new customer have joined on each date and how many customers have left on the that date. The challenge is to do this in a SINGLE SELECT. For more details please see the sample input and expected output.

Sample Input

NewCustomer

JoiningDate CustomerName
2018-06-04 00:00:00.000 Vikas
2018-06-05 00:00:00.000 Avtaar
2018-06-07 00:00:00.000 Pawan

OldCustomer

LeavingDate CustomerName
2018-06-04 00:00:00.000 Akshi
2018-06-04 00:00:00.000 Sahil
2018-06-06 00:00:00.000 Pranjal
2018-06-07 00:00:00.000 Mayank

Expected Output

Date #CustomerJoined #CustomerLeft
2018-06-04 00:00:00.000 1 2
2018-06-05 00:00:00.000 1 0
2018-06-06 00:00:00.000 0 1
2018-06-07 00:00:00.000 1 1

Script – DDL and INSERT Sample Data

 ```-- CREATE TABLE NewCustomer ( JoiningDate DATETIME ,CustomerName VARCHAR(20) ) GO INSERT INTO NewCustomer VALUES ('2018-06-04','Vikas'), ('2018-06-05','Avtaar'), ('2018-06-07','Pawan') GO CREATE TABLE OldCustomer ( LeavingDate DATETIME ,CustomerName VARCHAR(20) ) GO INSERT INTO OldCustomer VALUES ('2018-06-04','Akshi'), ('2018-06-04','Sahil'), ('2018-06-06','Pranjal'), ('2018-06-07','Mayank') GO SELECT * FROM OldCustomer GO -- ```

SOLUTION – 1

 ```-- SELECT ISNULL(LeavingDate,JoiningDate) Date ,COUNT(DISTINCT b.CustomerName) #CustomerJoined ,COUNT(DISTINCT a.CustomerName) #CustomerLeft FROM OldCustomer a FULL OUTER JOIN NewCustomer b ON leavingdate = joiningdate GROUP BY LeavingDate,joiningdate ORDER BY ISNULL(LeavingDate,JoiningDate) -- ```

Output – 1

 ```-- Date #CustomerJoined #CustomerLeft ----------------------- --------------- ------------- 2018-06-04 00:00:00.000 1 2 2018-06-05 00:00:00.000 1 0 2018-06-06 00:00:00.000 0 1 2018-06-07 00:00:00.000 1 1 (4 rows affected) -- ```

SOLUTION – 2 | WITH 2 SELECTS

 ```-- ;WITH CTE AS ( SELECT LeavingDate,COUNT(*) Cnts FROM OldCustomer GROUP BY LeavingDate ) SELECT ISNULL(c.LeavingDate,x.JoiningDate) Date , ISNULL(x.Cnts,0) #CustomerJoined , ISNULL(c.Cnts,0) #CustomerLeft FROM CTE c FULL JOIN ( SELECT JoiningDate,COUNT(*) Cnts FROM NewCustomer GROUP BY JoiningDate )x ON x.JoiningDate = c.LeavingDate -- ```

Output – 2

 ```-- Date #CustomerJoined #CustomerLeft ----------------------- --------------- ------------- 2018-06-04 00:00:00.000 1 2 2018-06-05 00:00:00.000 1 0 2018-06-06 00:00:00.000 0 1 2018-06-07 00:00:00.000 1 1 (4 rows affected) -- ```

Enjoy 🙂

Pawan Khowal

Pawan is a SQL Server Developer. If you need any help in writing code/puzzle or training please email at – pawankkmr”AT”gmail.com. Meanwhile please go throgh the top pages from his blog.

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