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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 🙂

Please add comment(s) if you have one or multiple solutions in mind. Thank You.

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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