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Addition Of Digits


Addition Of Digits

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Shadab Shah
Shadab Shah
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Hi,
This was the question ask to one of my friend during an interview. He was ask to perform the addition of the digits.
Suppose the number is 985 the output would be 22(9+8+5).
SomewhereSomehow
SomewhereSomehow
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Here is my solution
declare @i int = 985;

with nums(n) as(select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10))nums(n))
select sum(convert(int,substring(convert(varchar(10),@i),n,1)))
from nums where n <= len(convert(varchar(10),@i))




I am really sorry for my poor gramma. And I hope that value of my answers will outweigh the harm for your eyes.
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Shadab Shah
Shadab Shah
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Thanks that help :-)
Alan.B
Alan.B
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SomewhereSomehow (8/3/2012)
Here is my solution
declare @i int = 985;

with nums(n) as(select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10))nums(n))
select sum(convert(int,substring(convert(varchar(10),@i),n,1)))
from nums where n <= len(convert(varchar(10),@i))



This is good. A couple things to note:
First, the length of the input variable (@i) is limited to the size of the two varchar declarations. For example, say @i = 99999999999 (that's eleven 9's), you would get an overflow error when converting the expression to varchar...

No big deal; who cares?... Just change the varchars to varchar(20) or varchar(50) and declare @i as bigint. There. Problem solved!!!

Nope. Still have one thing to address and this [b]WILL NOT produce an error[/b]. Instead you will just an incorrect aggregation. For the agregation to be accurate you would have to add additional values to your CTE. Again, say @i = 99999999999 (11 9's) you would return a 90(incorrect) instead of 99(correct). To fix this you would have to add an (11). If @i was 20 characters long you would have to add (11),(12)...(20).

A better way to write this would be:


DECLARE @i BIGINT=99999999999;

with nums(n) as
(
SELECT 1
UNION ALL
SELECT n+1 FROM nums WHERE n<30
)
select sum(convert(int,substring(convert(varchar(30),@i),n,1)))
from nums where n <= len(convert(varchar(30),@i));


Now @i can be 18 characters long (limited to 18 because of the bigint). How about we change @i to varchar(50). Now it works for a number that's 50 chars long.


Instead of:
select n from (values (1),(2),(3),(4)...(50))nums(n)



We are using some recursion:

with nums(n) as
(
SELECT 1
UNION ALL
SELECT n+1 FROM nums WHERE n<LEN(@i)
)



... and then we pull it all together:

DECLARE @i varchar(50)='99999999999999999999999999999999999999999999999999';

with nums(n) as
(
SELECT 1
UNION ALL
SELECT n+1 FROM nums WHERE n<LEN(@i)
)
select sum(convert(int,substring(convert(varchar(50),@i),n,1)))
from nums where n <= len(convert(varchar(50),@i));
GO



-- Alan Burstein



Best practices for getting help on SQLServerCentral
Need to split a string? Try DelimitedSplit8K or DelimitedSplit8K_LEAD (SQL 2012+)
Need a pattern-based splitter? Try PatternSplitCM
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"I can't stress enough the importance of switching from a 'sequential files' mindset to 'set-based' thinking. After you make the switch, you can spend your time tuning and optimizing your queries instead of maintaining lengthy, poor-performing code. " -- Itzek Ben-Gan 2001
Alan.B
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Here is my solution:


DECLARE @i varchar(50)='122333444455555';

WITH val(x,n) AS
(
SELECT LEFT(@i,LEN(@i)),'0'
UNION ALL
SELECT LEFT(x,LEN(x)-1),RIGHT(x,1) FROM val WHERE LEN(x)>0
)
SELECT SUM(CAST(n AS int)) FROM val



What's cool is you can replace
SELECT SUM(CAST(n AS int)) FROM val 


with
SELECT * FROM val 

to see how it works.

Result set:

x n
-------------------------------------------------- ----
122333444455555 0
12233344445555 5
1223334444555 5
122333444455 5
12233344445 5
1223334444 5
122333444 4
12233344 4
1223334 4
122333 4
12233 3
1223 3
122 3
12 2
1 2
1



-- Alan Burstein



Best practices for getting help on SQLServerCentral
Need to split a string? Try DelimitedSplit8K or DelimitedSplit8K_LEAD (SQL 2012+)
Need a pattern-based splitter? Try PatternSplitCM
Need to remove or replace those unwanted characters? Try PatExclude8K and PatReplace8K.

"I can't stress enough the importance of switching from a 'sequential files' mindset to 'set-based' thinking. After you make the switch, you can spend your time tuning and optimizing your queries instead of maintaining lengthy, poor-performing code. " -- Itzek Ben-Gan 2001
Michael Valentine Jones
Michael Valentine Jones
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A little dynamic SQL with a string of 100 digits as input:



declare @val varchar(100) =
'20876543914578560430730723092317208765439145785604'+
'3073072309231720876543914578560430730723092317208'

declare @cmd varchar(300)

set @cmd = 'select [Sum] = '+
reverse(substring(reverse(replace(replace(replace(
replace(replace(replace(replace(replace(replace(replace(
convert(varchar(300),@val),'9','9+'),'8','8+'),'7','7+'),'6','6+')
,'5','5+'),'4','4+'),'3','3+'),'2','2+'),'1','1+'),'0','0+')),2,300))

print '@val = '+@val
print '@cmd = '+@cmd

exec (@cmd)




Results:


@val = 208765439145785604307307230923172087654391457856043073072309231720876543914578560430730723092317208
@cmd = select [Sum] = 2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8
Sum
-----------
403

(1 row(s) affected)


SomewhereSomehow
SomewhereSomehow
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XMLSQLNinja,
Using the same logic. Yor solution will not work if there will be 51 digits.
Try to understand - my solution was intended to work only with int (and 999 999 999 99 is not int) in the same way, as yours only with 50 digits (btw, why 50, not 49 or 53?).
And one note, specifying input as string - not good idea imho, it coul be easily broken if there will be not a digit char in the string. If we talk about numbers, the input should be only one of numeric types - this is good form.


I am really sorry for my poor gramma. And I hope that value of my answers will outweigh the harm for your eyes.
Blog: http://somewheresomehow.ru
Twitter: @SomewereSomehow
SomewhereSomehow
SomewhereSomehow
Old Hand
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CELKO,
Good idea, interesting approach! Smth tell's me that it would be also the fastest way of doing this!


I am really sorry for my poor gramma. And I hope that value of my answers will outweigh the harm for your eyes.
Blog: http://somewheresomehow.ru
Twitter: @SomewereSomehow
Mark Cowne
Mark Cowne
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Another way, up to BIGINTs only

DECLARE @num BIGINT = 985;

WITH Tens(Pos,Val) AS (
SELECT 1, CAST(1 AS BIGINT) UNION ALL
SELECT 2, CAST(10 AS BIGINT) UNION ALL
SELECT 3, CAST(100 AS BIGINT) UNION ALL
SELECT 4, CAST(1000 AS BIGINT) UNION ALL
SELECT 5, CAST(10000 AS BIGINT) UNION ALL
SELECT 6, CAST(100000 AS BIGINT) UNION ALL
SELECT 7, CAST(1000000 AS BIGINT) UNION ALL
SELECT 8, CAST(10000000 AS BIGINT) UNION ALL
SELECT 9, CAST(100000000 AS BIGINT) UNION ALL
SELECT 10,CAST(1000000000 AS BIGINT) UNION ALL
SELECT 11,CAST(10000000000 AS BIGINT) UNION ALL
SELECT 12,CAST(100000000000 AS BIGINT) UNION ALL
SELECT 13,CAST(1000000000000 AS BIGINT) UNION ALL
SELECT 14,CAST(10000000000000 AS BIGINT) UNION ALL
SELECT 15,CAST(100000000000000 AS BIGINT) UNION ALL
SELECT 16,CAST(1000000000000000 AS BIGINT) UNION ALL
SELECT 17,CAST(10000000000000000 AS BIGINT) UNION ALL
SELECT 18,CAST(100000000000000000 AS BIGINT) UNION ALL
SELECT 19,CAST(1000000000000000000 AS BIGINT))
SELECT SUM((@num / Val) % 10)
FROM Tens
WHERE Val<=@num;



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