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Home/ Questions/Q 419617
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T18:47:43+00:00 2026-05-12T18:47:43+00:00

I have a database for which I need to aggregate records into another smaller

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I have a database for which I need to aggregate records into another smaller set. This result set should contain the difference between maximum and minumum of specific columns of the original records where they add up to certain SUM, a closed interval constant C.

The constant C determines how the original records are aggregated and no entry in the resulting set ever exceeds it. Naturally I am supposed to run this in natural primary key order..

To illustrate: table has:

  • [key]
  • [a]
  • [b]
  • [minColumn]
  • [maxColumn]
  • [N]

…all are int datatype.

I am after a result set that has entries where the MAX(maxColumn) – MIN(minColumn) for that group such that when their difference is summed up it is less or equal to constant C.

Apart from the MAX(maxColumn) and MIN(minColumn) value I also need the FIRST record column [a] and LAST record column [b] values before creating a new entry in this result set. Finally, the N column should be SUMmed for all original records in a group.

Is there an efficient way to do this without cursors?

—–[Trivial Sample]————————————————————

I am attempting to group-by a slightly complicated form of a running sum, constant C.

There is only one table, columns are all of int type and sample data

declare @t table (
  PK int primary key
    , int a, int b, int minColumn, int maxColumn, int N 
)

insert @t values (1,5,6,100,200,1000)
insert @t values (2,7,8,210,300,2000)
insert @t values (3,9,10,420,600,3000)
insert @t values (4,11,12,640,800,4000)

Thus for:

key, a,   b, minColumn, maxColumn,    N
---------------------------------------
1,   5,   6,       100,       200, 1000 
2,   7,   8,       210,       300, 2000 
3,   9,  10,       420,       600, 3000 
4,   11, 12,       640,       800, 4000 

I need the result set to look like, for a constant C of 210 :

firstA | lastB | MIN_minColumn | MAX_maxColumn | SUM_N
5       8                  100             300    3000 
9       10                 420             600    3000 
11      12                 640             800    4000 

[ Adding the bounty and sample as discussed below]

For C = 381, It should contain 2 rows:

firstA | lastB | MIN_minColumn | MAX_maxColumn | SUM_N
5            8             100             300    3000 
9           12             420             800    7000

Hope this demonstrates the problem better.. and for a constant C say 1000 you would get 1 record in the result:

firstA | lastB | MIN_minColumn | MAX_maxColumn | SUM_N
5           12             100             800   10000
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-12T18:47:43+00:00Added an answer on May 12, 2026 at 6:47 pm
    DECLARE @c int
    SELECT @c = 210
    
    SELECT MIN(a) firstA,
           MAX(b) lastB, 
           MIN(minColumn) MIN_minColumn, 
           MAX(maxColumn) MAX_maxColumn, 
           SUM(N) SUM_N
    FROM @t t 
    JOIN (SELECT key, floor(sum/@c) as rank
            FROM (SELECT key, 
                         (SELECT SUM(t2.maxColumn - t2.minColumn) 
                            FROM @t t2 
                           WHERE t2.key <= t1.key 
                        GROUP BY t1.key) as sum
                   FROM @t t1) A
         ) B on B.key = t.key
    GROUP BY B.rank
    
    /*
    
    Table A: for each key, calculating SUM[maxColumn-minColumn] of all keys below it.
    Table B: for each key, using the sum in A, calculating a rank so that:
      sum = (rank + y)*@c where 0 <= y < 1. 
      ex: @c=210, rank(100) = 0, rank(200) = 0, rank(220) = 1, ...
    finally grouping by rank, you'll have what you want.
    
    */
    
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