Identifying the boundaries of N-groups - sql

Question

I have spent quite some time dealing with the following:
Imagine that you have N number of groups with multiple records each and every record has unique starting and ending points.
In other words:
ID|GroupName|StartingPoint|EndingPoint|seq(row_number)|desired_seq
__|_________|_____________|___________|_______________|____________
1 | Grp1 |2014-01-06 |2014-01-07 |1 |1
__|_________|_____________|___________|_______________|____________
2 | Grp1 |2014-01-07 | 2014-01-08|2 |2
__|_________|_____________|___________|_______________|____________
3 | Grp2 |2014-01-08 | 2014-01-09|1 |1
__|_________|_____________|___________|_______________|____________
4 | Grp1 |2014-01-09 | 2014-01-10|3 |1
__|_________|_____________|___________|_______________|____________
5 | Grp2 |2014-01-10 | 2014-01-11|2 |1
__|_________|_____________|___________|_______________|____________
As you can see, the starting point for every consecutive record is the same as the ending point of the previous.
Basically, I would like to obtain the minimumS and maximumS for each group based on the dates. Once a record with new group name appears, then consider it as a new group and reset the sequencing.
Single row_number() function is not sufficient enough for this task since it doesnt reflect the change in the group names.(I have included a seq column in the sample data which represents the values generated by row number)
Desired result based on the sample data:
1 Grp1 |2014-01-06 | 2014-01-08
2 Grp2 |2014-01-08 | 2014-01-09
3 Grp1 |2014-01-09 | 2014-01-10
4 Grp2 |2014-01-10 | 2014-01-11
What I have tried:
;with cte as(
select *
, row_number() over (partition by GroupName order by startingpoint) as seq
from table1
)
select *
into #temp2
from cte t1
left join cte t2 on t1.id=t2.id and t1.seq= t2.seq-1
select *
,(select startingPoint from #temp2 t2 where t1.id=t2.id and t2.seq= (select MIN(seq) from #temp2) as Oldest
(select startingPoint from #temp2 t2 where t1.id=t2.id and t2.seq= (select MAX(seq) from #temp2) as MostRecent
from #temp2 t1

Answer 1

This is a gaps-and-islands problem with subgrouping. The trick is grouping by the difference between two ROW_NUMBER() values, one partitioned and one unpartitioned.
WITH t AS (
SELECT
GroupName,
StartingPoint,
EndingPoint,
ROW_NUMBER() OVER(PARTITION BY GroupName ORDER BY StartingPoint)
- ROW_NUMBER() OVER(ORDER BY StartingPoint) AS SubGroupId
FROM #test
)
SELECT
ROW_NUMBER() OVER (ORDER BY MIN(StartingPoint)) AS SortOrderId,
GroupName AS GroupName,
MIN(StartingPoint) AS GroupStartingPoint,
MAX(EndingPoint) AS GroupEndingPoint
FROM t
GROUP BY GroupName, SubGroupId
ORDER BY SortOrderId

Answer 2

This is so much easier with the lag() functionality in SQL Server 2012. The way I approach these problems is to find where groups start, assigning a flag of 1 or 0 to each row. Then take a cumulative sum of the 1s to get a new group id.
In SQL Server 2008, you can do this with correlated subqueries (or joins):
with table1_flag as (
select t1.*,
isnull((select top 1 1
from table1 t2
where t2.groupname = t1.groupname and
t2.endingpoint = t1.startingpoint
), 0) as groupstartflag
from table1 t1
),
table1_flag_cum as (
select tf.*,
(select sum(groupstartflag)
from table1_flag tf2
where tf2.groupname = tf.groupname and
tf2.startingpoint <= tf.startingpoint
) as groupnum
from table1_flag tf
)
select groupnum, groupname,
min(startingpoint) as startingpoint, max(endingpoint) as endingpoint
from table1_flag_cum
group by groupnum, groupname;

Answer 3

Not sure, but maybe:
SELECT DISTINCT
GroupName,
MIN(StartingPoint) OVER (PARTITION BY GroupName ORDER BY Id),
MAX(EndingPoint) OVER (PARTITION BY GroupName ORDER BY Id)
FROM table1
Because partition does not lead to the reduction of number of rows there will be originally duplicated entries, which are removed with distinct.