select minimum date after maximum gap oracle - oracle

Question

I have data for a member like below
EFF_DT-Term_dt
1/1/13-7/31/14
1/1/15-3/31/15
5/1/15-5/31/15
6/1/15-12/31/15
1/1/16-12/31/16
Here there are 2 gaps - after 7/31/14 and 3/31/15. I want to select the row 5/1/15-5/31/15 as it is the minimum date after maximum gap. I tried using
select ( FIRST_VALUE(EFF_DT) OVER (PARTITION BY MemberID ORDER BY FLAG DESC) AS CUR_EFF_DT)
from
(
select EFF_DT,
CASE WHEN LAG(TERM_DT, 1) OVER (PARTITION BY MemberID ORDER BY TERM_DT) = EFF_DT - 1 THEN 0
ELSE sequence.nextval
END AS FLAG
from effective_dates_table).
This is giving correct result, but i don't want to use sequence Is there any other easiest way to do this?

Answer 1

Here's one way... compute the differences using the analytic LAG() function, then group by member_id and use the aggregate LAST() function.
NOTE: there may be more than one pair of rows with the same, greatest gap between term_dt and the following eff_dt. You must clarify which row should be selected if that happens. The solution below picks the earliest occurrence (if this happens). If you want the latest occurrence, change MIN to MAX. If you want something else, just say what the requirement is.
with
inputs ( member_id, eff_dt, term_dt ) as (
select 101, to_date('1/1/13', 'mm/dd/yy'), to_date('7/31/14' , 'mm/dd/yy') from dual union all
select 101, to_date('1/1/15', 'mm/dd/yy'), to_date('3/31/15' , 'mm/dd/yy') from dual union all
select 101, to_date('5/1/15', 'mm/dd/yy'), to_date('5/31/15' , 'mm/dd/yy') from dual union all
select 101, to_date('6/1/15', 'mm/dd/yy'), to_date('12/31/15', 'mm/dd/yy') from dual union all
select 101, to_date('1/1/16', 'mm/dd/yy'), to_date('12/31/16', 'mm/dd/yy') from dual
)
-- End of simulated inputs (for testing only, not part of the solution).
-- Use your actual table and column names in the SQL query below.
select member_id,
min(eff_dt) keep (dense_rank last order by diff nulls first) as eff_dt,
min(term_dt) keep (dense_rank last order by diff nulls first) as term_dt
from (
select member_id, eff_dt, term_dt,
eff_dt - lag(term_dt) over (partition by member_id order by eff_dt) as diff
from inputs
)
group by member_id
;
MEMBER_ID EFF_DT TERM_DT
--------- ------------------- -------------------
101 2015-01-01 00:00:00 2015-03-31 00:00:00