Find overlapping range in PL/SQL - sql

Question

Sample data below
id start end
a 1 3
a 5 6
a 8 9
b 2 4
b 6 7
b 9 10
c 2 4
c 6 7
c 9 10
I'm trying to come up with a query that will return all the overlap start-end inclusive between a, b, and c (but extendable to more). So the expected data will look like the following
start end
2 3
6 6
9 9
The only way I can picture this is with a custom aggregate function that tracks the current valid intervals then computes the new intervals during the iterate phase. However I can't see this approach being practical when working with large datasets. So if some bright mind out there have a query or some innate function that I'm not aware of I would greatly appreciate the help.

Answer 1

You can do this using aggregation and a join. Assuming no internal overlaps for "a" and "b":
select greatest(ta.start, tb.start) as start,
least(ta.end, tb.end) as end
from t ta join
t tb
on ta.start <= tb.end and ta.end >= tb.start and
ta.id = 'a' and tb.id = 'b';

Answer 2

This is a lot uglier and more complex than Gordon's solution, but I think it gives the expected answer better and should extend to work with more ids:
WITH NUMS(N) AS ( --GENERATE NUMBERS N FROM THE SMALLEST START VALUE TO THE LARGEST END VALUE
SELECT MIN("START") N FROM T
UNION ALL
SELECT N+1 FROM NUMS WHERE N < (SELECT MAX("END") FROM T)
),
SEQS(N,START_RANK,END_RANK) AS (
SELECT N,
CASE WHEN IS_START=1 THEN ROW_NUMBER() OVER (PARTITION BY IS_START ORDER BY N) ELSE 0 END START_RANK, --ASSIGN A RANK TO EACH RANGE START
CASE WHEN IS_END=1 THEN ROW_NUMBER() OVER (PARTITION BY IS_END ORDER BY N) ELSE 0 END END_RANK --ASSIGN A RANK TO EACH RANGE END
FROM (
SELECT N,
CASE WHEN NVL(LAG(N) OVER (ORDER BY N),N) + 1 <> N THEN 1 ELSE 0 END IS_START, --MARK N AS A RANGE START
CASE WHEN NVL(LEAD(N) OVER (ORDER BY N),N) -1 <> N THEN 1 ELSE 0 END IS_END /* MARK N AS A RANGE END */
FROM (
SELECT DISTINCT N FROM ( --GET THE SET OF NUMBERS N THAT ARE INCLUDED IN ALL ID RANGES
SELECT NUMS.*,T.*,COUNT(*) OVER (PARTITION BY N) N_CNT,COUNT(DISTINCT "ID") OVER () ID_CNT
FROM NUMS
JOIN T ON (NUMS.N >= T."START" AND NUMS.N <= T."END")
) WHERE N_CNT=ID_CNT
)
) WHERE IS_START + IS_END > 0
)
SELECT STARTS.N "START",ENDS.N "END" FROM SEQS STARTS
JOIN SEQS ENDS ON (STARTS.START_RANK=ENDS.END_RANK AND STARTS.N <= ENDS.N) ORDER BY "START"; --MATCH CORRESPONDING RANGE START/END VALUES
First we generate all the numbers between the smallest start value and the largest end value.
Then we find the numbers that are included in all the provided "id" ranges by joining our generated numbers to the ranges, and selecting each number "n" that appears once for each "id".
Then we determine whether each of these values "n" starts or ends a range. To determine that, for each N we say:
If the previous value of N does not exist or is not 1 less than current N, current N starts a range. If the next value of N does not exist or is not 1 greater than current N, current N ends a range.
Next, we assign a "rank" to each start and end value so we can match them up.
Finally, we self-join where the ranks match (and where the start <= the end) to get our result.
EDIT: After some searching, I came across this question which shows a better way to find the start/ends and refactored the query to:
WITH NUMS(N) AS ( --GENERATE NUMBERS N FROM THE SMALLEST START VALUE TO THE LARGEST END VALUE
SELECT MIN("START") N FROM T
UNION ALL
SELECT N+1 FROM NUMS WHERE N < (SELECT MAX("END") FROM T)
)
SELECT MIN(N) "START",MAX(N) "END" FROM (
SELECT N,ROW_NUMBER() OVER (ORDER BY N)-N GRP_ID
FROM (
SELECT DISTINCT N FROM ( --GET THE SET OF NUMBERS N THAT ARE INCLUDED IN ALL ID RANGES
SELECT NUMS.*,T.*,COUNT(*) OVER (PARTITION BY N) N_CNT,COUNT(DISTINCT "ID") OVER () ID_CNT
FROM NUMS
JOIN T ON (NUMS.N >= T."START" AND NUMS.N <= T."END")
) WHERE N_CNT=ID_CNT
)
)
GROUP BY GRP_ID ORDER BY "START";