Find longest consecutive time interval per object - mysql

Question

There are Merchants and they can submit Claims.
I need to find the longest time period during which a Merchant had at least 1 claim. So a time period (in fractions of a day, whatever) per merchant_id.
So, for example:
+-------------+-----------+----------------------+----------------------+
| merchant_id | claim_id | from | to |
+-------------+-----------+----------------------+----------------------+
| 1 | 11 | 2016-08-15 12:00:00 | 2016-08-17 12:00:00 |
| 1 | 22 | 2016-08-16 12:00:00 | 2016-08-18 12:00:00 |
| 1 | 33 | 2016-08-19 12:00:00 | 2016-08-20 12:00:00 |
| 2 | 66 | 2016-08-15 12:00:00 | 2016-08-17 12:00:00 |
| 2 | 67 | 2016-08-18 12:00:00 | 2016-08-19 12:00:00 |
+-------------+-----------+----------------------+----------------------+
For merchant_id = 1 it would be 3 days.
For merchant_id = 2 it would be 2 days.
How do I do that?

Answer 1

Doing this alone in MySQL is really complex. I've tried for a particular merchant_id. I am not still sure if this is 100% right without checking for different set of inputs.
But you can give this a try and later I can explain the logic behind.
SELECT
firstTable.merchant_id,
MAX(TIMESTAMPDIFF(DAY,firstTable.from,secondTable.to)) AS maxConsecutiveDays
FROM
(
SELECT
A.merchant_id,
A.from,
#rn1 := #rn1 + 1 AS row_number
FROM merchants A
CROSS JOIN (SELECT #rn1 := 0) var
WHERE A.merchant_id = 2
AND NOT EXISTS (
SELECT 1 FROM merchants B WHERE B.merchant_id = A.merchant_id AND A.idt <> B.idt AND A.`from` BETWEEN B.from AND B.to
)
ORDER BY A.from
) AS firstTable
INNER JOIN (
SELECT
A.merchant_id,
A.to,
#rn2 := #rn2 + 1 AS row_number
FROM merchants A
CROSS JOIN (SELECT #rn2 := 0) var
WHERE A.merchant_id = 2
AND NOT EXISTS (
SELECT 1 FROM merchants B WHERE B.merchant_id = A.merchant_id AND A.idt <> B.idt AND A.to BETWEEN B.from AND B.to
)
ORDER BY A.to
) AS secondTable
ON firstTable.row_number = secondTable.row_number;
WORKING DEMO
Algorithm:
Let's consider the following steps for a particular merchant_id
First find all the start points which are not inside in any of the
ranges. I call this independent start points. Let's say these start
points are stored in a set S.
Second now find all the end points which are not inside in any of
the ranges. These are independent end points and are stored in a set
E.
Sort the sets in ascending order of time.
Now give a rank to every element of a set starting from 1.
Join these two sets on matching rank number.
Now enumerate the two sets simultaneously and get the difference in
days. And later find the maximum of this difference.
The last step can be illustrated by the following code snippet:
int maxDiff = 0;
for(int i=0; i< E.size(); i++){
if((E.get(i) - S.get(i) > maxDiff){
maxDiff = E.get(i) - S.get(i);
}
}
And maxDiff is your output;
EDIT:
In order to get longest consecutive days for each merchant check this DEMO