SQL Server 2019

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Microsoft SQL Server 2019 (RTM-CU6) (KB4563110) - 15.0.4053.23 (X64) 
	Jul 25 2020 11:26:55 
	Copyright (C) 2019 Microsoft Corporation
	Express Edition (64-bit) on Windows Server 2019 Standard 10.0 <X64> (Build 17763: ) (Hypervisor)



CREATE table transactions (
 client_id INTEGER NOT NULL,
 is_cancelled TINYINT,
 date DATE NOT NULL
 );
INSERT INTO transactions (client_id, is_cancelled, date) 
VALUES
(1, 0, '2022-01-01'),
(1, 1, '2022-01-01'),
(1, 1, '2022-01-02'),
(1, 1, '2022-01-04'),
(1, 0, '2022-01-05'),
(1, 1, '2022-01-06'),

(2, 0, '2022-01-01'),
(2, 0, '2022-01-02'),
(2, 0, '2022-01-03'),
(2, 0, '2022-01-04'),
(2, 1, '2022-01-05'),
(2, 1, '2022-01-06'),
(2, 0, '2022-01-07'),
(2, 1, '2022-01-08'),

(3, 1, '2022-01-01'),
(3, 1, '2022-01-02'),
(3, 1, '2022-01-03'),
(3, 0, '2022-01-04'),
(3, 0, '2022-01-05'),
(3, 1, '2022-01-06'),
(3, 1, '2022-01-07'),
(3, 1, '2022-01-08')
22 rows affected

WITH PrevValues AS (
    SELECT
      client_id,
      LAG(is_cancelled) OVER (PARTITION BY client_id ORDER BY date) AS previous_cancelled,
      CASE WHEN is_cancelled = LAG(is_cancelled) OVER (PARTITION BY client_id ORDER BY date)
        THEN NULL ELSE 1 END
        as is_different,
      date,
      is_cancelled
    FROM transactions
),
Grouped AS (
    SELECT 
      client_id,
      date,
      is_different,
      COUNT(is_different) OVER (PARTITION BY client_id ORDER BY date ROWS UNBOUNDED PRECEDING) AS group_id
    FROM PrevValues
),
ByGroups AS (
    SELECT
      client_id,
      COUNT(*) as in_a_row
    FROM Grouped
    GROUP BY
      client_id,
      group_id
    HAVING COUNT(*) >= 3
)
SELECT
  client_id,
  MAX(in_a_row) as max_in_a_row,
  MIN(in_a_row) as min_in_a_row,
  COUNT(*) as num_groups
FROM ByGroups
GROUP BY
  client_id;


client_id
max_in_a_row
min_in_a_row
num_groups


1
3
3
1

2
4
4
1

3
3
3
2


Warning: Null value is eliminated by an aggregate or other SET operation.