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LeetCode-Solutions/MySQL/find-active-users.sql at master · thebluntcoder/LeetCode-Solutions · GitHub
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find-active-users.sql
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MySQL
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find-active-users.sql
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#
Time: O(nlogn)
#
Space: O(n)
WITH active_users_cte
AS
(
SELECT
user_id,
IF (DATEDIFF(created_at, LAG(created_at,
1
) OVER (PARTITION BY user_id
ORDER BY
created_at))
<=
7
,
1
,
0
)
AS
active
FROM
users
)
SELECT DISTINCT
user_id
FROM
active_users_cte
WHERE
active
=
1
;
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