LeetCode Medium 1355 IBM Interview SQL Question with Detailed Explanation

In this video I solve and explain a medium difficulty leetcode SQL question using MySQL query. This question has been asked in Apple, Facebook, Amazon, Google, Adobe, Microsoft, Adobe interviews or what we popularly call FAANG interviews.
I explain the related concept as well. This question includes points to keep in mind to develop SQL queries.

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[muddassirnazar]

Simple Solution:with cte as (select activity, count(*) as cnt from friends
group by activity)
select activity from cte
where cnt not in(select max(cnt) from cte)
and cnt not in(select min(cnt) from cte)

[expandingourselves]

with cte as
(select activity, count(id) as cnt
from friends group by 1 )

select activity from cte
where cnt NOT IN ( Select max(cnt) from cte ) and cnt NOT IN ( Select min(cnt) from cte )

[yashnikhare]

with data as (
select *, count(f.id) over(partition by a.activity order by a.id) as cnt
from Friends f join Activities a on f.activity_id=a.id )

select a.activity from data where cnt range between (min(cnt)+1) and (max(cnt)-1)

[vijaygupta]

can i use this Please let me know

with cte as
(
Select activity, count(id) as num_part from Friends
group by activity)
Select activity from cte
where num_part>(select min(num_part) from cte) and num_part<(select max(num_part) from cte)

[mickyman]

WITH activity_counts AS (
SELECT
activity,
COUNT(*) AS count
FROM
Friends
GROUP BY
activity
)
SELECT
activity AS results
FROM
activity_counts
WHERE
count != (SELECT MAX(count) FROM activity_counts)
AND count != (SELECT MIN(count) FROM activity_counts);

[mickyman]

WITH activity_counts AS (
SELECT
activity,
COUNT(*) AS count,
MAX(COUNT(*)) OVER () AS max_count,
MIN(COUNT(*)) OVER () AS min_count
FROM
Friends
GROUP BY
activity
)
SELECT
activity AS results
FROM
activity_counts
WHERE
count != max_count
AND count != min_count;