LeetCode Medium 1308 Interview SQL Question with Detailed Explanation | Practice SQL

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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[saisushanthdurvasulabitsian]

difference between ROWS and RANGE :
ROWS specifies the number of rows that precede or follow the current row and
RANGE specifies the range of values wrt the current row

[PrithaMajumder]

Sir, Thanks a lot for This Amazing Video 😊
Problem - 49 Solved ✅ in Practice SQL - LeetCode 👩🏻‍💻

[sohailshaikh]

If we just use ORDER BY a.day, a. gender inside the OVER clause... We will get the cumulative sum in each row.

Tried this and it worked without using 'ROWS BETWEEN....'

Tested this with MAX() window function... By default, OVER clause selects only preceding rows when ORDER BY is used in it.

[KathirVel-nbkf]

SELECT
gender,
day,
SUM(score_points) OVER (
PARTITION BY gender order by gender, day
) AS total
FROM Scores; this query works correctly sir

[fazilhussain]

because combination of primary, day is a primary key, so we don't need to add distinct gender

[chargeurself]

select
gender, day,
sum(score_points) over(partition by gender order by day) as total,
sum(score_points) over(PARTITION BY gender ) as total_overall
from scores
order by gender

Output:
gender, day, total, total_overall
F, 2019-12-30, 17, 80
F, 2019-12-31, 40, 80
F, 2020-01-01, 57, 80
F, 2020-01-07, 80, 80
M, 2019-12-18, 2, 36
M, 2019-12-25, 13, 36
M, 2019-12-30, 26, 36
M, 2019-12-31, 29, 36
M, 2020-01-07, 36, 36
###
if u don't give order by it will give u cumulative sum for that window else it will give overall sum for the specified window(partition by)

[divyanshusinghchauhan]

select *, sum(score_points) over(partition by gender order by day) as totalPoint from competition_scores;

[-Joshna]

with cte as
(select *, sum(score_points) over(partition by gender order by day)
as prev_col from Scores group by gender, day
order by gender, day)
select gender, day, prev_col as total from cte;

[ushasr]

select gender, day, sum(total) over (partition by gender order by gender, day) as total
from scores
This would also work! Thanks for great videos sir