I think the answer missed the case when A send B a friend request, and B send A a friend request, and both requests got approved. In this case, A or B really just gained one friend. But the answer seems to count this case twice.
Isn't union (remove duplicates) should be used instead of union all?
select id1 as id, count(id2) as num from (select requester_id as id1, accepter_id as id2 from request_accepted union select accepter_id as id1, requester_id as id2 from request_accepted) tmp1 group by id1 order by num desc limit 1
@Fangyyfrank I agree. The solution is wrong. Hope they can fix this soon.
@Fangyyfrank Use "UNION ALL". In Friend Requests I, it says it's possible that one user send multiple requests to another and get accepted multiple times. But you idea to use UNION is correct.
@Fangyyfrank Agree. Even looking through the test cases of OJ, union all is clearly wrong. There is mutual friend request and repeated friend request.
select requester_id as id1, accepter_id as id2
select accepter_id as id1, requester_id as id2
@Fangyyfrank Nice Thinking.
I noticed the note in the problem "The friend request could only been accepted once". Doesn't it mean when the case "A send B a friend request, and B send A a friend request" occurs, only one of A and B can accept it?
Like of course they can send friend request to each other, but once one of them accepts it (it will be recorded in the table) the other person won't be able to accept it again(It won't write this record to the table second time with same id pair). It is just like how Facebook works.
Please correct me if I am wrong, thanks.
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