Last updated on October 9th, 2024 at 10:30 pm
This Leetcode problem Friend Requests I Overall Acceptance Rate LeetCode Solution is done in SQL.
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Level of Question
Easy
Friend Requests I Overall Acceptance Rate LeetCode Solution
Table of Contents
Problem Statement
In social network like Facebook or Twitter, people send friend requests and accept others’ requests as well. Now given two tables as below:
sender_id | send_to_id | request_date |
---|---|---|
1 | 2 | 2016-06-01 |
1 | 3 | 2016-06-01 |
1 | 4 | 2016-06-01 |
2 | 3 | 2016-06-02 |
3 | 4 | 2016-06-09 |
requester_id | accepter_id | accept_date |
---|---|---|
1 | 2 | 2016-06-03 |
1 | 3 | 2016-06-08 |
2 | 3 | 2016-06-08 |
3 | 4 | 2016-06-09 |
3 | 4 | 2016-06-10 |
Write a query to find the overall acceptance rate of requests rounded to 2 decimals, which is the number of acceptance divide the number of requests.
For the sample data above, your query should return the following result.
accept_rate |
0.80 |
Note: The accepted requests are not necessarily from the table friend_request. In this case, you just need to simply count the total accepted requests (no matter whether they are in the original requests), and divide it by the number of requests to get the acceptance rate.
It is possible that a sender sends multiple requests to the same receiver, and a request could be accepted more than once. In this case, the ‘duplicated’ requests or acceptances are only counted once.
If there is no requests at all, you should return 0.00 as the accept_rate.
Explanation: There are 4 unique accepted requests, and there are 5 requests in
total. So the rate is 0.80.
1. Friend Requests I Overall Acceptance Rate LeetCode Solution MySQL
select round( if(requests = 0, 0, accepts / requests), 2 ) as accept_rate from ( select count(distinct sender_id, send_to_id) as requests from friend_request ) as r, ( select count( distinct requester_id, accepter_id ) as accepts from request_accepted ) as a;