[leetcode][Database][Hard]1972. First and Last Call On the Same Day

題目

Table: Calls

+--------------+----------+
| Column Name  | Type     |
+--------------+----------+
| caller_id    | int      |
| recipient_id | int      |
| call_time    | datetime |
+--------------+----------+
(caller_id, recipient_id, call_time) is the primary key for this table.
Each row contains information about the time of a phone call between caller_id and recipient_id.

Write an SQL query to report the IDs of the users whose first and last calls on any day were with the same person. Calls are counted regardless of being the caller or the recipient.

Return the result table in any order.

SQL Schema

Create table If Not Exists Calls (caller_id int, recipient_id int, call_time datetime)
Truncate table Calls
insert into Calls (caller_id, recipient_id, call_time) values ('8', '4', '2021-08-24 17:46:07')
insert into Calls (caller_id, recipient_id, call_time) values ('4', '8', '2021-08-24 19:57:13')
insert into Calls (caller_id, recipient_id, call_time) values ('5', '1', '2021-08-11 05:28:44')
insert into Calls (caller_id, recipient_id, call_time) values ('8', '3', '2021-08-17 04:04:15')
insert into Calls (caller_id, recipient_id, call_time) values ('11', '3', '2021-08-17 13:07:00')
insert into Calls (caller_id, recipient_id, call_time) values ('8', '11', '2021-08-17 22:22:22')

解題思考

• 建立 user_calls的 with clause，以提供最後輸出的表格使用。
  user_calls將每筆 calls 中的通話紀錄拆分成兩筆 record ，即該筆通話紀錄的兩位 user 分別以自己的角度，紀錄該筆通話的時間以及通話的對象
  A ← communicate → B : A → B and B → A
• 建立 rank_calls 的 with clause ，對 user_calls 中的通話紀錄進行排序。
  對每個 user_id 的所有 call_time 通話時間做 升序 和 降序，以便找出第一筆通話 first call 和最後一筆通話 last call 。
• 對 rank_calls 每個 user_id 的 first call 和 last call 進行統計，若不重複的通話對象只有一位，則可以找出 first call 和 last call 都是同一人的 user_id

解決方案

with
user_calls as (
    select caller_id as user_id, call_time, recipient_id from calls
    union
    select recipient_id as user_id, call_time, caller_id as recipient_id from calls
),
rank_calls as (
    select
        user_id, recipient_id,
        date(call_time) as day,
        dense_rank() over(partition by user_id, date(call_time) order by call_time asc) as rn,
        dense_rank() over(partition by user_id, date(call_time) order by call_time desc) as rk
    from user_calls
)

select distinct
    user_id
from rank_calls
where rn=1 or rk=1
group by user_id, day
having count(distinct recipient_id) = 1