SQL 문제풀이
leetcode - 1407. Top Travellers
Jerrytwo
2022. 7. 11. 14:31
난이도 : Easy
Table: Users
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| name | varchar |
+---------------+---------+
id is the primary key for this table.
name is the name of the user.
Table: Rides
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| user_id | int |
| distance | int |
+---------------+---------+
id is the primary key for this table.
user_id is the id of the user who traveled the distance "distance".
Write an SQL query to report the distance traveled by each user.
Return the result table ordered by travelled_distance in descending order, if two or more users traveled the same distance, order them by their name in ascending order.
The query result format is in the following example.
Example 1:
Input:
Users table:
+------+-----------+
| id | name |
+------+-----------+
| 1 | Alice |
| 2 | Bob |
| 3 | Alex |
| 4 | Donald |
| 7 | Lee |
| 13 | Jonathan |
| 19 | Elvis |
+------+-----------+
Rides table:
+------+----------+----------+
| id | user_id | distance |
+------+----------+----------+
| 1 | 1 | 120 |
| 2 | 2 | 317 |
| 3 | 3 | 222 |
| 4 | 7 | 100 |
| 5 | 13 | 312 |
| 6 | 19 | 50 |
| 7 | 7 | 120 |
| 8 | 19 | 400 |
| 9 | 7 | 230 |
+------+----------+----------+
Output:
+----------+--------------------+
| name | travelled_distance |
+----------+--------------------+
| Elvis | 450 |
| Lee | 450 |
| Bob | 317 |
| Jonathan | 312 |
| Alex | 222 |
| Alice | 120 |
| Donald | 0 |
+----------+--------------------+
Explanation:
Elvis and Lee traveled 450 miles, Elvis is the top traveler as his name is alphabetically smaller than Lee.
Bob, Jonathan, Alex, and Alice have only one ride and we just order them by the total distances of the ride.
Donald did not have any rides, the distance traveled by him is 0.
SELECT name
, Coalesce(dst, 0) travelled_distance
FROM Users
LEFT JOIN (
SELECT user_id
, SUM(distance) dst
FROM Rides
GROUP BY user_id
) sub ON Users.id = sub.user_id
ORDER BY travelled_distance DESC, name
Accepted (23.98%)
SELECT name
, IFNULL(SUM(distance), 0) travelled_distance
-- SUM(IFNULL(distance, 0)) >> 이것도 가능
FROM Users
LEFT JOIN Rides ON Users.id = Rides.user_id
GROUP BY user_id
ORDER BY travelled_distance DESC, name
Accepted (35.98%)