LeetCode //SQL - 197. Rising Temperature

197. Rising Temperature

Table: Weather

±--------------±--------+
| Column Name | Type |
±--------------±--------+
| id | int |
| recordDate | date |
| temperature | int |
±--------------±--------+
id is the column with unique values for this table.
There are no different rows with the same recordDate.
This table contains information about the temperature on a certain day.

Write a solution to find all dates’ Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The result format is in the following example.
 

Example 1:

Input:
Weather table:
±—±-----------±------------+
| id | recordDate | temperature |
±—±-----------±------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
±—±-----------±------------+
Output:
±—+
| id |
±—+
| 2 |
| 4 |
±—+
Explanation:
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

From: LeetCode
Link: 197. Rising Temperature

Solution:

Ideas:

1. Self Join: Join the Weather table with itself using aliases w1 and w2.
2. Date Comparison: Ensure w1.recordDate is exactly one day after w2.recordDate using DATE_ADD.
3. Temperature Comparison: Compare the temperatures and select the id where the temperature of the current day (w1.temperature) is greater than the temperature of the previous day (w2.temperature).

Code:

SELECT w1.id
FROM Weather w1
JOIN Weather w2
ON w1.recordDate = DATE_ADD(w2.recordDate, INTERVAL 1 DAY)
WHERE w1.temperature > w2.temperature;