Rising Temperature LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Rising Temperature in C++, Java & Python-LeetCode problems. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemRising Temperature – LeetCode Problem

Rising Temperature– LeetCode Problem

Problem:

Table: Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
id is the primary key for this table.
This table contains information about the temperature on a certain day.

Write an SQL query to find all dates’ Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The query 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).
Rising Temperature– LeetCode Solutions
Rising Temperature  Solution in SQL:

First Problem solution.

select id from
(
SELECT id,
case
when lag(Temperature) over(order by recordDate) < Temperature
and lag(recordDate) over (order by RecordDate) = recordDate - interval '1' day
THEN 1
ELSE 0 END SEL
FROM weather
)x
where sel = 1

Second Problem solution.

SELECT Id FROM
(
SELECT Id, (Temperature - LAG(Temperature) OVER (ORDER BY RecordDate)) as temp_diff,
       DATEDIFF(RecordDate, LAG(RecordDate) OVER (ORDER BY RecordDate)) as date_diff
FROM Weather
) as innerQ
WHERE temp_diff > 0 and date_diff = 1

Third Problem solution.

SELECT w.id FROM Weather w JOIN Weather p
ON w.recordDate = DATE_ADD(p.recordDate, INTERVAL 1 DAY) AND w.Temperature > p.Temperature

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