README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0100-0199/0197.Rising%20Temperature/README_EN.md	Database

197. Rising Temperature

中文文档

Description

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).

Solutions

Solution 1: Self-Join + DATEDIFF/SUBDATE Function

We can use self-join to compare each row in the Weather table with its previous row. If the temperature is higher and the date difference is one day, then it is the result we are looking for.

Python3

import pandas as pd


def rising_temperature(weather: pd.DataFrame) -> pd.DataFrame:
    weather.sort_values(by="recordDate", inplace=True)
    return weather[
        (weather.temperature.diff() > 0) & (weather.recordDate.diff().dt.days == 1)
    ][["id"]]

MySQL

# Write your MySQL query statement below
SELECT w1.id
FROM
    Weather AS w1
    JOIN Weather AS w2
        ON DATEDIFF(w1.recordDate, w2.recordDate) = 1 AND w1.temperature > w2.temperature;

Solution 2

MySQL

# Write your MySQL query statement below
SELECT w1.id
FROM
    Weather AS w1
    JOIN Weather AS w2
        ON SUBDATE(w1.recordDate, 1) = w2.recordDate AND w1.temperature > w2.temperature;