# Time:  O(n^2)
# Space: O(n)
#
# Given a Weather table, write a SQL query to find all dates' 
# Ids with higher temperature compared to its previous (yesterday's) dates.
# 
# +---------+------------+------------------+
# | Id(INT) | Date(DATE) | Temperature(INT) |
# +---------+------------+------------------+
# |       1 | 2015-01-01 |               10 |
# |       2 | 2015-01-02 |               25 |
# |       3 | 2015-01-03 |               20 |
# |       4 | 2015-01-04 |               30 |
# +---------+------------+------------------+
# For example, return the following Ids for the above Weather table:
# +----+
# | Id |
# +----+
# |  2 |
# |  4 |
# +----+
#

# Write your MySQL query statement below
SELECT wt1.Id 
FROM Weather wt1, Weather wt2
WHERE wt1.Temperature > wt2.Temperature AND 
      TO_DAYS(wt1.DATE)-TO_DAYS(wt2.DATE)=1;