How hard is LeetCode 2987. Find Expensive Cities?

LeetCode 2987. Find Expensive Cities is rated Easy on LeetCode.

What topics does LeetCode 2987. Find Expensive Cities cover?

LeetCode 2987. Find Expensive Cities is tagged Database on LeetCode.

Is LeetCode 2987. Find Expensive Cities a premium problem?

Yes. LeetCode 2987. Find Expensive Cities is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

Find Expensive Cities — LeetCode 2987 Python Solution

EasyLeetCode PremiumDatabase

Problem: #2987
Reading time: 2 min
Source: leetcode.com

Table schema

SQL

Table: Listings +-------------+---------+ | Column Name | Type | +-------------+---------+ | listing_id | int | | city | varchar | | price | int | +-------------+---------+ listing_id is column of unique values for this table. This table contains listing_id, city, and price.

Example

SQL

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| listing_id  | int     |
| city        | varchar |
| price       | int     |
+-------------+---------+
listing_id is column of unique values for this table.
This table contains listing_id, city, and price.

Python solution

Python

import duckdb
import pandas as pd

def solution(listings: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Listings", listings)
    return con.execute("""SELECT city
FROM Listings
GROUP BY city
HAVING AVG(price) > (SELECT AVG(price) FROM Listings)
ORDER BY 1;""").df()

Complexity

Measure	Complexity
Time	O(n log n) (typical)
Space	O(n) auxiliary

Frequently asked questions

How hard is LeetCode 2987. Find Expensive Cities?: LeetCode 2987. Find Expensive Cities is rated Easy on LeetCode.
What topics does LeetCode 2987. Find Expensive Cities cover?: LeetCode 2987. Find Expensive Cities is tagged Database on LeetCode.
Is LeetCode 2987. Find Expensive Cities a premium problem?: Yes. LeetCode 2987. Find Expensive Cities is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.