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.

EasyNo. 2987LeetCode Premium

Find Expensive Cities — LeetCode 2987 Python Solution

Topics: Database
Reading time: 2 min
Source: leetcode.com

Table schema

SQLTable schema

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

SQLExample

+-------------+---------+
| 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

PythonSolution

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

Time: O(n log n)(typical)
Space · auxiliary: O(n)

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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.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

Table schema

Example

Python solution

Complexity

Related problems

Frequently asked questions

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Approach

Approach

Approach