How hard is LeetCode 1459. Rectangles Area?

LeetCode 1459. Rectangles Area is rated Medium on LeetCode.

What topics does LeetCode 1459. Rectangles Area cover?

LeetCode 1459. Rectangles Area is tagged Database on LeetCode.

Is LeetCode 1459. Rectangles Area a premium problem?

Yes. LeetCode 1459. Rectangles Area is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 1459LeetCode Premium

Rectangles Area — LeetCode 1459 Python Solution

Topics: Database
Reading time: 3 min
Source: leetcode.com

Table schema

SQLTable schema

Table: Points +---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | x_value | int | | y_value | int | +---------------+---------+ id is the column with unique values for this table. Each point is represented as a 2D coordinate (x_value, y_value).

Example

SQLExample

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| x_value       | int     |
| y_value       | int     |
+---------------+---------+
id is the column with unique values for this table.
Each point is represented as a 2D coordinate (x_value, y_value).

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(points: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Points", points)
    return con.execute("""SELECT
    p1.id AS p1,
    p2.id AS p2,
    ABS(p1.x_value - p2.x_value) * ABS(p1.y_value - p2.y_value) AS area
FROM
    Points AS p1
    JOIN Points AS p2 ON p1.id < p2.id
WHERE p1.x_value != p2.x_value AND p1.y_value != p2.y_value
ORDER BY area DESC, p1, p2;""").df()

Complexity

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

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Frequently asked questions

How hard is LeetCode 1459. Rectangles Area?: LeetCode 1459. Rectangles Area is rated Medium on LeetCode.
What topics does LeetCode 1459. Rectangles Area cover?: LeetCode 1459. Rectangles Area is tagged Database on LeetCode.
Is LeetCode 1459. Rectangles Area a premium problem?: Yes. LeetCode 1459. Rectangles Area 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

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Frequently asked questions

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Approach

Approach

Approach