How hard is LeetCode 1159. Market Analysis II?

LeetCode 1159. Market Analysis II is rated Hard on LeetCode.

What topics does LeetCode 1159. Market Analysis II cover?

LeetCode 1159. Market Analysis II is tagged Database on LeetCode.

Is LeetCode 1159. Market Analysis II a premium problem?

Yes. LeetCode 1159. Market Analysis II is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

HardNo. 1159LeetCode Premium

Market Analysis II — LeetCode 1159 Python Solution

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

Table schema

SQLTable schema

Table: Users +----------------+---------+ | Column Name | Type | +----------------+---------+ | user_id | int | | join_date | date | | favorite_brand | varchar | +----------------+---------+ user_id is the primary key (column with unique values) of this table. This table has the info of the users of an online shopping website where users can sell and buy items.

Example

SQLExample

+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| user_id        | int     |
| join_date      | date    |
| favorite_brand | varchar |
+----------------+---------+
user_id is the primary key (column with unique values) of this table.
This table has the info of the users of an online shopping website where users can sell and buy items.

Complexity

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

Related problems
No.	Problem	Difficulty
0175	Combine Two Tables	Easy
0176	Second Highest Salary	Medium
0177	Nth Highest Salary	Medium
0178	Rank Scores	Medium
0180	Consecutive Numbers	Medium
0181	Employees Earning More Than Their Managers	Easy

Frequently asked questions

How hard is LeetCode 1159. Market Analysis II?: LeetCode 1159. Market Analysis II is rated Hard on LeetCode.
What topics does LeetCode 1159. Market Analysis II cover?: LeetCode 1159. Market Analysis II is tagged Database on LeetCode.
Is LeetCode 1159. Market Analysis II a premium problem?: Yes. LeetCode 1159. Market Analysis II 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

Complexity

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Frequently asked questions

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Approach

Approach

Approach