How hard is LeetCode 1355. Activity Participants?

LeetCode 1355. Activity Participants is rated Medium on LeetCode.

What topics does LeetCode 1355. Activity Participants cover?

LeetCode 1355. Activity Participants is tagged Database on LeetCode.

Is LeetCode 1355. Activity Participants a premium problem?

Yes. LeetCode 1355. Activity Participants is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.

MediumNo. 1355LeetCode Premium

Activity Participants — LeetCode 1355 Python Solution

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

Table schema

SQLTable schema

Table: Friends +---------------+---------+ | Column Name | Type | +---------------+---------+ | id | int | | name | varchar | | activity | varchar | +---------------+---------+ id is the id of the friend and the primary key for this table in SQL. name is the name of the friend.

Example

SQLExample

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| name          | varchar |
| activity      | varchar |
+---------------+---------+
id is the id of the friend and the primary key for this table in SQL.
name is the name of the friend.
activity is the name of the activity which the friend takes part in.

Python solution

PythonSolution

import duckdb
import pandas as pd

def solution(friends: pd.DataFrame, activities: pd.DataFrame) -> pd.DataFrame:
    con = duckdb.connect()
    con.register("Friends", friends)
    con.register("Activities", activities)
    return con.execute("""WITH
    t AS (
        SELECT activity, COUNT(1) AS cnt
        FROM Friends
        GROUP BY activity
    )
SELECT activity
FROM t
WHERE cnt > (SELECT MIN(cnt) FROM t) AND cnt < (SELECT MAX(cnt) FROM t);""").df()

Complexity

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

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Frequently asked questions

How hard is LeetCode 1355. Activity Participants?: LeetCode 1355. Activity Participants is rated Medium on LeetCode.
What topics does LeetCode 1355. Activity Participants cover?: LeetCode 1355. Activity Participants is tagged Database on LeetCode.
Is LeetCode 1355. Activity Participants a premium problem?: Yes. LeetCode 1355. Activity Participants 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.

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Transcribing...

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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."

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