EasyNo. 2669LeetCode Premium
Count Artist Occurrences On Spotify Ranking List — LeetCode 2669 Python Solution
- Topics
- Database
- Reading time
- 3 min
- Source
- leetcode.com
Table schema
Table: Spotify +-------------+---------+ | Column Name | Type | +-------------+---------+ | id | int | | track_name | varchar | | artist | varchar | +-------------+---------+ id is the primary key (column with unique values) for this table. Each row contains an id, track_name, and artist.Example
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| id | int |
| track_name | varchar |
| artist | varchar |
+-------------+---------+
id is the primary key (column with unique values) for this table.
Each row contains an id, track_name, and artist.Python solution
import duckdb
import pandas as pd
# Pass input tables as keyword arguments matching the SQL table names.
def solution(**tables) -> pd.DataFrame:
con = duckdb.connect()
for name, df in tables.items():
con.register(name, df)
return con.execute("""SELECT
artist,
COUNT(1) AS occurrences
FROM Spotify
GROUP BY artist
ORDER BY occurrences DESC, artist;""").df()Complexity
- Time
- O(n log n)(typical)
- Space · auxiliary
- O(n)
Related problems
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|---|---|---|
| 0175 | Combine Two Tables | Easy |
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| 0177 | Nth Highest Salary | Medium |
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| 0180 | Consecutive Numbers | Medium |
| 0181 | Employees Earning More Than Their Managers | Easy |
Frequently asked questions
- How hard is LeetCode 2669. Count Artist Occurrences On Spotify Ranking List?
LeetCode 2669. Count Artist Occurrences On Spotify Ranking List is rated Easy on LeetCode.
- What topics does LeetCode 2669. Count Artist Occurrences On Spotify Ranking List cover?
LeetCode 2669. Count Artist Occurrences On Spotify Ranking List is tagged Database on LeetCode.
- Is LeetCode 2669. Count Artist Occurrences On Spotify Ranking List a premium problem?
Yes. LeetCode 2669. Count Artist Occurrences On Spotify Ranking List is a LeetCode Premium problem, so the full statement and test cases require a paid LeetCode subscription.