Leetcode #1917: Leetcodify Friends Recommendations
In this guide, we solve Leetcode #1917 Leetcodify Friends Recommendations in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Table: Listens +-------------+---------+ | Column Name | Type | +-------------+---------+ | user_id | int | | song_id | int | | day | date | +-------------+---------+ This table may contain duplicates (In other words, there is no primary key for this table in SQL). Each row of this table indicates that the user user_id listened to the song song_id on the day day.
Quick Facts
- Difficulty: Hard
- Premium: Yes
- Tags: Database
Intuition
The task is relational in nature, which maps cleanly to DataFrame operations in Python.
By treating tables as DataFrames, joins and group-bys become concise and readable.
Approach
Load the inputs as DataFrames and apply the appropriate merge, filter, or group-by.
Select or rename the columns to match the required output.
Steps:
- Load inputs as DataFrames.
- Apply merge/groupby/filter operations.
- Select the output columns.
Example
+-------------+---------+
| Column Name | Type |
+-------------+---------+
| user_id | int |
| song_id | int |
| day | date |
+-------------+---------+
This table may contain duplicates (In other words, there is no primary key for this table in SQL).
Each row of this table indicates that the user user_id listened to the song song_id on the day day.
Python Solution
import duckdb
import pandas as pd
def solution(listens: pd.DataFrame, friendship: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Listens", listens)
con.register("Friendship", friendship)
return con.execute("""WITH
T AS (
SELECT user1_id, user2_id FROM Friendship
UNION
SELECT user2_id AS user1_id, user1_id AS user2_id FROM Friendship
)
SELECT DISTINCT l1.user_id, l2.user_id AS recommended_id
FROM
Listens AS l1,
Listens AS l2
WHERE
l1.day = l2.day
AND l1.song_id = l2.song_id
AND l1.user_id != l2.user_id
AND NOT EXISTS (
SELECT 1
FROM T AS t
WHERE l1.user_id = t.user1_id AND l2.user_id = t.user2_id
)
GROUP BY l1.day, l1.user_id, l2.user_id
HAVING COUNT(DISTINCT l1.song_id) >= 3;""").df()
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.