Leetcode #1811: Find Interview Candidates
In this guide, we solve Leetcode #1811 Find Interview Candidates 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: Contests +--------------+------+ | Column Name | Type | +--------------+------+ | contest_id | int | | gold_medal | int | | silver_medal | int | | bronze_medal | int | +--------------+------+ contest_id is the column with unique values for this table. This table contains the LeetCode contest ID and the user IDs of the gold, silver, and bronze medalists.
Quick Facts
- Difficulty: Medium
- 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 |
+--------------+------+
| contest_id | int |
| gold_medal | int |
| silver_medal | int |
| bronze_medal | int |
+--------------+------+
contest_id is the column with unique values for this table.
This table contains the LeetCode contest ID and the user IDs of the gold, silver, and bronze medalists.
It is guaranteed that any consecutive contests have consecutive IDs and that no ID is skipped.
Python Solution
import duckdb
import pandas as pd
def solution(contests: pd.DataFrame, users: pd.DataFrame) -> pd.DataFrame:
con = duckdb.connect()
con.register("Contests", contests)
con.register("Users", users)
return con.execute("""WITH
S AS (
SELECT contest_id, gold_medal AS user_id, 1 AS type
FROM Contests
UNION
SELECT contest_id, silver_medal AS user_id, 2 AS type
FROM Contests
UNION
SELECT contest_id, bronze_medal AS user_id, 3 AS type
FROM Contests
),
T AS (
SELECT
user_id,
(
contest_id - ROW_NUMBER() OVER (
PARTITION BY user_id
ORDER BY contest_id
)
) AS diff
FROM S
),
P AS (
SELECT user_id
FROM S
WHERE type = 1
GROUP BY user_id
HAVING COUNT(1) >= 3
UNION
SELECT DISTINCT user_id
FROM T
GROUP BY user_id, diff
HAVING COUNT(1) >= 3
)
SELECT name, mail
FROM
P AS p
LEFT JOIN Users AS u ON p.user_id = u.user_id;""").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.