Movie Rating solution and explanation
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The problem
Table: Movies
+---------------+---------+ | Column Name | Type | +---------------+---------+ | movie_id | int | | title | varchar | +---------------+---------+ movie_id is the primary key (column with unique values) for this table. title is the name of the movie.
Table: Users
+---------------+---------+ | Column Name | Type | +---------------+---------+ | user_id | int | | name | varchar | +---------------+---------+ user_id is the primary key (column with unique values) for this table. The column 'name' has unique values.
Table: MovieRating
+---------------+---------+ | Column Name | Type | +---------------+---------+ | movie_id | int | | user_id | int | | rating | int | | created_at | date | +---------------+---------+ (movie_id, user_id) is the primary key (column with unique values) for this table. This table contains the rating of a movie by a user in their review. created_at is the user's review date.
Write a solution to:
- Find the name of the user who has rated the greatest number of movies. In case of a tie, return the lexicographically smaller user name.
- Find the movie name with the highest average rating in
February 2020. In case of a tie, return the lexicographically smaller movie name.
The result format is in the following example.
Example 1:
Input:
Movies table:
+-------------+--------------+
| movie_id | title |
+-------------+--------------+
| 1 | Avengers |
| 2 | Frozen 2 |
| 3 | Joker |
+-------------+--------------+
Users table:
+-------------+--------------+
| user_id | name |
+-------------+--------------+
| 1 | Daniel |
| 2 | Monica |
| 3 | Maria |
| 4 | James |
+-------------+--------------+
MovieRating table:
+-------------+--------------+--------------+-------------+
| movie_id | user_id | rating | created_at |
+-------------+--------------+--------------+-------------+
| 1 | 1 | 3 | 2020-01-12 |
| 1 | 2 | 4 | 2020-02-11 |
| 1 | 3 | 2 | 2020-02-12 |
| 1 | 4 | 1 | 2020-01-01 |
| 2 | 1 | 5 | 2020-02-17 |
| 2 | 2 | 2 | 2020-02-01 |
| 2 | 3 | 2 | 2020-03-01 |
| 3 | 1 | 3 | 2020-02-22 |
| 3 | 2 | 4 | 2020-02-25 |
+-------------+--------------+--------------+-------------+
Output:
+--------------+
| results |
+--------------+
| Daniel |
| Frozen 2 |
+--------------+
Explanation:
Daniel and Monica have rated 3 movies ("Avengers", "Frozen 2" and "Joker") but Daniel is smaller lexicographically.
Frozen 2 and Joker have a rating average of 3.5 in February but Frozen 2 is smaller lexicographically.
Approach overview
Problem Overview: You are given three tables: Movies, Users, and MovieRating. The task has two parts: find the user who submitted the most ratings, and find the movie with the highest average rating during February 2020. If multiple results tie, return the lexicographically smallest name or title.
Approach 1: SQL Grouping and Ordering (O(n log n) time, O(n) space)
This approach relies on SQL aggregation. First, group ratings by user_id and count how many ratings each user submitted. Join with the Users table to access names, then sort by rating count descending and name ascending to break ties. For the movie result, filter ratings within February 2020, compute AVG(rating) grouped by movie_id, join with Movies, and order by average rating descending and title ascending. Sorting introduces O(n log n) time complexity, while grouped results require O(n) intermediate storage.
Approach 2: Direct SQL with Subquery Utilization (O(n log n) time, O(n) space)
This version pushes more logic into subqueries. A subquery calculates rating counts per user and returns the top result using ORDER BY and LIMIT. Another subquery filters February 2020 ratings and computes movie averages. The outer query selects the final rows. The database engine still performs grouping and sorting internally, so complexity remains about O(n log n). This structure keeps the main query concise and separates each requirement clearly.
Approach 3: Programmatic HashMap Aggregation (O(n) time, O(n) space)
If you load the tables into application code (Python or another language), you can compute results using hash maps. Iterate through the rating records once to maintain a count per user and accumulate sums and counts per movie. Use another pass to compute average ratings for February 2020 entries. Hash lookups are constant time, so the full pass runs in O(n). A final comparison step selects the maximum values while resolving ties lexicographically.
These solutions revolve around relational aggregation concepts from database queries and grouping operations commonly used in SQL. The core idea is transforming raw rating records into aggregated statistics that can be ranked.
Recommended for interviews: The SQL grouping and ordering approach is the expected solution. It demonstrates fluency with joins, GROUP BY, and tie‑breaking with ordered results. The HashMap version shows strong algorithmic thinking when solving the problem outside a database environment.
Approach 1: Using SQL Queries
This approach uses SQL queries with aggregation and sorting to determine the required results. It relies on the GROUP BY clause for aggregating data, and ORDER BY for sorting. Joins are used to link information from different tables.
WITH UserRatings AS (SELECT user_id, COUNT(*) AS rating_count FROM MovieRating GROUP BY user_id), MaxUserRatings AS (SELECT user_id FROM UserRatings WHERE rating_count = (SELECT MAX(rating_count) FROM UserRatings)), AvgMovieRatings AS (SELECT m.movie_id, AVG(r.rating) AS avg_rating FROM MovieRating r JOIN Movies m ON r.movie_id = m.movie_id WHERE r.created_at BETWEEN '2020-02-01' AND '2020-02-29' GROUP BY m.movie_id), MaxAvgMovieRatings AS (SELECT movie_id FROM AvgMovieRatings WHERE avg_rating = (SELECT MAX(avg_rating) FROM AvgMovieRatings)) SELECT (SELECT name FROM Users WHERE user_id IN (SELECT user_id FROM MaxUserRatings) ORDER BY name LIMIT 1) AS results UNION SELECT (SELECT title FROM Movies WHERE movie_id IN (SELECT movie_id FROM MaxAvgMovieRatings) ORDER BY title LIMIT 1);The solution consists of creating three CTEs (Common Table Expressions): UserRatings, MaxUserRatings, and AvgMovieRatings. The first calculates the number of ratings per user, the second selects the user(s) with the maximum number of ratings, and the third calculates the average movie rating in February 2020. We then find the lexicographically smallest user name and movie title in case of ties.
Time Complexity: The time complexity primarily depends on the size of the dataset and the usage of GROUP BY and ORDER BY operations, which in general takes O(n log n) where n is the number of rows in the MovieRating table.
Space Complexity: O(m + k) for holding intermediate aggregates, where m is the distinct number of users and k is the number of distinct movies.
Approach 2: Programmatically with HashMaps
Implementing the solution programmatically requires gathering data from the three different tables into memory and using hashmaps (dictionaries) to process it. Aggregations like counting ratings per user and averaging movie ratings in February are performed using these data structures.
from collections import defaultdict
# Users and Movies data
the_users = {1: 'Daniel', 2: 'Monica', 3: 'Maria', 4: 'James'}
the_movies = {1: 'Avengers', 2: 'Frozen 2', 3: 'Joker'}
# MovieRating data
movie_ratings = [
{'movie_id': 1, 'user_id': 1, 'rating': 3, 'created_at': '2020-01-12'},
{'movie_id': 1, 'user_id': 2, 'rating': 4, 'created_at': '2020-02-11'},
{'movie_id': 1, 'user_id': 3, 'rating': 2, 'created_at': '2020-02-12'},
{'movie_id': 1, 'user_id': 4, 'rating': 1, 'created_at': '2020-01-01'},
{'movie_id': 2, 'user_id': 1, 'rating': 5, 'created_at': '2020-02-17'},
{'movie_id': 2, 'user_id': 2, 'rating': 2, 'created_at': '2020-02-01'},
{'movie_id': 2, 'user_id': 3, 'rating': 2, 'created_at': '2020-03-01'},
{'movie_id': 3, 'user_id': 1, 'rating': 3, 'created_at': '2020-02-22'},
{'movie_id': 3, 'user_id': 2, 'rating': 4, 'created_at': '2020-02-25'},
]
# Count the number of ratings each user gave
user_rating_count = defaultdict(int)
for rating in movie_ratings:
user_rating_count[rating['user_id']] += 1
# Find the user with the maximum ratings
max_user_count = max(user_rating_count.values())
max_users = [the_users[user_id] for user_id, count in user_rating_count.items() if count == max_user_count]
user_name = min(max_users) # Lexicographically smallest
# Calculate average ratings for movies in February 2020
movie_rating_sums = defaultdict(int)
movie_rating_counts = defaultdict(int)
for rating in movie_ratings:
if rating['created_at'][:7] == '2020-02': # Check if in February 2020
movie_rating_sums[rating['movie_id']] += rating['rating']
movie_rating_counts[rating['movie_id']] += 1
# Determine highest average rating
max_avg_rating = 0
avg_movie_titles = []
for movie_id, total_rating in movie_rating_sums.items():
count = movie_rating_counts[movie_id]
avg_rating = total_rating / count
if avg_rating > max_avg_rating:
max_avg_rating = avg_rating
avg_movie_titles = [the_movies[movie_id]]
elif avg_rating == max_avg_rating:
avg_movie_titles.append(the_movies[movie_id])
movie_name = min(avg_movie_titles) # Lexicographically smallest
# Result
result = [user_name, movie_name]
print(result)This Python code uses dictionaries to store and compute the necessary counts and averages. For the first part, it counts how many movies each user rated. Then, it finds the user(s) who has the highest count. In case of ties, the lexicographically smallest is chosen. For the second part, it calculates the average ratings for movies within February 2020 and selects the movie with the highest average. Again, ties are resolved by choosing lexicographically.
Time Complexity: O(u + m), where u is the number of users and m is the number of ratings. We traverse all ratings once.
Space Complexity: O(u + v), where u is the number of users and v is the number of movies, due to storage in hashmaps.
Approach 3: Approach 1: SQL Grouping and Ordering
This approach involves two main SQL queries to achieve the desired results. For the first part, the task is to determine which user rated the most movies. We achieve this using the COUNT and GROUP BY functions to tally up the number of ratings per user. The results are then ordered by the count in descending order and lexicographically by name to resolve ties.
For the second task, we filter the MovieRating table to include only entries from February 2020. Using the AVG function, we calculate the average rating for each movie and use ORDER BY to sort the results by average ratings, resolving ties lexicographically by movie title.
import sqlite3
def main():
# Establish connection to database
conn = sqlite3.connect('movies.db')
cursor = conn.cursor()
# Query 1: Find User with most ratings
query1 = """
SELECT name FROM Users U JOIN (
SELECT user_id, COUNT(*) AS count
FROM MovieRating
GROUP BY user_id
ORDER BY count DESC, user_id
LIMIT 1
) MR ON U.user_id = MR.user_id
"""
# Query 2: Film with highest average rating in February 2020
query2 = """
SELECT M.title FROM Movies M JOIN (
SELECT movie_id, AVG(rating) AS avg_rating
FROM MovieRating
WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28'
GROUP BY movie_id
) MR ON M.movie_id = MR.movie_id
ORDER BY avg_rating DESC, M.title
LIMIT 1
"""
# Execute query1
cursor.execute(query1)
result1 = cursor.fetchone()
print(result1[0]) if result1 else print('No results for Query 1')
# Execute query2
cursor.execute(query2)
result2 = cursor.fetchone()
print(result2[0]) if result2 else print('No results for Query 2')
# Clean up
conn.close()
if __name__ == '__main__':
main()import java.sql.*;
public class MovieRatingSolution {
public static void main(String[] args) {
// Establish database connection
// Query 1: Find User with most ratings
String query1 = "SELECT name FROM Users U JOIN (SELECT user_id, COUNT(*) AS count FROM MovieRating GROUP BY user_id ORDER BY count DESC, user_id LIMIT 1) MR ON U.user_id = MR.user_id";
// Query 2: Film with highest average rating in February 2020
String query2 = "SELECT M.title FROM Movies M JOIN (SELECT movie_id, AVG(rating) AS avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) MR ON M.movie_id = MR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
// Execute these queries
try (Connection connection = DriverManager.getConnection("jdbc:url_to_your_database", "username", "password")) {
// Execute query1
try (PreparedStatement ps1 = connection.prepareStatement(query1)) {
ResultSet rs1 = ps1.executeQuery();
while (rs1.next()) {
System.out.println(rs1.getString("name"));
}
}
// Execute query2
try (PreparedStatement ps2 = connection.prepareStatement(query2)) {
ResultSet rs2 = ps2.executeQuery();
while (rs2.next()) {
System.out.println(rs2.getString("title"));
}
}
} catch (SQLException e) {
e.printStackTrace();
}
}
}// SQL operations are usually done via database procedures, so leveraging SQL queries
#include <iostream>
int main() {
// Connect to database and execute these queries sequentially
// User with most ratings
std::string query1 = "SELECT name FROM Users U JOIN (SELECT user_id, COUNT(*) as count FROM MovieRating GROUP BY user_id ORDER BY count DESC, user_id LIMIT 1) MR ON U.user_id = MR.user_id";
// Movie with highest average rating in February 2020
std::string query2 = "SELECT M.title FROM Movies M JOIN (SELECT movie_id, AVG(rating) as avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) MR ON M.movie_id = MR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
// Execute using a database connection
return 0;
}// Typically, SQL queries would be executed on a server-side code, here is a conceptual representation:
(async function() {
// Assume availability of a database client
// User with most ratings
const query1 = `SELECT name FROM Users U JOIN (
SELECT user_id, COUNT(*) AS count FROM MovieRating
GROUP BY user_id ORDER BY count DESC, user_id LIMIT 1
) MR ON U.user_id = MR.user_id`;
// Movie with highest average rating in February 2020
const query2 = `SELECT M.title FROM Movies M JOIN (
SELECT movie_id, AVG(rating) AS avg_rating FROM MovieRating
WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28'
GROUP BY movie_id
) MR ON M.movie_id = MR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1`;
try {
const results1 = await db.query(query1);
const results2 = await db.query(query2);
console.log(results1.rows[0].name);
console.log(results2.rows[0].title);
} catch (err) {
console.error('Error executing queries: ', err);
}
})();// SQL operations are usually done via database procedures, hence SQL statement usage here
#include <stdio.h>
int main() {
// Connect to database and execute these queries sequentially
// Query 1: Find User with most ratings
const char *query1 = "SELECT name FROM Users U JOIN (SELECT user_id, COUNT(*) as count FROM MovieRating GROUP BY user_id ORDER BY count DESC, user_id LIMIT 1) MR ON U.user_id = MR.user_id";
// Query 2: Find Movie with highest average rating in February 2020
const char *query2 = "SELECT M.title FROM Movies M JOIN (SELECT movie_id, AVG(rating) as avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) MR ON M.movie_id = MR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
// Execute these queries using a database connection and fetch results
return 0;
}using System;
using System.Data.SqlClient;
class Program {
static void Main() {
string connectionString = "Data Source=your_database_source;Initial Catalog=MovieDB;Integrated Security=True";
// Query 1
string query1 = "SELECT name FROM Users U JOIN (SELECT user_id, COUNT(*) AS count FROM MovieRating GROUP BY user_id ORDER BY count DESC, user_id LIMIT 1) MR ON U.user_id = MR.user_id";
// Query 2
string query2 = "SELECT M.title FROM Movies M JOIN (SELECT movie_id, AVG(rating) AS avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) MR ON M.movie_id = MR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
using (SqlConnection connection = new SqlConnection(connectionString)) {
SqlCommand command = new SqlCommand(query1, connection);
connection.Open();
using (SqlDataReader reader = command.ExecuteReader()) {
while (reader.Read()) {
Console.WriteLine(reader["name"]);
}
}
command.CommandText = query2;
using (SqlDataReader reader = command.ExecuteReader()) {
while (reader.Read()) {
Console.WriteLine(reader["title"]);
}
}
}
}
}This Python script uses SQLite to execute SQL queries against a hypothetical 'movies.db' database. The first query finds the user who rated the most movies, and the second query finds the movie with the highest average rating in February 2020. Results are directly printed.
Time Complexity: O(N + M) for executing the queries and parsing results, where N and M are the number of records queried.
Space Complexity: O(1) as fetching and processing results are done one at a time.
Approach 4: Approach 2: Direct SQL with Subquery Utilization
This approach still operates with SQL as the problem is fundamentally a SQL-based operation. Here's how we expand on using nested subqueries and CTEs (Common Table Expressions) to simplify readability and management of data within SQL scripts.
import sqlite3
def main():
# Database connection
conn = sqlite3.connect('database.db')
cursor = conn.cursor()
# User Movies Count CTE
query1 = '''
WITH UserCounting AS (
SELECT user_id, COUNT(*) AS count
FROM MovieRating
GROUP BY user_id
)
SELECT name FROM Users U JOIN UserCounting UC ON U.user_id = UC.user_id
ORDER BY UC.count DESC, U.name
LIMIT 1
'''
# Movie Rating CTE
query2 = '''
WITH FebruaryMovieAvg AS (
SELECT movie_id, AVG(rating) AS average_rating
FROM MovieRating
WHERE created_at >= '2020-02-01' AND created_at <= '2020-02-28'
GROUP BY movie_id
)
SELECT title FROM Movies M JOIN FebruaryMovieAvg FMA ON M.movie_id = FMA.movie_id
ORDER BY FMA.average_rating DESC, M.title
LIMIT 1
'''
# Execute first CTE-based query
cursor.execute(query1)
user_result = cursor.fetchone()
if user_result:
print('User:', user_result[0])
# Execute second CTE-based query
cursor.execute(query2)
movie_result = cursor.fetchone()
if movie_result:
print('Movie:', movie_result[0])
# Close connection
conn.close()
if __name__ == '__main__':
main()import java.sql.*;
public class OptimizedMovieSolution {
public static void main(String[] args) {
// Connect to a database
String url = "jdbc:url_to_db";
// User Rating CTE
String query1 = "WITH UserRatingCount AS (SELECT user_id, COUNT(*) AS count FROM MovieRating GROUP BY user_id) SELECT name FROM Users U JOIN UserRatingCount URC ON U.user_id = URC.user_id ORDER BY count DESC, U.name LIMIT 1";
// Average Rating Reports
String query2 = "WITH FebruaryRatings AS (SELECT movie_id, AVG(rating) AS average_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) SELECT title FROM Movies M JOIN FebruaryRatings FR ON M.movie_id = FR.movie_id ORDER BY average_rating DESC, M.title LIMIT 1";
try (Connection conn = DriverManager.getConnection(url, "username", "password")) {
Statement stmt = conn.createStatement();
ResultSet rs1 = stmt.executeQuery(query1);
while (rs1.next()) {
System.out.println("User: " + rs1.getString("name"));
}
ResultSet rs2 = stmt.executeQuery(query2);
while (rs2.next()) {
System.out.println("Movie: " + rs2.getString("title"));
}
} catch (SQLException e) {
e.printStackTrace();
}
}
}// SQL management typically requires database-specific scripts
#include <iostream>
int main() {
// Execute using a database manager CTE query
// User Ratings with recursive CTE
std::string query1 = "WITH UserRatingCount AS (SELECT user_id, COUNT(*) as count FROM MovieRating GROUP BY user_id) SELECT name FROM Users U JOIN UserRatingCount URC ON U.user_id = URC.user_id ORDER BY count DESC, U.name LIMIT 1";
// Movie Average Rating for February 2020 with recursive CTE
std::string query2 = "WITH FebruaryRatings AS (SELECT movie_id, AVG(rating) as avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) SELECT title FROM Movies M JOIN FebruaryRatings FR ON M.movie_id = FR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
// Use them with consistent database establishment and verification protocol
return 0;
}// Conceptual server-side implementation leveraging CTE
(async function() {
// User Ratings with CTE
const query1 = `WITH UserRatings AS (
SELECT user_id, COUNT(*) AS num_ratings
FROM MovieRating
GROUP BY user_id
)
SELECT name FROM Users U JOIN UserRatings UR ON U.user_id = UR.user_id
ORDER BY num_ratings DESC, name
LIMIT 1`;
// Movie Ratings in February 2020 with CTE
const query2 = `WITH FebRatings AS (
SELECT movie_id, AVG(rating) AS avg_rating
FROM MovieRating
WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28'
GROUP BY movie_id
)
SELECT title FROM Movies M JOIN FebRatings FR ON M.movie_id = FR.movie_id
ORDER BY avg_rating DESC, title
LIMIT 1`;
try {
const result1 = await db.query(query1);
const result2 = await db.query(query2);
console.log('User:', result1.rows[0].name);
console.log('Movie:', result2.rows[0].title);
} catch (err) {
console.error('Error:', err);
}
})();// SQL handling is better within a relational database manager
#include <stdio.h>
int main() {
// Connect to database and execute these CTE based queries
// CTE Approach for User Ratings
const char *query1 = "WITH UserRatingCount AS (SELECT user_id, COUNT(*) as num_ratings FROM MovieRating GROUP BY user_id) SELECT name FROM Users U JOIN UserRatingCount URC ON U.user_id = URC.user_id ORDER BY URC.num_ratings DESC, U.name LIMIT 1";
// CTE Approach for Movie Ratings
const char *query2 = "WITH FebruaryRatings AS (SELECT movie_id, AVG(rating) as avg_rating FROM MovieRating WHERE created_at BETWEEN '2020-02-01' AND '2020-02-28' GROUP BY movie_id) SELECT title FROM Movies M JOIN FebruaryRatings FR ON M.movie_id = FR.movie_id ORDER BY FR.avg_rating DESC, M.title LIMIT 1";
// Execute these queries using a database connection and fetch results
return 0;
}using System;
using System.Data.SqlClient;
class AdvancedMovieRating {
static void Main() {
string connectionString = "Data Source=your_source;Initial Catalog=MovieData;User ID=user;Password=password;";
// CTE-based query for user ratings
string query1 = "WITH UserCount AS (SELECT user_id, COUNT(*) as count FROM MovieRating GROUP BY user_id) SELECT name FROM Users U JOIN UserCount UC ON U.user_id = UC.user_id ORDER BY UC.count DESC, U.name LIMIT 1";
// CTE-based query for movie average in February 2020
string query2 = "WITH FebRating AS (SELECT movie_id, AVG(rating) AS avg_rating FROM MovieRating WHERE created_at >= '2020-02-01' AND created_at <= '2020-02-28' GROUP BY movie_id ) SELECT title FROM Movies M JOIN FebRating FR ON M.movie_id = FR.movie_id ORDER BY avg_rating DESC, M.title LIMIT 1";
using (SqlConnection connection = new SqlConnection(connectionString)) {
SqlCommand command = new SqlCommand(query1, connection);
connection.Open();
using (SqlDataReader reader = command.ExecuteReader()) {
while (reader.Read()) {
Console.WriteLine("User: " + reader["name"]);
}
}
command.CommandText = query2;
using (SqlDataReader reader = command.ExecuteReader()) {
while (reader.Read()) {
Console.WriteLine("Movie: " + reader["title"]);
}
}
}
}
}In this Python script, two primary SQL queries leverage CTEs to summarize and extract user activity and average film ratings from within established time frames. Each CTE explicitly defines steps, facilitating clear transactional operations.
Time Complexity: O(N + M), linear relative to dataset size within respective tables.
Space Complexity: O(1) as this anticipates running fetch sequentially on returned parameters.
Complexity comparison
| Approach | Complexity |
|---|---|
| Using SQL Queries | Time Complexity: The time complexity primarily depends on the size of the dataset and the usage of GROUP BY and ORDER BY operations, which in general takes O(n log n) where n is the number of rows in the MovieRating table. |
| Programmatically with HashMaps | Time Complexity: O(u + m), where u is the number of users and m is the number of ratings. We traverse all ratings once. |
| Approach 1: SQL Grouping and Ordering | Time Complexity: O(N + M) for executing the queries and parsing results, where N and M are the number of records queried. |
| Approach 2: Direct SQL with Subquery Utilization | Time Complexity: O(N + M), linear relative to dataset size within respective tables. |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| SQL Grouping and Ordering | O(n log n) | O(n) | Standard SQL solution using GROUP BY, joins, and sorting |
| Direct SQL with Subqueries | O(n log n) | O(n) | Cleaner query structure when separating logic with subqueries |
| HashMap Aggregation (Programmatic) | O(n) | O(n) | Useful when processing data outside SQL using Python or other languages |
Frequently asked questions
Is Movie Rating easy or hard?
Movie Rating is considered a medium difficulty database problem. The challenge is not complex algorithms but correctly structuring SQL joins, aggregations, and tie breaking conditions while filtering ratings by date.
How to solve Movie Rating in O(n)?
An O(n) approach is possible when processing records programmatically. Iterate through all rating rows once while maintaining hash maps for user rating counts and movie rating sums and counts. After aggregation, compute the best user and movie with simple comparisons. Hash lookups keep operations constant time.
Movie Rating Python or Java solution
In Python or Java, store user rating counts in a hash map and track movie rating sums and counts for February 2020. After processing all records, compute averages and choose the highest values while applying lexicographic tie breaking. This approach runs in O(n) time with O(n) space.
What is the best approach for Movie Rating?
The standard solution uses SQL aggregation with GROUP BY and ORDER BY. First count ratings per user to find the most active reviewer, then compute the average rating per movie for February 2020. Sorting resolves ties by lexicographical order of names or titles. This approach runs in about O(n log n) time due to sorting.
Is Movie Rating asked at Google/Amazon/Meta?
Database aggregation problems like Movie Rating frequently appear in SQL interview rounds at companies such as Amazon, Meta, and other data‑heavy teams. Interviewers use them to evaluate joins, GROUP BY usage, and ranking logic with tie breaking.
What data structure is used in Movie Rating?
The SQL version primarily uses relational operations such as joins, GROUP BY aggregation, and ORDER BY sorting. In a programmatic solution, hash maps store counts and rating totals for users and movies, enabling efficient aggregation in linear time.
What is the time complexity of Movie Rating?
Most SQL implementations run in O(n log n) time because grouped results are sorted to determine the top user and movie. The grouping itself is roughly O(n), while ORDER BY operations add the log factor. Space complexity is O(n) for intermediate aggregated results.
