LeetCode 620: Not Boring Movies Solution in Python – A Step-by-Step Guide

Imagine you’re a movie critic with a database of films—each with an ID, description, and rating—and your task is to pick out the exciting ones. You’re told to skip anything labeled "boring," keep only movies with odd-numbered IDs, and sort them by rating from best to worst. That’s the heart of LeetCode 620: Not Boring Movies, an easy-level problem that’s all about filtering and sorting data from a table. Using Python, we’ll explore two solutions: the Best Solution, a list filtering approach with sorting for simplicity and speed, and an Alternative Solution, a dictionary-based grouping method that’s more detailed but less necessary here. With detailed examples, beginner-friendly breakdowns—especially for the filtering method—and clear code, this guide will help you spotlight those fun films, whether you’re new to coding or brushing up. Let’s roll the film reel and start curating!

What Is LeetCode 620: Not Boring Movies?

In LeetCode 620: Not Boring Movies, you’re given a table cinema with columns id (integer), movie (title), description (string), and rating (float). Your task is to filter this table to return only movies that:

Are not "boring" (description ≠ "boring").
Have an odd id.
Are sorted by rating in descending order (highest first).

For example, from a list including "War" (id=1, rating=8.9) and "Boring Tale" (id=2, rating=7.5), you’d pick "War" if its ID were odd and skip "Boring Tale." This problem tests your ability to filter and sort data, typically posed as an SQL query with WHERE and ORDER BY, but we’ll solve it in Python for a coding twist.

Problem Statement

Input: A list of records (or table rows) with [id, movie, description, rating].
Output: A list of [id, movie, description, rating] for movies meeting criteria, sorted by rating descending.
Rules:

Exclude movies where description = "boring".
Include only movies with odd id (id % 2 = 1).
Sort by rating from highest to lowest.

Constraints

Table has at least 1 row.
1 ≤ id ≤ 10⁵.
movie and description are strings ≤ 100 chars.
0.0 ≤ rating ≤ 10.0.

Examples

Input:

Output:

Input:

Output: [] (No non-boring, odd-ID movies)

Input:

Output: [[5, "Fun Flick", "comedy", 7.9]]

These examples set the scene—let’s solve it!

Understanding the Problem: Filtering Fun Films

To solve LeetCode 620: Not Boring Movies in Python, we need to process a list of movie records, filter out "boring" movies, keep only those with odd IDs, and sort the rest by rating from highest to lowest. A naive approach might loop multiple times, but we can streamline it. We’ll use:

Best Solution (List Filtering with Sorting): O(n log n) time, O(n) space—fast and intuitive (n = number of rows).
Alternative Solution (Dictionary-Based Grouping): O(n log n) time, O(n) space—detailed but overkill.

Let’s start with the list filtering solution, breaking it down for beginners.

Best Solution: List Filtering with Sorting

Why This Is the Best Solution

The list filtering with sorting approach is the top choice for LeetCode 620 because it’s efficient—O(n log n) time due to sorting—and straightforward, applying filters and sorting in a single flow that mirrors the problem’s requirements. It uses Python’s built-in tools, making it scalable within constraints (≤ 10⁵ rows) and easy to grasp. It’s like curating a movie list with a quick checklist and ranking!

How It Works

Think of this as picking winners from a film festival:

Step 1: Filter Movies:

Keep only rows where description ≠ "boring" and id % 2 = 1.

Step 2: Sort by Rating:

Sort the filtered list by rating in descending order.

Step 3: Return Result:

Output the filtered, sorted list.

It’s like a movie critic’s filter—screen and rank!

Step-by-Step Example

Example:

id | movie       | description | rating
1  | War         | action      | 8.9
2  | Boring Tale | boring      | 7.5
3  | Skyfall     | action      | 8.8

Step 1: Filter:

[1, "War", "action", 8.9]: Not boring, id=1 (odd), keep.
[2, "Boring Tale", "boring", 7.5]: Boring, skip.
[3, "Skyfall", "action", 8.8]: Not boring, id=3 (odd), keep.
Result: [[1, "War", "action", 8.9], [3, "Skyfall", "action", 8.8]].

Step 2: Sort by rating:

8.9 > 8.8, so: [[1, "War", "action", 8.9], [3, "Skyfall", "action", 8.8]].

Output: [[1, "War", "action", 8.9], [3, "Skyfall", "action", 8.8]].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

class Solution:
    def notBoringMovies(self, cinema: List[List]) -> List[List]:
        # Step 1: Filter non-boring movies with odd IDs
        filtered = []
        for movie in cinema:
            id_val, title, desc, rating = movie
            if desc != "boring" and id_val % 2 == 1:
                filtered.append(movie)

        # Step 2: Sort by rating in descending order
        filtered.sort(key=lambda x: x[3], reverse=True)

        # Step 3: Return result
        return filtered

Lines 4-8: Filter:

Check description ≠ "boring" and id is odd.
Add qualifying movies to filtered.

Line 11: Sort:

Use key to sort by rating (index 3), reverse=True for descending.

Time Complexity: O(n log n)—sorting dominates.
Space Complexity: O(n)—store filtered list.

This is like a movie curator—filter and rank fast!

Alternative Solution: Dictionary-Based Grouping

Why an Alternative Approach?

The dictionary-based grouping approach organizes movies by rating—O(n log n) time, O(n) space. It’s more detailed, allowing grouping before filtering, but it’s overkill for this problem since we don’t need intermediate grouping. It’s like cataloging movies by score before picking the fun ones!

How It Works

Picture this as a movie catalog:

Step 1: Group by rating into a dictionary.
Step 2: Filter non-boring, odd-ID movies.
Step 3: Sort ratings descending, build result.
Step 4: Return list.

It’s like a detailed movie archive—sorted and filtered!

Step-by-Step Example

Example:

id | movie       | description | rating
1  | War         | action      | 8.9
2  | Boring Tale | boring      | 7.5
3  | Skyfall     | action      | 8.8

Step 1: Group by rating:

{8.9: [[1, "War", "action", 8.9]], 7.5: [[2, "Boring Tale", "boring", 7.5]], 8.8: [[3, "Skyfall", "action", 8.8]]}.

Step 2: Filter:

8.9: Keep [1, "War", ...] (odd, not boring).
7.5: Skip (boring).
8.8: Keep [3, "Skyfall", ...] (odd, not boring).

Step 3: Sort ratings: [8.9, 8.8], result: [[1, "War", ...], [3, "Skyfall", ...]].
Output: [[1, "War", "action", 8.9], [3, "Skyfall", "action", 8.8]].

Code for Dictionary Approach

from collections import defaultdict

class Solution:
    def notBoringMovies(self, cinema: List[List]) -> List[List]:
        # Step 1: Group by rating
        rating_dict = defaultdict(list)
        for movie in cinema:
            rating_dict[movie[3]].append(movie)

        # Step 2: Filter non-boring, odd-ID movies
        filtered = []
        for rating in rating_dict:
            for movie in rating_dict[rating]:
                id_val, _, desc, _ = movie
                if desc != "boring" and id_val % 2 == 1:
                    filtered.append(movie)

        # Step 3: Sort by rating descending
        filtered.sort(key=lambda x: x[3], reverse=True)

        # Step 4: Return result
        return filtered

Lines 6-8: Group by rating.
Lines 11-15: Filter within groups.
Line 18: Sort by rating.
Time Complexity: O(n log n)—sorting.
Space Complexity: O(n)—dict and list.

It’s a movie archive—detailed but unnecessary!

Comparing the Two Solutions

List Filtering (Best):

Pros: O(n log n) time, O(n) space, simple and direct.
Cons: Less flexible for grouping.

Dictionary (Alternative):

Pros: O(n log n) time, O(n) space, detailed grouping.
Cons: Overkill for this task.

List filtering wins for simplicity.

Additional Examples and Edge Cases

Input: [[1, "Fun", "comedy", 8.0]]

Output: [[1, "Fun", "comedy", 8.0]].

Input: [[2, "Dull", "boring", 7.0]]

Output: [].

Both handle these well.

Complexity Breakdown

List Filtering: Time O(n log n), Space O(n).
Dictionary: Time O(n log n), Space O(n).

List filtering is optimal here.

Key Takeaways

List Filtering: Quick filter—smart!
Dictionary: Grouped detail—clear!
Data: Sorting is fun.
Python Tip: Lists rock—see Python Basics.

Final Thoughts: Spotlight Fun Films

LeetCode 620: Not Boring Movies in Python is a fun data-filtering challenge. List filtering with sorting offers simplicity and speed, while dictionary grouping adds detail. Want more? Try LeetCode 175: Combine Two Tables or LeetCode 627: Swap Salary. Ready to curate? Head to Solve LeetCode 620 on LeetCode and pick those not-boring movies today!