LeetCode #1341 — MEDIUM

Movie Rating

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

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.
Each movie has a unique title.

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.

Step 01

Brute Force Baseline

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers": {"Movies": ["movie_id", "title"], "Users": ["user_id", "name"], "MovieRating": ["movie_id", "user_id", "rating", "created_at"]}, "rows": {"Movies": [[1, "Avengers"], [2, "Frozen 2"], [3, "Joker"]], "Users": [[1, "Daniel"], [2, "Monica"], [3, "Maria"], [4, "James"]], "MovieRating": [[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"]]}}

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1341: Movie Rating
// Auto-generated Java example from rust.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (rust):
// // Accepted solution for LeetCode #1341: Movie Rating
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1341: Movie Rating
// # Write your MySQL query statement below
// (
//     SELECT name AS results
//     FROM
//         Users
//         JOIN MovieRating USING (user_id)
//     GROUP BY user_id
//     ORDER BY COUNT(1) DESC, name
//     LIMIT 1
// )
// UNION ALL
// (
//     SELECT title
//     FROM
//         MovieRating
//         JOIN Movies USING (movie_id)
//     WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
//     GROUP BY movie_id
//     ORDER BY AVG(rating) DESC, title
//     LIMIT 1
// );
// "#
// }

// Accepted solution for LeetCode #1341: Movie Rating
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1341: Movie Rating
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1341: Movie Rating
// # Write your MySQL query statement below
// (
//     SELECT name AS results
//     FROM
//         Users
//         JOIN MovieRating USING (user_id)
//     GROUP BY user_id
//     ORDER BY COUNT(1) DESC, name
//     LIMIT 1
// )
// UNION ALL
// (
//     SELECT title
//     FROM
//         MovieRating
//         JOIN Movies USING (movie_id)
//     WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
//     GROUP BY movie_id
//     ORDER BY AVG(rating) DESC, title
//     LIMIT 1
// );
// "#
// }

# Accepted solution for LeetCode #1341: Movie Rating
# Auto-generated Python example from rust.
def example_solution() -> None:
    return
# Reference (rust):
# // Accepted solution for LeetCode #1341: Movie Rating
# pub fn sql_example() -> &'static str {
#     r#"
# -- Accepted solution for LeetCode #1341: Movie Rating
# # Write your MySQL query statement below
# (
#     SELECT name AS results
#     FROM
#         Users
#         JOIN MovieRating USING (user_id)
#     GROUP BY user_id
#     ORDER BY COUNT(1) DESC, name
#     LIMIT 1
# )
# UNION ALL
# (
#     SELECT title
#     FROM
#         MovieRating
#         JOIN Movies USING (movie_id)
#     WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
#     GROUP BY movie_id
#     ORDER BY AVG(rating) DESC, title
#     LIMIT 1
# );
# "#
# }

// Accepted solution for LeetCode #1341: Movie Rating
pub fn sql_example() -> &'static str {
    r#"
-- Accepted solution for LeetCode #1341: Movie Rating
# Write your MySQL query statement below
(
    SELECT name AS results
    FROM
        Users
        JOIN MovieRating USING (user_id)
    GROUP BY user_id
    ORDER BY COUNT(1) DESC, name
    LIMIT 1
)
UNION ALL
(
    SELECT title
    FROM
        MovieRating
        JOIN Movies USING (movie_id)
    WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
    GROUP BY movie_id
    ORDER BY AVG(rating) DESC, title
    LIMIT 1
);
"#
}

// Accepted solution for LeetCode #1341: Movie Rating
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1341: Movie Rating
// pub fn sql_example() -> &'static str {
//     r#"
// -- Accepted solution for LeetCode #1341: Movie Rating
// # Write your MySQL query statement below
// (
//     SELECT name AS results
//     FROM
//         Users
//         JOIN MovieRating USING (user_id)
//     GROUP BY user_id
//     ORDER BY COUNT(1) DESC, name
//     LIMIT 1
// )
// UNION ALL
// (
//     SELECT title
//     FROM
//         MovieRating
//         JOIN Movies USING (movie_id)
//     WHERE DATE_FORMAT(created_at, '%Y-%m') = '2020-02'
//     GROUP BY movie_id
//     ORDER BY AVG(rating) DESC, title
//     LIMIT 1
// );
// "#
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.