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.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
Table: Cinema
+----------------+----------+ | Column Name | Type | +----------------+----------+ | id | int | | movie | varchar | | description | varchar | | rating | float | +----------------+----------+ id is the primary key (column with unique values) for this table. Each row contains information about the name of a movie, its genre, and its rating. rating is a 2 decimal places float in the range [0, 10]
Write a solution to report the movies with an odd-numbered ID and a description that is not "boring".
Return the result table ordered by rating in descending order.
The result format is in the following example.
Example 1:
Input: Cinema table: +----+------------+-------------+--------+ | id | movie | description | rating | +----+------------+-------------+--------+ | 1 | War | great 3D | 8.9 | | 2 | Science | fiction | 8.5 | | 3 | irish | boring | 6.2 | | 4 | Ice song | Fantacy | 8.6 | | 5 | House card | Interesting | 9.1 | +----+------------+-------------+--------+ Output: +----+------------+-------------+--------+ | id | movie | description | rating | +----+------------+-------------+--------+ | 5 | House card | Interesting | 9.1 | | 1 | War | great 3D | 8.9 | +----+------------+-------------+--------+ Explanation: We have three movies with odd-numbered IDs: 1, 3, and 5. The movie with ID = 3 is boring so we do not include it in the answer.
Problem summary: Table: Cinema +----------------+----------+ | Column Name | Type | +----------------+----------+ | id | int | | movie | varchar | | description | varchar | | rating | float | +----------------+----------+ id is the primary key (column with unique values) for this table. Each row contains information about the name of a movie, its genre, and its rating. rating is a 2 decimal places float in the range [0, 10] Write a solution to report the movies with an odd-numbered ID and a description that is not "boring". Return the result table ordered by rating in descending order. The result format is in the following example.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"cinema":["id", "movie", "description", "rating"]},"rows":{"cinema":[[1, "War", "great 3D", 8.9], [2, "Science", "fiction", 8.5], [3, "irish", "boring", 6.2], [4, "Ice song", "Fantacy", 8.6], [5, "House card", "Interesting", 9.1]]}}Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #620: Not Boring Movies
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #620: Not Boring Movies
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #620: Not Boring Movies
// # Write your MySQL query statement below
// SELECT *
// FROM Cinema
// WHERE description != 'boring' AND id & 1 = 1
// ORDER BY 4 DESC;
// "#
// }
// Accepted solution for LeetCode #620: Not Boring Movies
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #620: Not Boring Movies
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #620: Not Boring Movies
// # Write your MySQL query statement below
// SELECT *
// FROM Cinema
// WHERE description != 'boring' AND id & 1 = 1
// ORDER BY 4 DESC;
// "#
// }
# Accepted solution for LeetCode #620: Not Boring Movies
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #620: Not Boring Movies
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #620: Not Boring Movies
# # Write your MySQL query statement below
# SELECT *
# FROM Cinema
# WHERE description != 'boring' AND id & 1 = 1
# ORDER BY 4 DESC;
# "#
# }
// Accepted solution for LeetCode #620: Not Boring Movies
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #620: Not Boring Movies
# Write your MySQL query statement below
SELECT *
FROM Cinema
WHERE description != 'boring' AND id & 1 = 1
ORDER BY 4 DESC;
"#
}
// Accepted solution for LeetCode #620: Not Boring Movies
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #620: Not Boring Movies
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #620: Not Boring Movies
// # Write your MySQL query statement below
// SELECT *
// FROM Cinema
// WHERE description != 'boring' AND id & 1 = 1
// ORDER BY 4 DESC;
// "#
// }
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.