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: Users
+-------------+---------+ | Column Name | Type | +-------------+---------+ | user_id | int | | user_name | varchar | +-------------+---------+ user_id is the primary key (column with unique values) for this table. Each row of this table contains the name and the id of a user.
Table: Register
+-------------+---------+ | Column Name | Type | +-------------+---------+ | contest_id | int | | user_id | int | +-------------+---------+ (contest_id, user_id) is the primary key (combination of columns with unique values) for this table. Each row of this table contains the id of a user and the contest they registered into.
Write a solution to find the percentage of the users registered in each contest rounded to two decimals.
Return the result table ordered by percentage in descending order. In case of a tie, order it by contest_id in ascending order.
The result format is in the following example.
Example 1:
Input: Users table: +---------+-----------+ | user_id | user_name | +---------+-----------+ | 6 | Alice | | 2 | Bob | | 7 | Alex | +---------+-----------+ Register table: +------------+---------+ | contest_id | user_id | +------------+---------+ | 215 | 6 | | 209 | 2 | | 208 | 2 | | 210 | 6 | | 208 | 6 | | 209 | 7 | | 209 | 6 | | 215 | 7 | | 208 | 7 | | 210 | 2 | | 207 | 2 | | 210 | 7 | +------------+---------+ Output: +------------+------------+ | contest_id | percentage | +------------+------------+ | 208 | 100.0 | | 209 | 100.0 | | 210 | 100.0 | | 215 | 66.67 | | 207 | 33.33 | +------------+------------+ Explanation: All the users registered in contests 208, 209, and 210. The percentage is 100% and we sort them in the answer table by contest_id in ascending order. Alice and Alex registered in contest 215 and the percentage is ((2/3) * 100) = 66.67% Bob registered in contest 207 and the percentage is ((1/3) * 100) = 33.33%
Problem summary: Table: Users +-------------+---------+ | Column Name | Type | +-------------+---------+ | user_id | int | | user_name | varchar | +-------------+---------+ user_id is the primary key (column with unique values) for this table. Each row of this table contains the name and the id of a user. Table: Register +-------------+---------+ | Column Name | Type | +-------------+---------+ | contest_id | int | | user_id | int | +-------------+---------+ (contest_id, user_id) is the primary key (combination of columns with unique values) for this table. Each row of this table contains the id of a user and the contest they registered into. Write a solution to find the percentage of the users registered in each contest rounded to two decimals. Return the result table ordered by percentage in descending order. In case of a tie, order it by contest_id in ascending order. The result format is in the
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
{"headers":{"Users":["user_id","user_name"],"Register":["contest_id","user_id"]},"rows":{"Users":[[6,"Alice"],[2,"Bob"],[7,"Alex"]],"Register":[[215,6],[209,2],[208,2],[210,6],[208,6],[209,7],[209,6],[215,7],[208,7],[210,2],[207,2],[210,7]]}}queries-quality-and-percentage)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// Auto-generated Java example from rust.
class Solution {
public void exampleSolution() {
}
}
// Reference (rust):
// // Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// # Write your MySQL query statement below
// SELECT
// contest_id,
// ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
// FROM Register
// GROUP BY 1
// ORDER BY 2 DESC, 1;
// "#
// }
// Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// Auto-generated Go example from rust.
func exampleSolution() {
}
// Reference (rust):
// // Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// # Write your MySQL query statement below
// SELECT
// contest_id,
// ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
// FROM Register
// GROUP BY 1
// ORDER BY 2 DESC, 1;
// "#
// }
# Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
# Auto-generated Python example from rust.
def example_solution() -> None:
return
# Reference (rust):
# // Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
# pub fn sql_example() -> &'static str {
# r#"
# -- Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
# # Write your MySQL query statement below
# SELECT
# contest_id,
# ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
# FROM Register
# GROUP BY 1
# ORDER BY 2 DESC, 1;
# "#
# }
// Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
pub fn sql_example() -> &'static str {
r#"
-- Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
# Write your MySQL query statement below
SELECT
contest_id,
ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
FROM Register
GROUP BY 1
ORDER BY 2 DESC, 1;
"#
}
// Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// Auto-generated TypeScript example from rust.
function exampleSolution(): void {
}
// Reference (rust):
// // Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// pub fn sql_example() -> &'static str {
// r#"
// -- Accepted solution for LeetCode #1633: Percentage of Users Attended a Contest
// # Write your MySQL query statement below
// SELECT
// contest_id,
// ROUND(COUNT(1) * 100 / (SELECT COUNT(1) FROM Users), 2) AS percentage
// FROM Register
// GROUP BY 1
// ORDER BY 2 DESC, 1;
// "#
// }
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.