Solution Explanation for LeetCode Problem 2026: Low-Quality Problems

This problem requires identifying LeetCode problems with a like percentage less than 60%. The solution involves calculating the like percentage for each problem and filtering based on this threshold.

Approach:

The most straightforward approach uses SQL queries to directly manipulate the Problems table. We calculate the like percentage within the query and then filter rows based on this percentage.

SQL Solution (MySQL):

SELECT problem_id
FROM Problems
WHERE likes / (likes + dislikes) < 0.6
ORDER BY problem_id;

Explanation:

1. SELECT problem_id: This selects the problem_id column, which represents the ID of each problem.

2. FROM Problems: This specifies that we are querying the Problems table.

3. WHERE likes / (likes + dislikes) < 0.6: This is the core of the solution. It calculates the like percentage (likes / (likes + dislikes)) for each problem and filters the results to include only those where the percentage is strictly less than 0.6 (60%). Note that we're implicitly converting the result of the division to a floating-point number for comparison.

4. ORDER BY problem_id: This sorts the results in ascending order of problem_id, as required by the problem statement.

Time Complexity Analysis:

The time complexity of the SQL query is O(N), where N is the number of rows in the Problems table. This is because the database system will need to iterate through each row to compute the like percentage and apply the filter. The ORDER BY clause adds a small overhead for sorting, but it doesn't change the overall time complexity. The exact efficiency depends on the database system's query optimization techniques, but a linear scan is the most likely operation.

Space Complexity Analysis:

The space complexity is O(M), where M is the number of low-quality problems that satisfy the condition. The space is used to store the results before they are returned. In the worst case, M could be equal to N, but typically, it would be a smaller subset of the total problems. The database system's internal workings might also use additional temporary space, but this is usually handled internally and not considered part of the algorithm's space complexity.

No other programming languages are needed to solve this problem as it is purely a database query question. The provided SQL code is a complete and efficient solution.