1148. Article Views I - Detailed Explanation

Problem Statement

You are given a table named Views with the following columns:

• article_id: The unique identifier for the article.
• author_id: The identifier for the article’s author.
• article_title: The title of the article.
• views: The number of views for a particular record.

Some articles may have multiple records in the table. Your task is to write a SQL query that returns a summary of article views by aggregating the total views for each article. The result should include the article's id, the author's id, the article's title, and the total number of views. Finally, the output should be sorted in ascending order by article_id.

Example

Given the following data in the Views table:

Your SQL query should output:

Hints

1. Aggregation with GROUP BY:
   When you have multiple rows for the same article, you need to aggregate them using the SUM() function on the views column. Group the rows by article_id, author_id, and article_title.

2. Sorting the Results:
   Use the ORDER BY clause to sort the final result in ascending order based on article_id.

Approach

Step 1: Understand the Data

The table Views may contain several records for the same article. Each record represents a number of views for that article. To obtain the total views for an article, you must add up the values in the views column for each unique article.

Step 2: Grouping Records

Use the GROUP BY clause to group the records by the columns that uniquely identify an article. In this case, group by article_id, author_id, and article_title. This ensures that all rows corresponding to a specific article are combined.

Step 3: Summing Views

For each group, calculate the total views using the SUM() function on the views column.

Step 4: Sorting the Output

Finally, sort the results by article_id using the ORDER BY clause so that the output is in ascending order.

SQL Query

SELECT 
    article_id, 
    author_id, 
    article_title, 
    SUM(views) AS views
FROM 
    Views
GROUP BY 
    article_id, 
    author_id, 
    article_title
ORDER BY 
    article_id;

Explanation

• SELECT clause:
  We select the columns article_id, author_id, and article_title to identify each article. The SUM(views) AS views computes the total number of views for each article.

• FROM clause:
  Indicates the source table, which is Views.

• GROUP BY clause:
  Groups the rows by article_id, author_id, and article_title so that the aggregation function SUM() works on each group separately.

• ORDER BY clause:
  Sorts the resulting rows by article_id in ascending order.

Common Pitfalls

• Missing GROUP BY Columns:
  If you aggregate views without including all the identifying columns in the GROUP BY clause, the SQL query will result in an error or incorrect grouping.

• Incorrect Aggregation:
  Failing to use the SUM() function correctly might result in not aggregating the views as required.

Python Code

Edge Cases

1. Empty Table:
   When the Views table has no records, the SQL query should return an empty result set.

2. Single Record:
   If there is only one row in the table, the query should simply return that single record with its view count.

3. Duplicate Records for an Article:
   If multiple records exist for the same article (i.e., identical values for article_id, author_id, and article_title), the query must correctly sum the views from all such rows and return one aggregated record.

4. Articles with Zero Views:
   Articles that have records with 0 views should appear in the output with a total view count of 0 after aggregation.

5. Boundary View Values:
   If the view counts are extremely large, the query should still aggregate them properly without overflow issues (assuming the database handles large integers correctly).

6. Special Characters in Article Titles:
   Article titles might contain spaces or special characters. The query should correctly group records even if article_title values are non-standard or contain special symbols.

Related LeetCode Problems

• Second Highest Salary (LeetCode 176)
  Focuses on using aggregation and subqueries to derive a specific ranking from the data.

• Employees Earning More Than Their Managers (LeetCode 181)
  Utilizes self-joins and aggregation functions to compare data between rows.