Understanding the Problem and Background

In this article, we will delve into a technical problem involving BigQuery, a cloud-based data warehousing service by Google. The question revolves around finding the maximum overlap of array elements across rows for each user in a table.

BigQuery is a fully managed enterprise data warehouse service that makes it easy to analyze large datasets without requiring significant technical expertise or infrastructure knowledge. It allows users to easily move between Hadoop, cloud storage, and other tools and programming languages.

The problem presented involves a table with a user, row ID, and an array of elements (data). The task is to write a query that finds the maximum overlap of these array elements across rows for each unique user in the table.

Breaking Down the Problem

To solve this problem, we need to identify common elements between two rows belonging to the same user. We will do this by joining two instances of the temporary table created from the original data.

We must consider how to efficiently compare the common elements and how to determine which row contains more common elements with another row for a given user.

Approaches to Solving the Problem

There are several ways to approach this problem. One method involves comparing each element in an array between two rows belonging to the same user and counting the occurrences of those common elements. However, this approach may not be efficient if there are many repeated values or arrays.

Another approach is to use a hash-based solution, where we use an array of hashes as keys for our temporary tables.

The Solution

We will implement the solution using BigQuery Standard SQL and utilize a join operation between two temporary tables created from the original data.

Creating Temporary Tables

To create these temporary tables, we first need to split the data column into individual elements by using the UNNEST function in BigQuery. This allows us to compare each element between rows belonging to the same user.

with temp as (
  select row_id, user, el
  from your_table, unnest(data) el
)

In this step, we create a temporary table (temp) that includes all elements in the data column and their corresponding row_id and user.

Joining Temporary Tables

To find common elements between two rows belonging to the same user, we join the temporary tables on both the user column and the el column.

join temp t2
on t1.user = t2.user
and t1.row_id < t2.row_id
and t1.el = t2.el

In this step, we join two instances of the temporary table (temp) on both the user and el columns. This allows us to find common elements between rows belonging to the same user.

Grouping and Aggregating

We need to group the joined results by user, t1.row_id, and t2.row_id to ensure we count only unique overlaps.

group by user, t1.row_id, t2.row_id

In this step, we group the joined results by all three columns. This ensures that each row in our final result set is counted correctly.

Finding the Maximum Overlap

We use the ROW_NUMBER() function with an over clause to rank rows within each user partition by their overlap count in descending order.

qualify row_number() over(partition by user order by count(*) desc) = 1    

In this step, we qualify our results so that only the first row for each unique user is returned. This means we’re left with one row per user containing the maximum overlap of array elements across rows.

Final Query

Now, let’s combine all these steps into a single query and discuss its output:

with temp as (
  select row_id, user, el
  from your_table, unnest(data) el
)
select t1.user, array_agg(t1.el) as max_intersection
from temp t1
join temp t2
on t1.user = t2.user
and t1.row_id < t2.row_id
and t1.el = t2.el
group by user, t1.row_id, t2.row_id
qualify row_number() over(partition by user order by count(*) desc) = 1    

Conclusion

In this article, we discussed a technical problem involving finding the maximum overlap of array elements across rows for each unique user in a BigQuery table. We also walked through an example query that implements the solution and explained how it works.

The provided code is well-structured and should be easy to understand and replicate. We hope you found this tutorial informative and helpful. If you have any questions or need further clarification, please don’t hesitate to ask.

Last modified on 2025-03-18