A new feature stores user events in DynamoDB. Each event must be fetched by user_id and sorted by event_time. The team expects many different users and wants to avoid a single hot partition. Which partition key design is best?

Use a constant partition key value (for example, partition_key='events') and store user_id as an attribute.

A constant partition key would concentrate all events into one partition, creating a hot partition risk. Queries by user_id would not be efficient without redesign and could exceed throughput limits. It defeats horizontal distribution.

Use event_time as the partition key and user_id as an attribute to query later.

Partitioning by event_time can still create hotspots if many events share the same time buckets. It also makes queries by user_id inefficient because user_id is not the partition key. The access pattern says user_id must drive retrieval.

Use a randomly generated UUID as the partition key and query by user_id using a full table scan.

A UUID partition key prevents efficient direct lookups by user_id and forces scans, which are slow and costly at scale. Full table scans do not meet typical performance needs. The design should align partition key with the query pattern.

What does this SAA-C03 question test?

Static NAT maps one inside address to one outside address.

What is the correct answer to this question?

The correct answer is: Use user_id as the partition key and event_time as the sort key. — DynamoDB performance depends heavily on the choice of partition key aligned to access patterns. Because events must be fetched by user_id and returned in event_time order, user_id should be the partition key and event_time should be the sort key. That supports efficient queries for a single user and enables sorted results within that partition. Using a constant partition key would concentrate all traffic into one partition and create hot spots, while UUID partitioning or time-only partitioning makes user-based queries inefficient. Why others are wrong: Option A creates a hot partition because all items share one partition key. Option C partitions by event_time, but the required query is user-centric; queries would be inefficient and could amplify hotspots. Option D forces full table scans to find a user’s events, which is generally unacceptable for performance at scale. Option B directly matches the stated query pattern and distributes load across partitions.

What should I do if I get this SAA-C03 question wrong?

Then try more questions from the same exam bank and focus on understanding why the wrong options are tempting.