Partition ordering guarantees in Kafka - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to produce messages to a specific partition in Kafka.

Kafka

producer.send(new ProducerRecord<>("topic", [1], "key", "value"));

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A0

Btrue

Cnull

D"partition"

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string instead of an integer for partition number.

Partition number must be an integer like 0 to send messages to a specific partition.

2fill in blank

medium

Complete the code to ensure message ordering within a partition by setting the correct key.

Kafka

producer.send(new ProducerRecord<>("topic", [1], "value"));

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Anull

B"timestamp"

C"userID"

D"partitionKey"

Attempts:

3 left

💡 Hint

Common Mistakes

Using null key causes round-robin partitioning, losing ordering guarantees.

Using a consistent key like "userID" ensures messages with the same key go to the same partition, preserving order.

3fill in blank

hard

Fix the error in the code to consume messages in order from a single partition.

Kafka

consumer.assign(Collections.singleton(new TopicPartition("topic", [1])));

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Anull

B0

C"partition"

D"0"

Attempts:

3 left

💡 Hint

Common Mistakes

Passing partition as a string instead of integer.

The partition number must be an integer like 0, not a string or null.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps partitions to their last committed offsets.

Kafka

offsets = {tp.partition(): [1] for tp in consumer.assignment() if tp.partition() [2] 0}

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Aconsumer.committed(tp)

Bconsumer.position(tp)

C>

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using consumer.position() instead of committed offsets.

Using '==' instead of '>'.

We get the committed offset for each partition and filter partitions with number greater than 0.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps partition numbers to their latest offsets if offset is greater than 100.

Kafka

latest_offsets = { [1]: [2] for tp in consumer.assignment() if [3] > 100}

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Atp.partition()

Bconsumer.position(tp)

Dtp.offset()

Attempts:

3 left

💡 Hint

Common Mistakes

Using tp.offset() which is invalid.

Filtering with wrong offset value.

We map partition numbers to their current positions (latest offsets) and filter those with offsets greater than 100.