Partition assignment in Kafka - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Kafka partition assignment code snippet?

Consider a Kafka consumer group with 3 consumers and a topic with 6 partitions. The default RangeAssignor is used. What partitions will consumer 2 (index 1) be assigned?

Kafka

consumers = ['consumer0', 'consumer1', 'consumer2']
partitions = list(range(6))

# RangeAssignor assigns partitions in contiguous blocks
# Each consumer gets partitions in a range

assignments = {}
num_consumers = len(consumers)
num_partitions = len(partitions)

partitions_per_consumer = num_partitions // num_consumers
remainder = num_partitions % num_consumers

start = 0
for i, consumer in enumerate(consumers):
    extra = 1 if i < remainder else 0
    end = start + partitions_per_consumer + extra
    assignments[consumer] = partitions[start:end]
    start = end

print(assignments['consumer1'])

A [2, 3, 4]

B [1, 2]

C [2, 3]

D [3, 4]

💡 Hint

Remember that RangeAssignor distributes partitions in contiguous blocks and the first consumers get the extra partitions if partitions don't divide evenly.

With 6 partitions and 3 consumers, each consumer gets 2 partitions. Since 6 % 3 = 0, no extra partitions are assigned. The partitions are split as: consumer0 -> [0,1], consumer1 -> [2,3], consumer2 -> [4,5]. So consumer1 gets partitions [2,3]. The correct answer is C: [2,3].

❓ Predict Output

intermediate

What error does this Kafka custom partition assignment code raise?

This code tries to assign partitions manually but has a bug. What error will it raise?

Kafka

consumers = ['c1', 'c2']
partitions = [0, 1, 2]

assignments = {}
for i, consumer in enumerate(consumers):
    assignments[consumer] = partitions[i*2:(i+1)*2]

print(assignments)

A No error, output: {'c1': [0, 1], 'c2': [2]}

B IndexError: list index out of range

C KeyError: 'c2'

D TypeError: unsupported operand type(s)

💡 Hint

Check how slicing works in Python when the slice end index exceeds list length.

Python slicing does not raise IndexError if the slice end index is beyond the list length. It just returns the available elements. So consumer 'c2' gets partition [2]. No error occurs.

❓ Predict Output

advanced

What is the output of this Kafka StickyAssignor partition assignment simulation?

Given 4 consumers and 7 partitions, the StickyAssignor tries to balance partitions evenly. After one rebalance, what partitions does consumer 'c3' get?

Kafka

consumers = ['c0', 'c1', 'c2', 'c3']
partitions = list(range(7))

# StickyAssignor tries to assign partitions evenly
# Each consumer gets either 1 or 2 partitions

assignments = {
    'c0': [0, 1],
    'c1': [2, 3],
    'c2': [4, 5],
    'c3': [6]
}

print(assignments['c3'])

A [6]

B [5, 6]

C [4, 6]

D [3, 6]

💡 Hint

StickyAssignor tries to keep previous assignments and balance partitions as evenly as possible.

With 7 partitions and 4 consumers, three consumers get 2 partitions and one gets 1 partition. Here, 'c3' gets only partition 6.

🧠 Conceptual

advanced

Which option best describes Kafka's CooperativeStickyAssignor behavior?

Kafka's CooperativeStickyAssignor allows incremental rebalancing. Which statement is true about its partition assignment?

A It reassigns all partitions to all consumers on every rebalance.

B It assigns partitions based on consumer CPU usage.

C It assigns partitions randomly without considering previous assignments.

D It only moves partitions that need to be reassigned, minimizing partition movement.

💡 Hint

Think about how incremental rebalancing helps reduce disruption.

CooperativeStickyAssignor moves only the partitions that must be reassigned, reducing the number of partitions moved during rebalances.

❓ Predict Output

expert

What is the output of this Kafka partition assignment code with a custom assignor?

This code simulates a custom assignor that assigns partitions to consumers in a round-robin fashion. What is the assignment for consumer 'c1'?

Kafka

consumers = ['c0', 'c1', 'c2']
partitions = list(range(8))

assignments = {c: [] for c in consumers}

for i, p in enumerate(partitions):
    consumer = consumers[i % len(consumers)]
    assignments[consumer].append(p)

print(assignments['c1'])

A [1, 3, 5, 7]

B [1, 4, 7]

C [1, 4, 6]

D [1, 5, 7]

💡 Hint

Round-robin assigns partitions one by one cycling through consumers.

Partitions assigned to 'c1' are those where partition index modulo 3 equals 1: partitions 1, 4, and 7.