Cache eviction policies (LRU, LFU, TTL) in HLD - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to identify the cache eviction policy that removes the least recently used item.

HLD

if cache.is_full():
    cache.evict_policy = "[1]"

Drag options to blanks, or click blank then click option'

ATTL

BLFU

CLRU

DFIFO

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing LRU with LFU or TTL.

The LRU (Least Recently Used) policy evicts the item that was used the longest time ago.

2fill in blank

medium

Complete the code to set the cache eviction policy that removes the least frequently used item.

HLD

if cache.needs_eviction():
    cache.policy = "[1]"

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ARandom

BLRU

CTTL

DLFU

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing LFU with LRU or TTL.

LFU (Least Frequently Used) evicts the item with the lowest access frequency.

3fill in blank

hard

Fix the error in the code to correctly implement a time-based cache eviction policy.

HLD

cache.set_eviction_policy("[1]")
cache.expiry_time = 300  # seconds

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ATTL

BLRU

CLFU

DFIFO

Attempts:

3 left

💡 Hint

Common Mistakes

Using LRU or LFU where TTL is needed.

TTL (Time To Live) evicts items after a set time expires.

4fill in blank

hard

Fill both blanks to complete the dictionary comprehension that filters cache entries by TTL and sorts by usage frequency.

HLD

filtered_cache = {k: v for k, v in cache.items() if v.expiry > [1]
sorted_cache = sorted(filtered_cache.items(), key=lambda item: item[1].[2])

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Acurrent_time

Blast_used

Cfrequency

Dexpiry

Attempts:

3 left

💡 Hint

Common Mistakes

Using expiry instead of current_time for filtering.

Sorting by last_used instead of frequency.

We filter entries where expiry is greater than current time, then sort by frequency for LFU.

5fill in blank

hard

Fill all three blanks to complete the cache eviction function that uses LRU policy with TTL expiration.

HLD

def evict_cache(cache):
    now = [1]
    valid_items = {k: v for k, v in cache.items() if v.expiry > now}
    lru_key = min(valid_items, key=lambda k: valid_items[k].[2])
    del cache[lru_key]
    return cache

current_time = [3]

Drag options to blanks, or click blank then click option'

Atime.time()

Blast_accessed

Cdatetime.now()

Dexpiry

Attempts:

3 left

💡 Hint

Common Mistakes

Using expiry instead of last_accessed for LRU.

Using datetime.now() inconsistently.

The function uses current time from time.time(), filters valid items by expiry, finds the least recently accessed key, and deletes it.

Practice

1. Which cache eviction policy removes the data that was used the longest time ago?

easy

A. Random Replacement

B. Least Frequently Used (LFU)

C. Time To Live (TTL)

D. Least Recently Used (LRU)

Solution

1. Step 1: Understand LRU policy

   LRU removes the cache entry that was accessed the longest time ago, prioritizing recent usage.

2. Step 2: Compare with other policies

   LFU removes least used items by frequency, TTL removes expired items by time, Random Replacement is arbitrary.

3. Final Answer:

   Least Recently Used (LRU) -> Option D

4. Quick Check:

   Oldest access time = LRU [OK]

Hint: LRU = remove oldest used data [OK]

Common Mistakes:

• Confusing LRU with LFU
• Thinking TTL removes based on usage
• Assuming random removal is LRU

2. Which of the following is a correct description of the TTL cache eviction policy?

easy

A. Removes data after a fixed expiration time

B. Removes the least recently used data

C. Removes the least frequently used data

D. Removes data randomly when cache is full

Solution

1. Step 1: Define TTL policy

   TTL (Time To Live) removes cache entries after their set expiration time passes.

2. Step 2: Contrast with other policies

   LRU removes oldest used, LFU removes least used by frequency, random is arbitrary.

3. Final Answer:

   Removes data after a fixed expiration time -> Option A

4. Quick Check:

   TTL = expiration time removal [OK]

Hint: TTL means data expires after set time [OK]

Common Mistakes:

• Mixing TTL with LRU or LFU
• Thinking TTL depends on usage frequency
• Assuming TTL removes randomly

3. Consider a cache with the following access pattern: keys A, B, A, C, B, A. Using LFU eviction, which key will be evicted when the cache is full and a new key D is added?

medium

A. Key C

B. Key A

C. Key B

D. Key D

Solution

1. Step 1: Count frequency of each key

   Key A accessed 3 times, B accessed 2 times, C accessed 1 time.

2. Step 2: Identify LFU eviction candidate

   LFU removes the least frequently used key, which is C with frequency 1.

3. Final Answer:

   Key C -> Option A

4. Quick Check:

   Lowest frequency = C [OK]

Hint: LFU evicts lowest frequency key [OK]

Common Mistakes:

• Evicting most recent key instead of least frequent
• Confusing frequency with recency
• Evicting key D which is new

4. A cache using TTL eviction is not removing expired entries as expected. Which of the following is the most likely cause?

medium

A. The TTL value is set too low

B. The cache uses LRU instead of TTL

C. The cache does not check expiration before access

D. The cache size is too large

Solution

1. Step 1: Understand TTL eviction mechanism

   TTL eviction requires checking if entries have expired before returning or during cleanup.

2. Step 2: Identify why expired entries remain

   If cache does not check expiration on access or cleanup, expired entries stay indefinitely.

3. Final Answer:

   The cache does not check expiration before access -> Option C

4. Quick Check:

   Expiration check missing = no TTL eviction [OK]

Hint: TTL needs expiration checks to remove entries [OK]

Common Mistakes:

• Assuming low TTL causes no eviction
• Confusing TTL with LRU behavior
• Thinking cache size affects TTL eviction

5. You are designing a cache for a news website where articles are frequently accessed shortly after publishing but rarely after a few days. Which eviction policy is best suited to keep fresh content while removing stale data efficiently?

hard

A. Least Recently Used (LRU)

B. Time To Live (TTL)

C. Random Replacement

D. Least Frequently Used (LFU)

Solution

1. Step 1: Analyze data usage pattern

   Articles are popular shortly after publishing, then become stale after days.

2. Step 2: Match eviction policy to pattern

   TTL removes data after expiration time, ideal for removing stale articles automatically.

3. Step 3: Compare with other policies

   LRU and LFU focus on usage, but TTL ensures freshness by time. Random is inefficient.

4. Final Answer:

   Time To Live (TTL) -> Option B

5. Quick Check:

   Freshness by expiration time = TTL [OK]

Hint: Use TTL for time-based freshness eviction [OK]

Common Mistakes:

• Choosing LRU or LFU ignoring data staleness
• Assuming random eviction is efficient
• Not considering time-based expiration