Design a key-value store in HLD - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define the main component of a key-value store system.

HLD

class [1]Store:
    def __init__(self):
        self.store = {}

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

ACache

BDatabase

CKeyValue

DFile

Attempts:

3 left

💡 Hint

Common Mistakes

Using generic names like Database or Cache which are broader concepts.

The main component is called a KeyValueStore because it stores data as key-value pairs.

2fill in blank

medium

Complete the code to add a method that stores a value by key.

HLD

def put(self, [1], value):
    self.store[key] = value

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

Adata

Bitem

Cvalue

Dkey

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'item' or 'data' which do not match the internal usage.

The method parameter should be named 'key' to represent the key where the value is stored.

3fill in blank

hard

Fix the error in the method that retrieves a value by key.

HLD

def get(self, [1]):
    return self.store.get(key, None)

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

Akey

Bvalue

Citem

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using parameter names that do not match the variable used inside the method.

The parameter must be named 'key' to match the usage inside the method.

4fill in blank

hard

Fill both blanks to implement a method that deletes a key-value pair safely.

HLD

def delete(self, [1]):
    if [2] in self.store:
        del self.store[key]

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

Akey

Bvalue

Ditem

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names causing errors or undefined variables.

Both blanks should be 'key' to check and delete the key from the store.

5fill in blank

hard

Fill all three blanks to implement a method that returns all keys with values greater than a threshold.

HLD

def keys_above(self, [1]):
    return [k for k, v in self.store.items() if v [2] [3]]

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

Athreshold

B>

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong comparison operators or mismatched variable names.

The method takes a threshold, and returns keys where values are greater than the threshold.

Practice

1. What is the primary purpose of a key-value store in system design?

easy

A. To perform complex relational queries

B. To store large binary files efficiently

C. To save data as pairs for quick lookup

D. To manage user authentication and sessions

Solution

1. Step 1: Understand key-value store basics

   A key-value store saves data as pairs where each key maps to a value for fast retrieval.

2. Step 2: Compare with other storage types

   Unlike relational databases, key-value stores do not support complex queries or file storage.

3. Final Answer:

   To save data as pairs for quick lookup -> Option C

4. Quick Check:

   Key-value store = data pairs [OK]

Hint: Key-value stores focus on pairs, not complex queries [OK]

Common Mistakes:

• Confusing key-value store with relational database
• Thinking it handles large files natively
• Assuming it manages user sessions directly

2. Which of the following is the correct operation to add or update a value in a key-value store?

easy

A. exists(key)

B. put(key, value)

C. delete(key)

D. get(key)

Solution

1. Step 1: Identify operation purpose

   Adding or updating a value requires an operation that sets the value for a key.

2. Step 2: Match operation names

   "put" is commonly used to insert or update key-value pairs; "get" retrieves, "delete" removes, "exists" checks presence.

3. Final Answer:

   put(key, value) -> Option B

4. Quick Check:

   Put = add/update [OK]

Hint: Put means add or update a key-value pair [OK]

Common Mistakes:

• Using get to add data
• Confusing delete with update
• Using exists to insert values

3. Given this pseudo-code for a key-value store:

store = {}
store.put('a', 1)
store.put('b', 2)
store.put('a', 3)
value = store.get('a')

What is the value of value after these operations?

medium

A. 3

B. 2

C. 1

D. None

Solution

1. Step 1: Track put operations

   First, key 'a' is set to 1, then 'b' to 2, then 'a' is updated to 3, overwriting previous value.

2. Step 2: Retrieve the value for 'a'

   The last value assigned to 'a' is 3, so store.get('a') returns 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Last put for 'a' = 3 [OK]

Hint: Last put for a key overwrites previous value [OK]

Common Mistakes:

• Assuming first value stays after update
• Confusing keys 'a' and 'b'
• Thinking get returns None if key exists

4. Consider this code snippet for a key-value store:

store = {}
def get_value(key):
    if key in store:
        return store[key]
    else:
        return None

store.put('x', 10)
print(get_value('x'))

What is the main issue preventing this code from working correctly?

medium

A. The put method is not defined for the dictionary

B. The get_value function returns None incorrectly

C. The key 'x' is not added to the store

D. The print statement syntax is wrong

Solution

1. Step 1: Check dictionary operations

   Python dictionaries do not have a put method; they use assignment like store[key] = value.

2. Step 2: Identify error cause

   Calling store.put('x', 10) will cause an AttributeError because put is undefined.

3. Final Answer:

   The put method is not defined for the dictionary -> Option A

4. Quick Check:

   Dicts use assignment, not put [OK]

Hint: Dictionaries use assignment, not put() method [OK]

Common Mistakes:

• Assuming put exists on dict
• Ignoring error from undefined method
• Thinking get_value logic is faulty

5. You want to design a scalable key-value store that handles millions of requests per second. Which design choice best supports this goal?

hard

A. Use a single in-memory dictionary on one server

B. Store all data on a single disk-based database

C. Use a relational database with complex joins

D. Partition data across multiple servers using consistent hashing

Solution

1. Step 1: Understand scalability needs

   Handling millions of requests requires distributing load and data to avoid bottlenecks.

2. Step 2: Evaluate design options

   A single in-memory dictionary or disk-based DB limits capacity; relational DB with joins is slow for key-value access. Consistent hashing partitions data evenly across servers, enabling horizontal scaling.

3. Final Answer:

   Partition data across multiple servers using consistent hashing -> Option D

4. Quick Check:

   Consistent hashing = scalable partitioning [OK]

Hint: Distribute data with consistent hashing for scalability [OK]

Common Mistakes:

• Relying on single server limits throughput
• Using disk-based DB slows access
• Choosing relational DB for simple key-value