Denormalization trade-offs in MongoDB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to insert a denormalized document with embedded address.

MongoDB

db.customers.insertOne({ name: "Alice", address: [1] })

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

A[ "123 Main St", "Springfield" ]

B{ street: "123 Main St", city: "Springfield" }

C"123 Main St, Springfield"

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using an array or string instead of an object for the address.

Setting address to null which loses the data.

Embedding the address as an object allows denormalization by storing related data together.

2fill in blank

medium

Complete the code to update the embedded address city field in a denormalized document.

MongoDB

db.customers.updateOne({ name: "Alice" }, { $set: { "address.[1]": "Shelbyville" } })

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

Astreet

Bcountry

Czipcode

Dcity

Attempts:

3 left

💡 Hint

Common Mistakes

Updating the wrong field like street or zipcode.

Not using dot notation to access nested fields.

To update the city inside the embedded address, use the key "city" in the dot notation.

3fill in blank

hard

Fix the error in the query to find customers with a specific city in their embedded address.

MongoDB

db.customers.find({ "address": [1] })

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

A{ city: "Shelbyville" }

B"Shelbyville"

C["Shelbyville"]

Dnull

Attempts:

3 left

💡 Hint

Common Mistakes

Using a string or array instead of an object for the embedded document.

Using null which matches no documents.

The query must match the embedded document structure, so use an object with the city field.

4fill in blank

hard

Fill both blanks to create a query that finds customers with an embedded address city of "Shelbyville".

MongoDB

db.customers.find({ "[1]": "[2]" })

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

Aaddress.city

BShelbyville

Caddress.street

DSpringfield

Attempts:

3 left

💡 Hint

Common Mistakes

Querying the wrong nested field like address.street.

Using the wrong city name.

Use dot notation "address.city" to query the nested city field and match the value "Shelbyville".

5fill in blank

hard

Fill all three blanks to update the embedded address zipcode to "12345" for customer "Alice".

MongoDB

db.customers.updateOne({ name: "[1]" }, { $set: { "address.[2]": "[3]" } })

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

AAlice

Bzipcode

C12345

Dcity

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong customer name.

Updating the wrong field like city instead of zipcode.

Setting an incorrect zipcode value.

Update the zipcode field inside the embedded address for the customer named "Alice" to "12345".

Practice

1. What is a main advantage of denormalization in MongoDB?

easy

A. Faster read operations by storing related data together

B. Simpler updates with no data duplication

C. Reduced storage space by avoiding repeated data

D. Automatic data consistency without extra effort

Solution

1. Step 1: Understand denormalization purpose

   Denormalization stores related data together to speed up reading.

2. Step 2: Compare options with this purpose

   Only Faster read operations by storing related data together mentions faster reads by storing data together, which matches denormalization benefits.

3. Final Answer:

   Faster read operations by storing related data together -> Option A

4. Quick Check:

   Denormalization = Faster reads [OK]

Hint: Denormalization speeds up reads by grouping data [OK]

Common Mistakes:

• Confusing denormalization with normalization benefits
• Thinking denormalization reduces storage space
• Assuming denormalization simplifies updates

2. Which MongoDB document structure shows denormalization?

easy

A. { _id: 1, name: 'Alice' }, { _id: 101, total: 50 }

B. { _id: 1, name: 'Alice', orders: [{id: 101, total: 50}] }

C. { _id: 1, name: 'Alice' }, orders: 101

D. { _id: 1, orders: [101] }

Solution

1. Step 1: Identify denormalized data structure

   Denormalization stores related data inside one document, like embedding orders inside a user document.

2. Step 2: Check options for embedded data

   { _id: 1, name: 'Alice', orders: [{id: 101, total: 50}] } embeds orders array inside user document, showing denormalization.

3. Final Answer:

   { _id: 1, name: 'Alice', orders: [{id: 101, total: 50}] } -> Option B

4. Quick Check:

   Embedded orders = Denormalization [OK]

Hint: Look for embedded arrays inside documents [OK]

Common Mistakes:

• Choosing separate documents as denormalized
• Confusing references with embedded data
• Ignoring array embedding as denormalization

3. Given this denormalized document:

{ _id: 1, name: 'Bob', address: { city: 'NY', zip: '10001' } }

What happens if Bob moves to 'LA' and you update only the city in this document?

medium

A. Update fails due to embedded document structure

B. All references to Bob's city update automatically

C. Only this document's city changes; duplicates elsewhere stay old

D. MongoDB merges old and new city values

Solution

1. Step 1: Understand update impact in denormalization

   Denormalization duplicates data, so updating one copy doesn't affect others.

2. Step 2: Analyze update effect on embedded address

   Updating city in one document changes only that document; duplicates elsewhere remain unchanged.

3. Final Answer:

   Only this document's city changes; duplicates elsewhere stay old -> Option C

4. Quick Check:

   Update one copy, duplicates unchanged [OK]

Hint: Updates affect only one copy in denormalized data [OK]

Common Mistakes:

• Assuming all duplicates update automatically
• Thinking embedded documents prevent updates
• Believing MongoDB merges conflicting values

4. You have duplicated user phone numbers in multiple documents due to denormalization. Which is a common problem you might face?

medium

A. Updating phone number requires changing all documents manually

B. MongoDB automatically syncs all duplicates on update

C. Phone numbers cannot be stored in embedded documents

D. Denormalization prevents querying phone numbers

Solution

1. Step 1: Identify update challenges with duplicated data

   Duplicated data means updates must be applied to every copy to keep consistency.

2. Step 2: Evaluate options about update behavior

   Only Updating phone number requires changing all documents manually correctly states that manual updates to all duplicates are needed.

3. Final Answer:

   Updating phone number requires changing all documents manually -> Option A

4. Quick Check:

   Manual update needed for duplicates [OK]

Hint: Duplicates need manual updates to stay consistent [OK]

Common Mistakes:

• Believing MongoDB auto-syncs duplicates
• Thinking denormalization forbids embedded fields
• Assuming denormalization blocks queries

5. You want to design a MongoDB schema for a blog where posts have many comments. You expect many reads but few comment edits. Which denormalization approach is best?

hard

A. Duplicate comments in both posts and a separate collection

B. Store comments in a separate collection and reference post IDs

C. Store only comment IDs inside posts without comment data

D. Embed all comments inside each post document

Solution

1. Step 1: Analyze read vs update frequency

   Many reads and few updates favor denormalization to speed up reads.

2. Step 2: Choose schema matching denormalization benefits

   Embedding comments inside posts (Embed all comments inside each post document) stores related data together for fast reads.

3. Step 3: Consider update cost

   Few comment edits mean duplicated data update cost is low, so embedding is suitable.

4. Final Answer:

   Embed all comments inside each post document -> Option D

5. Quick Check:

   Many reads + few updates = Embed data [OK]

Hint: Embed when reads >> updates for faster access [OK]

Common Mistakes:

• Choosing references despite many reads
• Duplicating data unnecessarily increasing update cost
• Ignoring update frequency in design