Mapped type with template literals in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a mapped type that adds 'get' prefix to each key.

Typescript

type Getters<T> = {
  [K in keyof T as `get$[1]`]: () => T[K];
};

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

Astring

BUppercase<K>

CLowercase<K>

DCapitalize<string & K>

Attempts:

3 left

💡 Hint

Common Mistakes

Using Uppercase or Lowercase changes all letters, not just the first.

Forgetting to intersect with string & K causes type errors.

The Capitalize utility capitalizes the first letter of the key, so the template literal forms keys like 'getName'.

2fill in blank

medium

Complete the mapped type to create setter methods with 'set' prefix and capitalized keys.

Typescript

type Setters<T> = {
  [K in keyof T as `set$[1]`]: (value: T[K]) => void;
};

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

ACapitalize<string & K>

BUppercase<string & K>

CLowercase<K>

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using Uppercase capitalizes all letters, which is not desired here.

Not intersecting with string & K causes type errors.

Using Capitalize capitalizes the first letter of the key, forming keys like 'setAge'.

3fill in blank

hard

Fix the error in the mapped type to correctly create keys with 'is' prefix and capitalized keys.

Typescript

type BooleanFlags<T> = {
  [K in keyof T as `is$[1]`]: boolean;
};

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

AUppercase<K>

BCapitalize<string & K>

CLowercase<string & K>

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using Uppercase capitalizes all letters, which is not correct here.

Not intersecting with string & K causes errors.

The Capitalize utility capitalizes the first letter of the key, so keys become like 'isActive'.

4fill in blank

hard

Fill both blanks to create a mapped type that adds 'on' prefix and capitalizes keys for event handlers.

Typescript

type EventHandlers<T> = {
  [K in keyof T as `on$[1]`]: (event: T[K]) => void;
};

// Usage example:
// const handlers: EventHandlers<{ click: MouseEvent, keypress: KeyboardEvent }>;

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

ACapitalize<string & K>

BUppercase<K>

CLowercase<K>

Dstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using Uppercase instead of Capitalize changes all letters.

Not using string in the template literal causes type errors.

The first blank uses Capitalize to capitalize the key's first letter. The second blank is string as a fallback type in the template literal.

5fill in blank

hard

Fill all three blanks to create a mapped type that adds 'with' prefix, capitalizes keys, and appends 'Value' suffix.

Typescript

type WithValue<T> = {
  [K in keyof T as `with$[1]$[2]`]: T[K];
};

// Example usage:
// type Example = WithValue<{ name: string, age: number }>;
// Result keys: 'withNameValue', 'withAgeValue'

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

ACapitalize<string & K>

BValue

Cstring

DUppercase<K>

Attempts:

3 left

💡 Hint

Common Mistakes

Using Uppercase instead of Capitalize changes all letters.

Forgetting to add the suffix 'Value'.

Not including string in the template literal causes errors.

The first blank capitalizes the key's first letter. The second blank adds the suffix 'Value'. The third blank is string to ensure the template literal works correctly.

Practice

1.

What does a mapped type with template literals in TypeScript allow you to do?

Example: { [K in keyof T as `get${Capitalize}`]: () => T[K] }

easy

A. Create new variables at runtime

B. Rename keys dynamically in object types

C. Change values of an object at runtime

D. Execute functions based on key names

Solution

1. Step 1: Understand mapped types

   Mapped types create new object types by transforming keys or values of an existing type.

2. Step 2: Role of template literals

   Template literals let you rename keys dynamically by combining strings and key names.

3. Final Answer:

   Rename keys dynamically in object types -> Option B

4. Quick Check:

   Mapped type + template literals = dynamic key renaming [OK]

Hint: Mapped types + template literals = dynamic key renaming [OK]

Common Mistakes:

• Confusing runtime behavior with type transformations
• Thinking it changes object values instead of keys
• Assuming it creates variables or runs code

2.

Which of the following is the correct syntax to rename keys using a mapped type with template literals?

type RenameKeys<T> = { [K in keyof T as ???]: T[K] }

easy

A. `new_${string & K}`

B. `new_${K}`

C. `new_${number & K}`

D. `new_${any & K}`

Solution

1. Step 1: Understand key remapping syntax

   Use as with template literals and intersect key type with string to ensure keys are strings.

2. Step 2: Check each option

   `new_${string & K}` uses string & K which is correct; others use incorrect types like number or any.

3. Final Answer:

   `new_${string & K}` -> Option A

4. Quick Check:

   Use string & K for safe key renaming [OK]

Hint: Use string & K to rename keys safely in template literals [OK]

Common Mistakes:

• Using K directly without string intersection
• Using number or any instead of string & K
• Missing backticks for template literals

3.

What is the resulting type of the following code?

type User = { name: string; age: number };
type PrefixedUser = { [K in keyof User as `get${Capitalize}`]: () => User[K] };

medium

A. { name: () => string; age: () => number }

B. { getName: string; getAge: number }

C. { getname: () => string; getage: () => number }

D. { getName: () => string; getAge: () => number }

Solution

1. Step 1: Analyze key transformation

   Keys 'name' and 'age' are capitalized and prefixed with 'get', becoming 'getName' and 'getAge'.

2. Step 2: Analyze value transformation

   Values become functions returning the original property types: () => string and () => number.

3. Final Answer:

   { getName: () => string; getAge: () => number } -> Option D

4. Quick Check:

   Mapped keys capitalized + function values = { getName: () => string; getAge: () => number } [OK]

Hint: Capitalize keys and prefix with 'get' for function types [OK]

Common Mistakes:

• Not capitalizing keys
• Returning values instead of functions
• Using original keys without prefix

4.

Identify the error in this mapped type using template literals:

type Flags = { isActive: boolean; isAdmin: boolean };
type PrefixedFlags = { [K in keyof Flags as `flag_${K}`]: Flags[K] };

medium

A. Key remapping must use Capitalize utility

B. Cannot use template literals in mapped types

C. Missing string intersection with K in template literal

D. Flags[K] is invalid syntax

Solution

1. Step 1: Check key remapping syntax

   Template literal keys must intersect K with string: string & K to ensure keys are strings.

2. Step 2: Identify missing intersection

   The code uses `flag_${K}` without string & K, causing a type error.

3. Final Answer:

   Missing string intersection with K in template literal -> Option C

4. Quick Check:

   Use string & K in template literals to fix error [OK]

Hint: Always intersect K with string in template literals [OK]

Common Mistakes:

• Omitting string & K intersection
• Thinking Capitalize is mandatory
• Believing template literals can't be used here

5.

Given this type, how would you create a mapped type that prefixes keys with on and capitalizes them, but only for keys starting with handle?

type Handlers = { handleClick: () => void; handleHover: () => void; reset: () => void };

hard

A. type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] };

B. type Prefixed = { [K in keyof Handlers as `on${Capitalize}`]: Handlers[K] };

C. type Prefixed = { [K in keyof Handlers as K extends string ? `on${Capitalize<K>}` : never]: Handlers[K] };

D. type Prefixed = { [K in keyof Handlers as K extends `on${string}` ? `handle${Capitalize}` : never]: Handlers[K] };

Solution

1. Step 1: Filter keys starting with 'handle'

   Use conditional type K extends `handle${string}` to select only keys starting with 'handle'.

2. Step 2: Rename keys with 'on' prefix and capitalize

   Use template literal `on${Capitalize}` to rename keys accordingly.

3. Step 3: Assign original value types

   Keep the original function types from Handlers[K].

4. Final Answer:

   type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; -> Option A

5. Quick Check:

   Conditional key remapping with prefix and capitalize = type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; [OK]

Hint: Use conditional types to filter keys before renaming [OK]

Common Mistakes:

• Not filtering keys by prefix
• Applying rename to all keys
• Incorrect conditional syntax