Non-distributive conditional types in Typescript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a distributive conditional type

What is the output type of the following TypeScript code?

Typescript

type A<T> = T extends string ? "yes" : "no";
type Result = A<string | number>;

A string | number

B "yes" | "no"

C "no"

D "yes"

💡 Hint

Conditional types distribute over unions by default.

Because conditional types distribute over unions, A becomes A | A, which is "yes" | "no".

❓ Predict Output

intermediate

Output of a non-distributive conditional type using tuple wrapping

What is the output type of this TypeScript code?

Typescript

type B<T> = [T] extends [string] ? "yes" : "no";
type Result = B<string | number>;

A "yes"

B "yes" | "no"

C "no"

D string | number

💡 Hint

Wrapping the type in a tuple disables distribution.

Because T is wrapped in a tuple, the conditional type does not distribute. Since string | number is not assignable to string, the result is "no".

🔧 Debug

advanced

Identify the error in a non-distributive conditional type

Which option causes a TypeScript error when used as a non-distributive conditional type?

Typescript

type C<T> = T extends string ? number : boolean;

A type D<T> = [T] extends string ? number : boolean;

B type D<T> = T extends string ? number : boolean;

C type D<T> = [T] extends [string] ? number : boolean;

D type D<T> = [T] extends [string] ? number : boolean | string;

💡 Hint

Check the syntax of tuple wrapping in conditional types.

Option A is invalid because [T] extends string is a type mismatch; the left side is a tuple type but the right side is a string type, causing a TypeScript error.

🧠 Conceptual

advanced

Why use tuple wrapping in conditional types?

What is the main reason to wrap a type parameter in a tuple in a conditional type?

A To make the conditional type distribute over unions

B To enforce the type parameter to be a tuple

C To convert the type parameter into an array type

D To prevent the conditional type from distributing over unions

💡 Hint

Think about how distribution works in conditional types.

Wrapping the type parameter in a tuple disables the default distribution behavior of conditional types over union types.

❓ Predict Output

expert

Output of a complex non-distributive conditional type with nested unions

What is the type of Result after running this TypeScript code?

Typescript

type E<T> = [T] extends ["a" | "b"] ? "match" : "no match";
type Result = E<"a" | "c">;

A "no match"

B "match" | "no match"

C "match"

D never

💡 Hint

Remember that tuple wrapping disables distribution and the whole union is checked at once.

Because the entire union "a" | "c" is checked at once and it is not assignable to "a" | "b", the result is "no match".

Practice

1. What does a non-distributive conditional type in TypeScript do differently compared to a distributive conditional type?

easy

A. It converts all types to string before evaluation.

B. It always distributes the condition over each member of the union type.

C. It only works with primitive types, not unions.

D. It evaluates the condition on the entire union type as a whole, not on each member separately.

Solution

1. Step 1: Understand distributive conditional types

   Distributive conditional types apply the condition to each member of a union type separately.

2. Step 2: Understand non-distributive conditional types

   Non-distributive conditional types evaluate the condition on the entire union type as a single entity, not distributing over members.

3. Final Answer:

   It evaluates the condition on the entire union type as a whole, not on each member separately. -> Option D

4. Quick Check:

   Non-distributive = whole union check [OK]

Hint: Non-distributive means no splitting union, check whole type [OK]

Common Mistakes:

• Thinking conditional always distributes over unions
• Confusing non-distributive with type narrowing
• Assuming it only works on primitives

2. Which syntax correctly defines a non-distributive conditional type in TypeScript?

easy

A. type NonDist<T> = T | string ? true : false;

B. type NonDist<T> = T extends string ? true : false;

C. type NonDist<T> = [T] extends [string] ? true : false;

D. type NonDist<T> = T & string ? true : false;

Solution

1. Step 1: Recognize non-distributive syntax

   Wrapping the type parameter in a tuple, like [T], prevents distribution over unions.

2. Step 2: Compare options

   type NonDist<T> = [T] extends [string] ? true : false; uses [T] extends [string], which is the correct non-distributive syntax. type NonDist<T> = T extends string ? true : false; distributes over unions. Options C and D are invalid conditional syntax.

3. Final Answer:

   type NonDist<T> = [T] extends [string] ? true : false; -> Option C

4. Quick Check:

   Tuple wrapping disables distribution [OK]

Hint: Wrap type in tuple [T] to prevent distribution [OK]

Common Mistakes:

• Not wrapping type in tuple, causing distribution
• Using invalid conditional syntax
• Confusing union with intersection in condition

3. Given the code:

type Test<T> = T extends string ? 'yes' : 'no';
type Result = Test<'a' | number>;

What is the type of Result?

medium

A. 'yes' | 'no'

B. 'yes'

C. 'no'

D. Error

Solution

1. Step 1: Understand distributive conditional types

   Since T is a union 'a' | number, the condition distributes over each member.

2. Step 2: Evaluate each member

   'a' extends string is true, so 'yes'; number extends string is false, so 'no'. Combined result is 'yes' | 'no'.

3. Final Answer:

   'yes' | 'no' -> Option A

4. Quick Check:

   Distributes over union = 'yes' | 'no' [OK]

Hint: Conditional distributes over union unless wrapped [OK]

Common Mistakes:

• Assuming whole union checked at once
• Expecting only 'yes' or only 'no'
• Confusing union with intersection

4. Identify the error in this non-distributive conditional type:

type NonDist<T> = T extends [string] ? true : false;

medium

A. It incorrectly checks if T extends a tuple containing string, not preventing distribution.

B. It correctly prevents distribution by wrapping T in a tuple.

C. It causes a syntax error because of missing brackets.

D. It always returns false regardless of T.

Solution

1. Step 1: Analyze the condition

   The condition checks if T extends [string], a tuple type, not wrapping T.

2. Step 2: Understand distribution prevention

   To prevent distribution, T itself must be wrapped in a tuple: [T] extends [string]. This code does not do that, so distribution is not prevented.

3. Final Answer:

   It incorrectly checks if T extends a tuple containing string, not preventing distribution. -> Option A

4. Quick Check:

   Wrap T, not string, to prevent distribution [OK]

Hint: Wrap T in tuple, not the type to check [OK]

Common Mistakes:

• Wrapping the wrong side of extends
• Confusing tuple with type wrapper
• Assuming syntax error instead of logic error

5. How can you write a non-distributive conditional type that checks if a union type T is exactly string | number (no more, no less)?

hard

A. type Exact<T> = T extends string | number ? true : false;

B. type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false;

C. type Exact<T> = T extends [string | number] ? true : false;

D. type Exact<T> = [T] extends [string] | [number] ? true : false;

Solution

1. Step 1: Prevent distribution with tuple wrapping

   Wrapping T in a tuple [T] prevents distribution over unions.

2. Step 2: Use mutual assignability checks

   Check [T] extends [string | number] and [string | number] extends [T] to ensure exact match, no extra or missing types.

3. Final Answer:

   type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false; -> Option B

4. Quick Check:

   Tuple wrap + mutual extends = exact type check [OK]

Hint: Wrap T and check both extends directions for exact match [OK]

Common Mistakes:

• Not preventing distribution, causing partial matches
• Checking only one extends direction
• Using incorrect tuple wrapping