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".