500+ TypeScript Interview Questions with Answers 2026

Object-Oriented Programming (18%): Concrete and abstract Classes, prototype and class Inheritance, runtime Polymorphism, access modifiers for Encapsulation (public, private, protected), and structural Abstraction.

Advanced TypeScript Concepts (15%): Generic constraints, custom Type Guards (is operator), complex Conditional Types, Mapped Types with modifiers, and Template Literal Types.

Error Handling and Debugging (12%): Custom Error subclasses, structural Try-Catch Blocks, compiler-level debugging, structured runtime Logging, and defensive error-handling best practices.

TypeScript and JavaScript Interoperability (10%): Declaration files (.d.ts), handling untyped third-party libraries, Module Systems (ESM vs CommonJS), Type Conversion, and strict structural interop patterns.

Design Patterns and Principles (8%): Implementing typesafe Creational, Structural, and Behavioral patterns (Singleton, Factory, Observer), strict SOLID Principles, and Separated Interface Patterns.

Testing and Best Practices (7%): Strongly typed Unit Testing, Integration Testing paradigms, configuration of modern Testing Frameworks, linting rules, and scalable Code Organization.

TypeScript Ecosystem and Tools (10%): Advanced tsconfig.json compiler options (strict, noImplicitAny, exactOptionalPropertyTypes), TypeScript Compiler (tsc) mechanics, ESLint, Prettier formatters, and custom IDE Integrations.

B) never

C) string | number

D) unknown

E) any

F) boolean

Why it is correct: When conditional types check a naked type parameter (T), they automatically distribute across union types. The evaluation splits into three distinct operations: (string extends boolean ? never : string) | (number extends boolean ? never : number) | (boolean extends boolean ? never : boolean). This simplifies cleanly to string | number | never. Since never represents the empty set in a union, it drops out completely, leaving string | number.

Why alternative options are incorrect:

Option B is incorrect: The type only resolves to never if every single constituent member of the union satisfies the conditional extension block.

Option D is incorrect: unknown represents a safe top type, which is not generated by this conditional evaluation pattern.

Option E is incorrect: any disables strict compiler checking entirely and does not result from this deterministic filtering logic.

Option F is incorrect: boolean is explicitly mapped to never by the true branch of the conditional type operation.

B) type Clean<T> = { -readonly [P in keyof T]-?: NonNullable<T[P]> };

C) type Clean<T> = { +readonly [P in keyof T]+?: T[P] };

D) type Clean<T> = { [P in keyof T]: T[P] | null };

E) type Clean<T> = { [P keyof T]: T[P] };

F) type Clean<T> = { mut [P in keyof T]: Required<T[P]> };

Why it is correct: TypeScript allows mapped types to strip modifiers using the prefix operator -. Using -readonly explicitly removes the immutable constraint from the key, and using -? completely removes the optional flag, making the property strictly required. Wrapping the value in the built-in NonNullable<T> utility strips out both null and undefined.

Why alternative options are incorrect:

Option C is incorrect: The + operator explicitly enforces or adds modifiers, which acts as the exact opposite of what the developer requires.

Option D is incorrect: This does not address the readonly status and intentionally introduces potential null bugs to the object map.

Option E is incorrect: This syntax lacks the mandatory in keyword, which causes an immediate compiler parsing error.

Option F is incorrect: There is no mut keyword in the TypeScript type system syntax; modifier removal relies exclusively on the minus operator.

B) It blocks the assignment because method parameters are strictly invariant across all assignment operations.

C) It allows the assignment cleanly because function parameter types evaluate contravariantly under strict checks.

D) It permits the code to run but forces a implicit type conversion down to any at runtime.

E) It triggers a fatal runtime warning because union parameters cannot accept standalone primitive types.

F) It allows the assignment only if the developer updates the tsconfig.json to switch off basic module tracking.

Why it is correct: Under strictFunctionTypes, function arguments evaluate contravariantly (the subtyping direction is reversed). A function that can process a broader range of inputs (string | number) can safely substitute for a function expecting a narrower set of inputs (string). The target function will only ever pass a string, which the broader source function is fully equipped to handle.

Why alternative options are incorrect:

Option B is incorrect: Parameter checks are not invariant; they follow contravariant rules to balance flexibility and mathematical safety.

Option D is incorrect: TypeScript provides structural type safety entirely at compile time; it never injects implicit fallback types at runtime.

Option E is incorrect: Type checking occurs exclusively during compilation; no runtime type validations are appended to basic operations.

Option F is incorrect: Module resolution configurations do not modify or override standard contravariant type assignment checks.

You can retake the exams as many times as you want

This is a huge original question bank

You get support from instructors if you have questions

Each question has a detailed explanation

Mobile-compatible with the Udemy app