Default parameters with types in Typescript - Time & Space Complexity

Time Complexity: Default parameters with types

O(n)

Understanding Time Complexity

We want to see how the time it takes to run a function changes when it uses default parameters with types.

Specifically, does adding default values affect how long the function takes as input grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function greet(name: string = "Guest", times: number = 1): void {
  for (let i = 0; i < times; i++) {
    console.log(`Hello, ${name}!`);
  }
}

// Calling greet("Alice", 3) prints "Hello, Alice!" 3 times

This function prints a greeting message multiple times, using default values if no arguments are given.

Identify Repeating Operations

• Primary operation: The for-loop that runs to print the greeting.
• How many times: It runs exactly times times, which depends on input.

How Execution Grows With Input

Each time we increase the times parameter, the loop runs more times, so the work grows directly with times.

Input Size (times)	Approx. Operations
10	10 prints
100	100 prints
1000	1000 prints

Pattern observation: The number of operations grows in a straight line as times increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to run the function grows directly in proportion to the number of times it prints the message.

Common Mistake

[X] Wrong: "Default parameters make the function slower or faster depending on their types."

[OK] Correct: Default parameters only set values if none are given; they do not add loops or extra work, so they don't affect how the time grows with input size.

Interview Connect

Understanding how default parameters affect time helps you explain function behavior clearly and shows you can analyze code beyond just logic.

Self-Check

What if the function called another function inside the loop that also had a loop? How would the time complexity change?

Practice

1.

What is the main purpose of default parameters in TypeScript functions?

easy

A. To provide fallback values when arguments are missing

B. To enforce strict typing on function return values

C. To declare variables inside functions

D. To create anonymous functions

Solution

1. Step 1: Understand default parameters

   Default parameters give a value to a function argument if no value is passed.

2. Step 2: Identify the purpose

   This helps avoid errors when arguments are missing by providing fallback values.

3. Final Answer:

   To provide fallback values when arguments are missing -> Option A

4. Quick Check:

   Default parameters = fallback values [OK]

Hint: Default parameters fill missing arguments automatically [OK]

Common Mistakes:

• Confusing default parameters with return type enforcement
• Thinking default parameters declare variables
• Mixing default parameters with anonymous functions

2.

Which of the following is the correct syntax for a TypeScript function with a default parameter of type number?

function multiply(x: number, y: number = 2) { return x * y; }

easy

A. function multiply(x: number, y = 2: number) { return x * y; }

B. function multiply(x = 2: number, y: number) { return x * y; }

C. function multiply(x: number, y: number = 2) { return x * y; }

D. function multiply(x:, y: number = 2) { return x * y; }

Solution

1. Step 1: Check parameter type and default value syntax

   In TypeScript, the type annotation comes before the default value: param: type = default.

2. Step 2: Verify each option

   function multiply(x: number, y: number = 2) { return x * y; } correctly uses y: number = 2. Others have syntax errors or wrong order.

3. Final Answer:

   function multiply(x: number, y: number = 2) { return x * y; } -> Option C

4. Quick Check:

   Type before default value = correct syntax [OK]

Hint: Type annotation before default value: param: type = value [OK]

Common Mistakes:

• Placing default value before type annotation
• Omitting type annotation on default parameter
• Assigning default to wrong parameter

3.

What is the output of the following TypeScript code?

function greet(name: string = "Guest") {
  return `Hello, ${name}!`;
}

console.log(greet());
console.log(greet("Alice"));

medium

A. Hello, undefined!\nHello, Alice!

B. Hello, Guest!\nHello, Alice!

C. Hello, !\nHello, Alice!

D. Error: Missing argument for name

Solution

1. Step 1: Analyze default parameter usage

   Calling greet() without argument uses default "Guest".

2. Step 2: Check outputs for both calls

   First call prints "Hello, Guest!"; second call prints "Hello, Alice!".

3. Final Answer:

   Hello, Guest!\nHello, Alice! -> Option B

4. Quick Check:

   Default parameter used when argument missing [OK]

Hint: Default parameter fills missing argument in function call [OK]

Common Mistakes:

• Expecting undefined instead of default value
• Thinking missing argument causes error
• Ignoring template literal output format

4.

Identify the error in this TypeScript function and choose the correct fix:

function calculateArea(width: number = 5, height: number) {
  return width * height;
}

medium

A. Move the parameter with default value to the end: function calculateArea(height: number, width: number = 5) {...}

B. Remove the default value from width: function calculateArea(width: number, height: number) {...}

C. Add a default value to height: function calculateArea(width: number = 5, height: number = 10) {...}

D. No error; function is correct as is

Solution

1. Step 1: Understand parameter order rules

   In TypeScript, parameters with default values must come after parameters without defaults.

2. Step 2: Fix parameter order

   Moving width with default to the end fixes the error.

3. Final Answer:

   Move the parameter with default value to the end: function calculateArea(height: number, width: number = 5) {...} -> Option A

4. Quick Check:

   Default parameters must be last [OK]

Hint: Put default parameters after required ones [OK]

Common Mistakes:

• Placing default parameters before required ones
• Ignoring TypeScript parameter order rules
• Assuming no error when order is wrong

5.

Consider this function that merges two objects with default parameters:

function mergeOptions(options: { verbose?: boolean } = { verbose: false }, defaults: { verbose: boolean } = { verbose: true }) {
  return { ...defaults, ...options };
}

console.log(mergeOptions());
console.log(mergeOptions({ verbose: true }));

What is the output of the two console.log calls?

hard

A. Error due to conflicting default parameters

B. { verbose: true }\n{ verbose: true }

C. { verbose: true }\n{ verbose: false }

D. { verbose: false }\n{ verbose: true }

Solution

1. Step 1: Understand default parameter values

   Both options and defaults have default objects with verbose flags.

2. Step 2: Analyze object merging

   Using spread, { ...defaults, ...options } merges with options overriding defaults.

3. Step 3: Evaluate calls

   First call uses defaults: options = { verbose: false }, defaults = { verbose: true }, merged result is { verbose: false } because options overwrites defaults.

4. Step 4: Check second call

   Second call passes { verbose: true } as options, so merged result is { verbose: true }.

5. Final Answer:

   { verbose: false }\n{ verbose: true } -> Option D

6. Quick Check:

   Options override defaults in merge [OK]

Hint: Spread merges with later objects overriding earlier ones [OK]

Common Mistakes:

• Confusing which object overrides which in spread
• Assuming default parameters merge automatically
• Ignoring optional property in type