Variable-length keyword arguments (**kwargs) in Python - Time & Space Complexity

Time Complexity: Variable-length keyword arguments (**kwargs)

O(n)

Understanding Time Complexity

When using variable-length keyword arguments, the program handles an unknown number of named inputs.

We want to know how the time to process these inputs changes as more keywords are added.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def print_keys(**kwargs):
    for key in kwargs:
        print(key)

print_keys(a=1, b=2, c=3)

This function prints all the keys passed as keyword arguments.

Identify Repeating Operations

• Primary operation: Looping over all keys in the keyword arguments.
• How many times: Once for each key passed in kwargs.

How Execution Grows With Input

As the number of keyword arguments grows, the loop runs more times, so the work grows steadily.

Input Size (n)	Approx. Operations
10	10 print operations
100	100 print operations
1000	1000 print operations

Pattern observation: The work increases directly with the number of keyword arguments.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line with the number of keyword arguments.

Common Mistake

[X] Wrong: "Using **kwargs makes the function run instantly no matter how many arguments."

[OK] Correct: The function still needs to look at each keyword to process it, so more keywords mean more work.

Interview Connect

Understanding how variable keyword arguments affect time helps you explain function behavior clearly and shows you can think about code efficiency.

Self-Check

"What if the function also processed the values of kwargs inside the loop? How would the time complexity change?"

Practice

1. What does **kwargs allow you to do in a Python function?

easy

A. Pass a variable number of keyword arguments as a dictionary

B. Pass a variable number of positional arguments as a tuple

C. Return multiple values from a function

D. Define a function without any parameters

Solution

1. Step 1: Understand **kwargs usage

   **kwargs collects extra keyword arguments into a dictionary inside the function.

2. Step 2: Compare with other options

   Pass a variable number of positional arguments as a tuple describes *args, not **kwargs. Options A and C are unrelated to **kwargs.

3. Final Answer:

   Pass a variable number of keyword arguments as a dictionary -> Option A

4. Quick Check:

   **kwargs = keyword args dict [OK]

Hint: Remember: **kwargs collects named arguments as a dict [OK]

Common Mistakes:

• Confusing *args with **kwargs
• Thinking **kwargs collects positional arguments
• Believing **kwargs returns multiple values

2. Which of the following is the correct way to define a function that accepts variable keyword arguments?

easy

A. def func(*kwargs):

B. def func(**kwargs):

C. def func(**kwargs, *args):

D. def func(kwargs**):

Solution

1. Step 1: Recall correct syntax for keyword arguments

   The correct syntax to accept variable keyword arguments is **kwargs.

2. Step 2: Check each option

   def func(*kwargs): uses single star which is for positional args. def func(**kwargs, *args): is invalid syntax because **kwargs must come after any *args. def func(kwargs**): is invalid syntax.

3. Final Answer:

   def func(**kwargs): -> Option B

4. Quick Check:

   Double star before kwargs means keyword args [OK]

Hint: Use double star ** before kwargs in function definition [OK]

Common Mistakes:

• Using single star * instead of double star **
• Placing **kwargs incorrectly in parameters
• Writing invalid syntax like kwargs**

3. What will be the output of the following code?

def greet(**kwargs):
    if 'name' in kwargs:
        return f"Hello, {kwargs['name']}!"
    else:
        return "Hello, stranger!"

print(greet(name='Alice'))
print(greet(age=30))

medium

A. Error: KeyError

B. Hello, Alice!\nHello, None!

C. Hello, stranger!\nHello, stranger!

D. Hello, Alice!\nHello, stranger!

Solution

1. Step 1: Analyze function behavior with kwargs

   The function checks if 'name' is a key in kwargs. If yes, it returns a greeting with that name; otherwise, it returns "Hello, stranger!".

2. Step 2: Evaluate each print statement

   First call passes name='Alice', so output is "Hello, Alice!". Second call passes age=30, no 'name' key, so output is "Hello, stranger!".

3. Final Answer:

   Hello, Alice!\nHello, stranger! -> Option D

4. Quick Check:

   Check key in kwargs dict to decide greeting [OK]

Hint: Check if 'name' key exists in kwargs before using it [OK]

Common Mistakes:

• Assuming missing key returns None instead of else case
• Expecting KeyError without checking key presence
• Confusing positional and keyword arguments

4. Identify the error in the following code and choose the correct fix:

def show_info(**kwargs):
    print(kwargs['age'])

show_info(name='Bob')

medium

A. Change **kwargs to *args

B. Remove the print statement

C. Add a default value for 'age' using kwargs.get('age', default)

D. Call show_info with age parameter

Solution

1. Step 1: Identify the error cause

   The function tries to print kwargs['age'], but the call only passes name='Bob'. This causes a KeyError because 'age' key is missing.

2. Step 2: Choose the fix

   Using kwargs.get('age', default) safely returns a default value if 'age' is missing, avoiding the error.

3. Final Answer:

   Add a default value for 'age' using kwargs.get('age', default) -> Option C

4. Quick Check:

   Use kwargs.get() to avoid KeyError on missing keys [OK]

Hint: Use kwargs.get('key', default) to avoid missing key errors [OK]

Common Mistakes:

• Not handling missing keys causing KeyError
• Confusing *args and **kwargs
• Ignoring the need to pass required keys

5. You want to write a function build_profile that accepts a mandatory username and any number of additional keyword arguments describing user info. Which of the following implementations correctly returns a dictionary with all this data?

hard

A. def build_profile(username, **kwargs):
profile = {'username': username}
profile.update(kwargs)
return profile

B. def build_profile(**kwargs, username):
kwargs['username'] = username
return kwargs

C. def build_profile(username, **kwargs):
profile = kwargs
profile['username'] = username
return profile

D. def build_profile(username, *args):
profile = dict(args)
profile['username'] = username
return profile

Solution

1. Step 1: Understand function parameter order

   Mandatory parameters must come before **kwargs. def build_profile(**kwargs, username): kwargs['username'] = username return kwargs is invalid syntax because **kwargs must be last.

2. Step 2: Check dictionary construction

   def build_profile(username, **kwargs): profile = kwargs profile['username'] = username return profile modifies kwargs directly, which can cause unexpected side effects. def build_profile(username, **kwargs): profile = {'username': username} profile.update(kwargs) return profile creates a new dict with username, then updates with kwargs safely.

3. Step 3: Evaluate *args usage

   def build_profile(username, *args): profile = dict(args) profile['username'] = username return profile uses *args which collects positional arguments, not keyword arguments, so it won't work as intended.

4. Final Answer:

   def build_profile(username, **kwargs): profile = {'username': username} profile.update(kwargs) return profile -> Option A

5. Quick Check:

   Use username param first, then update dict with kwargs [OK]

Hint: Put mandatory params before **kwargs and update dict safely [OK]

Common Mistakes:

• Placing **kwargs before mandatory parameters
• Modifying kwargs dict directly
• Using *args instead of **kwargs for keyword data