Enclosing scope in Python - Time & Space Complexity

Time Complexity: Enclosing scope

O(n)

Understanding Time Complexity

Let's explore how the time it takes to run code changes when we use variables from an enclosing scope inside a function.

We want to see how this affects the speed as the input grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def multiplier(factor):
    def multiply(number):
        return number * factor
    return multiply

times3 = multiplier(3)
result = times3(10)

This code creates a function that remembers a number from outside and uses it to multiply another number.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: There are no loops or repeated steps here; the multiply function just does one multiplication.
• How many times: The multiplication happens once per call to the inner function.

How Execution Grows With Input

Explain the growth pattern intuitively.

Input Size (n)	Approx. Operations
10	10 multiplications if called 10 times
100	100 multiplications if called 100 times
1000	1000 multiplications if called 1000 times

Pattern observation: The time grows directly with how many times you call the inner function, not because of the enclosing scope.

Final Time Complexity

Time Complexity: O(n)

This means the time grows in a straight line with the number of times you use the inner function.

Common Mistake

[X] Wrong: "Using variables from outside the function makes the code slower for each call."

[OK] Correct: The inner function just uses the remembered value directly; it doesn't add extra loops or steps, so speed depends on how many times you call it, not on the enclosing scope.

Interview Connect

Understanding how functions remember outside values helps you write clear and efficient code, a skill that shows you know how to manage data and speed well.

Self-Check

"What if the inner function had a loop inside it? How would the time complexity change?"

Practice

1. What does enclosing scope mean in Python functions?

easy

A. Inner functions can access variables from outer functions

B. Variables inside a function are always global

C. Functions cannot access variables outside their own block

D. Only global variables can be used inside functions

Solution

1. Step 1: Understand function nesting

   In Python, functions can be defined inside other functions, creating an inner and outer function relationship.

2. Step 2: Access rules for variables

   Inner functions can use variables defined in their outer functions, which is called the enclosing scope.

3. Final Answer:

   Inner functions can access variables from outer functions -> Option A

4. Quick Check:

   Enclosing scope = inner function accesses outer variables [OK]

Hint: Inner functions see outer variables unless shadowed [OK]

Common Mistakes:

• Thinking all variables inside functions are global
• Believing inner functions cannot use outer variables
• Confusing local and global scopes

2. Which of the following is the correct syntax to modify an outer variable inside an inner function?

easy

A. Just assign the variable directly without any keyword

B. Use global keyword inside the inner function

C. Use nonlocal keyword inside the inner function

D. Use outer keyword inside the inner function

Solution

1. Step 1: Understand variable modification rules

   To change a variable from an outer (enclosing) function inside an inner function, Python requires the nonlocal keyword.

2. Step 2: Differentiate from global

   global is for variables at the module level, not enclosing functions.

3. Final Answer:

   Use nonlocal keyword inside the inner function -> Option C

4. Quick Check:

   Modify outer variable = use nonlocal [OK]

Hint: Change outer function vars with 'nonlocal' inside inner function [OK]

Common Mistakes:

• Using 'global' instead of 'nonlocal' for outer function variables
• Trying to assign without any keyword causing UnboundLocalError
• Using a non-existent 'outer' keyword

3. What is the output of this code?

def outer():
    x = 5
    def inner():
        return x + 3
    return inner()
print(outer())

medium

A. 3

B. 5

C. Error

D. 8

Solution

1. Step 1: Trace variable usage

   Variable x is defined in outer() as 5. The inner function returns x + 3, which is 8.

2. Step 2: Function call and return

   outer() calls inner() and returns its result, so print(outer()) prints 8.

3. Final Answer:

   8 -> Option D

4. Quick Check:

   Inner uses outer x=5, returns 5+3=8 [OK]

Hint: Inner function can read outer variables directly [OK]

Common Mistakes:

• Expecting an error due to variable scope
• Confusing inner return value with outer variable
• Thinking inner cannot access outer variables

4. What is wrong with this code?

def outer():
    x = 10
    def inner():
        x = x + 5
        return x
    return inner()
print(outer())

medium

A. It will print 15

B. UnboundLocalError because inner tries to modify x without nonlocal

C. SyntaxError due to wrong indentation

D. NameError because x is not defined anywhere

Solution

1. Step 1: Analyze variable assignment in inner()

   Inside inner(), x = x + 5 tries to modify x. Python treats x as local but it is used before assignment.

2. Step 2: Understand error cause

   This causes an UnboundLocalError because x is referenced before assignment locally. The fix is to declare nonlocal x.

3. Final Answer:

   UnboundLocalError because inner tries to modify x without nonlocal -> Option B

4. Quick Check:

   Modify outer var inside inner needs nonlocal [OK]

Hint: Assigning outer var inside inner needs 'nonlocal' keyword [OK]

Common Mistakes:

• Assuming it prints 15 without error
• Confusing UnboundLocalError with NameError
• Ignoring the need for 'nonlocal' when modifying outer vars

5. Consider this code:

def counter():
    count = 0
    def increment():
        nonlocal count
        count += 1
        return count
    return increment
c = counter()
print(c())
print(c())
print(c())

What will be the output?

hard

A. [1, 2, 3]

B. [0, 1, 2]

C. [1, 1, 1]

D. Error due to missing global keyword

Solution

1. Step 1: Understand closure with nonlocal

   The counter() function returns increment(), which modifies count using nonlocal. This keeps count persistent across calls.

2. Step 2: Trace calls to c()

   Each call to c() increases count by 1 and returns it. So outputs are 1, then 2, then 3.

3. Final Answer:

   [1, 2, 3] -> Option A

4. Quick Check:

   Closure with nonlocal increments count each call [OK]

Hint: Use nonlocal to keep and update outer variable state [OK]

Common Mistakes:

• Expecting count to reset each call
• Confusing nonlocal with global
• Thinking it will raise an error without global