Nonlocal keyword in Python - Time & Space Complexity

Time Complexity: Nonlocal keyword

O(n)

Understanding Time Complexity

Let's explore how the nonlocal keyword affects the speed of a Python function.

We want to see how the number of steps changes as the input grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def outer(n):
    count = 0
    def inner():
        nonlocal count
        for i in range(n):
            count += 1
    inner()
    return count

This code counts from 0 up to n using a nested function that changes a variable from the outer function.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for loop inside inner() that runs n times.
• How many times: Exactly once per call to outer(n), but the loop runs n times.

How Execution Grows With Input

As n grows, the loop runs more times, so the work grows in a straight line with n.

Input Size (n)	Approx. Operations
10	10 steps
100	100 steps
1000	1000 steps

Pattern observation: Doubling n doubles the work done.

Final Time Complexity

Time Complexity: O(n)

This means the time it takes grows directly with the size of the input n.

Common Mistake

[X] Wrong: "Using nonlocal makes the function slower because it adds extra work."

[OK] Correct: The nonlocal keyword only changes where the variable lives; it does not add extra loops or steps. The main time cost is still the loop running n times.

Interview Connect

Understanding how nested functions and variable scopes affect performance helps you write clear and efficient code, a skill valued in many coding challenges.

Self-Check

What if we removed the nonlocal keyword and instead returned the count from inner()? How would the time complexity change?

Practice

1. What does the nonlocal keyword do in Python?

easy

A. Declares a variable as global across all modules.

B. Allows an inner function to modify a variable from its outer function.

C. Creates a new local variable inside the inner function.

D. Prevents any changes to variables in the outer function.

Solution

1. Step 1: Understand variable scopes

   Variables inside a function are local by default, and inner functions cannot change outer variables unless specified.

2. Step 2: Role of nonlocal

   The nonlocal keyword allows the inner function to access and modify variables from the nearest enclosing function scope.

3. Final Answer:

   Allows an inner function to modify a variable from its outer function. -> Option B

4. Quick Check:

   nonlocal changes outer function variable = A [OK]

Hint: Nonlocal lets inner functions change outer variables [OK]

Common Mistakes:

• Confusing nonlocal with global keyword
• Thinking nonlocal creates new variables
• Assuming nonlocal works outside functions

2. Which of the following is the correct syntax to use nonlocal inside a nested function?

easy

A. local variable_name

B. global variable_name

C. outer variable_name

D. nonlocal variable_name

Solution

1. Step 1: Recall the syntax for nonlocal

   The keyword nonlocal is followed by the variable name to indicate it refers to an outer function's variable.

2. Step 2: Compare options

   Only nonlocal variable_name uses the correct keyword and syntax: nonlocal variable_name.

3. Final Answer:

   nonlocal variable_name -> Option D

4. Quick Check:

   Correct nonlocal syntax = C [OK]

Hint: Use 'nonlocal' followed by variable name inside inner function [OK]

Common Mistakes:

• Using 'global' instead of 'nonlocal'
• Writing 'local' or 'outer' which are invalid keywords
• Forgetting to write variable name after nonlocal

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

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

medium

A. 10

B. None

C. Error: no binding for nonlocal 'x'

D. 5

Solution

1. Step 1: Trace variable assignment

   Variable x is set to 5 in outer(). The inner function declares nonlocal x and sets x = 10.

2. Step 2: Effect of nonlocal on x

   The nonlocal keyword allows inner() to modify x in outer(). So after calling inner(), x becomes 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   nonlocal changes outer x to 10 = D [OK]

Hint: nonlocal lets inner change outer variable value [OK]

Common Mistakes:

• Thinking x remains 5 because inner is separate
• Expecting a syntax error for nonlocal usage
• Assuming inner creates a new local x

4. Find the error in this code snippet:

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

medium

A. increment() should not return count

B. count should be global, not local

C. Missing 'nonlocal count' inside increment()

D. No error, code runs fine

Solution

1. Step 1: Identify variable scope issue

   Inside increment(), count = count + 1 tries to read and write count. Without nonlocal, Python treats count as local, but it is used before assignment.

2. Step 2: Fix with nonlocal

   Adding nonlocal count tells Python to use the count from counter(), allowing modification.

3. Final Answer:

   Missing 'nonlocal count' inside increment() -> Option C

4. Quick Check:

   Modify outer variable needs nonlocal = A [OK]

Hint: Add nonlocal before modifying outer variable inside inner function [OK]

Common Mistakes:

• Using global instead of nonlocal
• Ignoring variable scope causing UnboundLocalError
• Returning count without incrementing properly

5. Consider this code:

def make_accumulator():
    total = 0
    def add(value):
        nonlocal total
        total += value
        return total
    return add
acc = make_accumulator()
print(acc(5))
print(acc(3))
print(acc(-2))

What is the output of this code?

hard

A. 5\n8\n6

B. 5\n3\n-2

C. 0\n5\n8

D. Error: nonlocal used incorrectly

Solution

1. Step 1: Understand closure with nonlocal

   The function make_accumulator() returns add, which remembers total. The nonlocal total lets add update total each call.

2. Step 2: Trace calls to acc

   First call: total=0+5=5, prints 5.
   Second call: total=5+3=8, prints 8.
   Third call: total=8+(-2)=6, prints 6.

3. Final Answer:

   5 8 6 -> Option A

4. Quick Check:

   Accumulator sums values using nonlocal total = B [OK]

Hint: nonlocal keeps state in nested function closures [OK]

Common Mistakes:

• Expecting total to reset each call
• Confusing nonlocal with global
• Thinking output is separate values, not cumulative