Closures and variable binding in Ruby - Time & Space Complexity

Time Complexity: Closures and variable binding

O(n)

Understanding Time Complexity

When working with closures in Ruby, it's important to understand how the program runs as the input grows.

We want to see how the number of steps changes when closures capture variables and run repeatedly.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def make_multipliers(n)
  multipliers = []
  n.times do |i|
    multipliers << ->(x) { x * i }
  end
  multipliers
end

multipliers = make_multipliers(5)
multipliers.each { |m| puts m.call(10) }

This code creates a list of closures that multiply by numbers from 0 to n-1, then calls each closure with 10.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping n times to create closures and then looping n times to call them.
• How many times: Each loop runs n times, so total operations grow with n.

How Execution Grows With Input

As n grows, the number of closures created and calls made both increase linearly.

Input Size (n)	Approx. Operations
10	About 20 (10 creates + 10 calls)
100	About 200 (100 creates + 100 calls)
1000	About 2000 (1000 creates + 1000 calls)

Pattern observation: The total steps grow roughly twice as fast as n, which is still a straight line growth.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows directly in proportion to the number of closures created and called.

Common Mistake

[X] Wrong: "Closures cause the code to run slower exponentially because they capture variables."

[OK] Correct: Creating closures and calling them still happens in a simple loop, so time grows linearly, not exponentially.

Interview Connect

Understanding how closures work with variable binding and their time cost helps you explain your code clearly and reason about performance in real projects.

Self-Check

"What if we called each closure multiple times inside another loop? How would the time complexity change?"

Practice

1. What is a closure in Ruby?

easy

A. A variable that changes its value automatically

B. A method that can only be called once

C. A function that remembers variables from its creation context

D. A class that cannot be inherited

Solution

1. Step 1: Understand the definition of closure

   A closure is a function or block that keeps access to variables from where it was created, even if called later.

2. Step 2: Match the definition to options

   A function that remembers variables from its creation context correctly describes this behavior. Other options describe unrelated concepts.

3. Final Answer:

   A function that remembers variables from its creation context -> Option C

4. Quick Check:

   Closure = function remembering variables [OK]

Hint: Closures remember variables where they were made [OK]

Common Mistakes:

• Thinking closures are just methods
• Confusing closures with classes
• Believing closures change variable values automatically

2. Which of the following is the correct way to create a closure using a lambda in Ruby?

easy

A. closure = lambda { |x| x * 2 }

B. closure = def(x) x * 2 end

C. closure = function(x) { x * 2 }

D. closure = -> x x * 2

Solution

1. Step 1: Identify Ruby lambda syntax

   In Ruby, lambdas can be created with lambda { |args| ... } or ->(args) { ... }.

2. Step 2: Check each option

   closure = lambda { |x| x * 2 } uses lambda { |x| x * 2 }, which is valid. closure = def(x) x * 2 end uses invalid syntax for a lambda. closure = function(x) { x * 2 } uses JavaScript-like syntax. closure = -> x x * 2 misses parentheses in the arrow syntax, which is invalid.

3. Final Answer:

   closure = lambda { |x| x * 2 } -> Option A

4. Quick Check:

   Correct lambda syntax = closure = lambda { |x| x * 2 } [OK]

Hint: Use lambda { |args| ... } for Ruby closures [OK]

Common Mistakes:

• Using JavaScript or Python syntax
• Omitting parentheses in arrow lambdas
• Trying to define lambdas with def keyword

3. What will be the output of this Ruby code?

def make_closures
  closures = []
  3.times do |i|
    closures << lambda { i }
  end
  closures
end

closures = make_closures
closures.each { |c| puts c.call }

medium

A. 2
2
2

B. 0
1
2

C. 0
0
0

D. Error: undefined variable i

Solution

1. Step 1: Understand variable binding in block

   Ruby lambdas capture block variables like i by value in this context, so each lambda remembers its own i value.

2. Step 2: Evaluate each lambda call

   The loop sets i to 0, 1, 2 sequentially. Each lambda captures the current i value, so outputs are 0, 1, 2 respectively.

3. Final Answer:

   0 1 2 -> Option B

4. Quick Check:

   Lambdas capture distinct values per iteration [OK]

Hint: Ruby block vars in closures captured by value per iteration [OK]

Common Mistakes:

• Thinking lambdas capture shared variable (all 2)
• Thinking last value is 3 (off-by-one)
• Expecting error or all zeros

4. Find the error in this Ruby code that tries to create closures:

closures = []
3.times do |i|
  closures << lambda { j }
  j = i
end
closures.each { |c| puts c.call }

medium

A. Variable j is used before assignment inside the lambda

B. Lambda syntax is incorrect

C. times method cannot be used with blocks

D. closures array is not initialized

Solution

1. Step 1: Identify variable usage inside lambda

   The lambda uses variable j before it is assigned any value in the block, causing an error when called.

2. Step 2: Understand variable binding timing

   Since j is assigned after lambda creation, the lambda captures j as nil or undefined at call time, causing an error.

3. Final Answer:

   Variable j is used before assignment inside the lambda -> Option A

4. Quick Check:

   Use variables after assignment in closures [OK]

Hint: Assign variables before using them in closures [OK]

Common Mistakes:

• Assuming lambda syntax is wrong
• Thinking times method is invalid
• Ignoring variable initialization order

5. How can you modify this Ruby code to create closures that each remember a different value of i even if i changes later?

closures = []
3.times do |i|
  closures << lambda { i }
end
closures.each { |c| puts c.call }

hard

A. Call closures inside the times block

B. Change lambda to proc

C. Use global variable $i instead of i

D. Use a local variable inside the block to capture i: val = i; closures << lambda { val }

Solution

1. Step 1: Understand variable capture in closures

   Closures capture variables by reference, so if i changes later, all lambdas see the last value.

2. Step 2: Use a local variable to capture current value

   Assign val = i inside the block, then create lambda using val. This captures the value at that iteration.

3. Final Answer:

   Use a local variable inside the block to capture i: val = i; closures << lambda { val } -> Option D

4. Quick Check:

   Capture current value with local variable [OK]

Hint: Assign loop variable to local var inside block for closure [OK]

Common Mistakes:

• Thinking proc changes variable capture behavior
• Using global variables which share state
• Calling closures too early inside loop