Immutable data with freeze in Ruby - Time & Space Complexity

Time Complexity: Immutable data with freeze

O(n)

Understanding Time Complexity

We want to see how using freeze to make data unchangeable affects how long a program takes to run.

Specifically, we ask: does freezing data change the work done when we use it?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

array = [1, 2, 3, 4, 5]
array.freeze

array.each do |num|
  puts num
end

This code freezes an array to make it unchangeable, then prints each number in it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each element of the array with each.
• How many times: Once for each element in the array (n times).

How Execution Grows With Input

As the array gets bigger, the program prints more numbers, so it does more work.

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

Pattern observation: The work grows directly with the number of items. Freezing does not add extra repeated work.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line with the number of items in the array, even when it is frozen.

Common Mistake

[X] Wrong: "Freezing the array makes the program slower because it adds extra work every time we use the array."

[OK] Correct: Freezing only happens once and does not repeat during the loop. Using the frozen array is just as fast as using a normal array.

Interview Connect

Understanding how immutable data affects performance helps you write safer code without surprises about speed. It shows you can think about both correctness and efficiency together.

Self-Check

"What if we replaced each with a nested loop over the array? How would the time complexity change?"

Practice

1. What does the freeze method do to an object in Ruby?

easy

A. Prevents any further modifications to the object

B. Deletes the object from memory

C. Duplicates the object

D. Makes the object mutable

Solution

1. Step 1: Understand the purpose of freeze

   The freeze method is used to make an object immutable, meaning it cannot be changed after freezing.

2. Step 2: Identify the effect on the object

   Once frozen, any attempt to modify the object will raise an error, so it prevents further changes.

3. Final Answer:

   Prevents any further modifications to the object -> Option A

4. Quick Check:

   freeze makes object immutable = Prevents changes [OK]

Hint: Freeze means no changes allowed after calling it [OK]

Common Mistakes:

• Thinking freeze deletes the object
• Confusing freeze with duplication
• Assuming freeze makes object mutable

2. Which of the following is the correct way to freeze a string name in Ruby?

easy

A. freeze(name)

B. name.frozen = true

C. name.freeze()

D. freeze name

Solution

1. Step 1: Recall Ruby method call syntax

   In Ruby, to call a method on an object, use dot notation like object.method().

2. Step 2: Identify the correct freeze syntax

   The correct way to freeze the string name is name.freeze(). Other options are invalid syntax or incorrect usage.

3. Final Answer:

   name.freeze() -> Option C

4. Quick Check:

   Method call with dot and parentheses = Correct freeze syntax [OK]

Hint: Use dot notation: object.freeze() to freeze [OK]

Common Mistakes:

• Using freeze as a standalone function
• Trying to assign frozen = true
• Omitting dot before freeze

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

arr = [1, 2, 3]
arr.freeze
arr << 4
puts arr.inspect

medium

A. [1, 2, 3, 4]

B. RuntimeError: can't modify frozen Array

C. [1, 2, 3]

D. SyntaxError

Solution

1. Step 1: Understand freezing the array

   The array arr is frozen using arr.freeze, making it immutable.

2. Step 2: Analyze the modification attempt

   Trying to append 4 with arr << 4 will raise a RuntimeError because the array is frozen and cannot be changed.

3. Final Answer:

   RuntimeError: can't modify frozen Array -> Option B

4. Quick Check:

   Frozen object + modification = RuntimeError [OK]

Hint: Modifying frozen object causes RuntimeError [OK]

Common Mistakes:

• Assuming frozen array can still be changed
• Expecting silent failure or no error
• Confusing output with syntax error

4. Identify the error in this Ruby code snippet:

str = "hello"
str.freeze
str[0] = 'H'
puts str

medium

A. RuntimeError: can't modify frozen String

B. SyntaxError due to invalid string modification

C. No error, output is "Hello"

D. Output is "hello" without changes

Solution

1. Step 1: Recognize string freezing effect

   The string str is frozen, so it cannot be changed after str.freeze.

2. Step 2: Understand the attempted modification

   Trying to change the first character with str[0] = 'H' raises a RuntimeError because the string is frozen.

3. Final Answer:

   RuntimeError: can't modify frozen String -> Option A

4. Quick Check:

   Frozen string + modification = RuntimeError [OK]

Hint: Frozen string cannot be changed, causes RuntimeError [OK]

Common Mistakes:

• Thinking string changes silently fail
• Expecting output "Hello" without error
• Confusing syntax error with runtime error

5. Given this code, what is the best way to safely modify a frozen array without error?

original = [10, 20, 30].freeze
# Your code here to add 40

hard

A. original.freeze(false)

B. original << 40

C. original.push(40)

D. original = original.dup << 40

Solution

1. Step 1: Understand frozen array cannot be changed

   The array original is frozen, so direct modification like << or push will cause errors.

2. Step 2: Use duplication to create a modifiable copy

   Using original.dup creates a new unfrozen copy. Then appending 40 to this copy is allowed.

3. Step 3: Assign the modified copy back

   Assigning the new array with 40 added back to original variable safely updates it without error.

4. Final Answer:

   original = original.dup << 40 -> Option D

5. Quick Check:

   Duplicate frozen object before modifying = Safe change [OK]

Hint: Duplicate frozen object before changing it [OK]

Common Mistakes:

• Trying to modify frozen array directly
• Using non-existent freeze(false) method
• Assuming push works on frozen array