Function ownership interaction in Rust - Time & Space Complexity

Time Complexity: Function ownership interaction

O(n)

Understanding Time Complexity

When functions take ownership of data in Rust, it affects how many times data is moved or copied.

We want to see how this ownership passing impacts the number of operations as input grows.

Scenario Under Consideration

Analyze the time complexity of the following Rust code snippet.

fn consume_vector(v: Vec) {
    for item in v {
        println!("{}", item);
    }
}

fn main() {
    let data = vec![1, 2, 3, 4, 5];
    consume_vector(data);
}
    

This code moves ownership of a vector into a function and prints each item.

Identify Repeating Operations

Look for loops or repeated actions that take time.

• Primary operation: Looping over each item in the vector inside consume_vector.
• How many times: Once for each element in the vector (n times).

How Execution Grows With Input

As the vector gets bigger, the function prints more items, so work grows with the number of elements.

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; doubling items doubles work.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Passing ownership to a function copies the data, so it takes extra time beyond the loop."

[OK] Correct: In Rust, ownership transfer is a move, not a copy, so it does not add extra repeated work proportional to input size.

Interview Connect

Understanding how ownership affects function calls helps you write efficient Rust code and explain your reasoning clearly.

Self-Check

What if the function borrowed the vector instead of taking ownership? How would the time complexity change?

Practice

1. What happens when a function in Rust takes ownership of a variable passed to it?

easy

A. The function borrows the variable temporarily.

B. The original variable can no longer be used after the function call.

C. The variable is copied and both original and function have ownership.

D. The variable is automatically cloned inside the function.

Solution

1. Step 1: Understand ownership transfer

   When a function takes ownership, the original variable's ownership moves to the function.

2. Step 2: Effect on original variable

   After the function call, the original variable is no longer valid and cannot be used.

3. Final Answer:

   The original variable can no longer be used after the function call. -> Option B

4. Quick Check:

   Ownership moves = original invalid [OK]

Hint: Ownership moves to function, original variable becomes unusable [OK]

Common Mistakes:

• Thinking ownership is copied, not moved
• Confusing borrowing with ownership transfer
• Assuming automatic cloning happens

2. Which of the following function signatures correctly borrows a string slice without taking ownership?

easy

A. fn print_name(name: &str) { println!("{}", name); }

B. fn print_name(name: &String) { println!("{}", name); }

C. fn print_name(name: String) { println!("{}", name); }

D. fn print_name(name: String) -> String { name }

Solution

1. Step 1: Identify borrowing syntax

   Borrowing a string slice uses &str type, which is a reference to string data.

2. Step 2: Match function signature

   fn print_name(name: &str) { println!("{}", name); } uses &str, correctly borrowing without taking ownership.

3. Final Answer:

   fn print_name(name: &str) { println!("{}", name); } -> Option A

4. Quick Check:

   Borrow string slice = &str [OK]

Hint: Use &str to borrow string slices, not String type [OK]

Common Mistakes:

• Using String type takes ownership
• Using &String borrows but is less flexible than &str
• Returning String does not borrow

3. What is the output of the following Rust code?

fn take_ownership(s: String) {
    println!("Inside: {}", s);
}

fn main() {
    let s = String::from("hello");
    take_ownership(s);
    println!("Outside: {}", s);
}

medium

A. Inside: hello
Outside: hello

B. Inside: hello
Outside:

C. Inside: hello
Compile error: use of moved value 's'

D. Runtime error: use after free

Solution

1. Step 1: Analyze ownership transfer in function call

   The function take_ownership takes ownership of s, moving it inside the function.

2. Step 2: Check usage after ownership moved

   After the function call, s is no longer valid in main, so using it causes a compile error.

3. Final Answer:

   Inside: hello Compile error: use of moved value 's' -> Option C

4. Quick Check:

   Ownership moved = compile error on use after move [OK]

Hint: Using moved variable after function call causes compile error [OK]

Common Mistakes:

• Assuming variable is still valid after move
• Expecting runtime error instead of compile error
• Thinking variable is copied automatically

4. Identify the error in this Rust code and choose the fix:

fn add_one(num: &i32) -> i32 {
    *num + 1
}

fn main() {
    let x = 5;
    let y = add_one(x);
    println!("{}", y);
}

medium

A. No error, code runs correctly.

B. Change function parameter to num: i32 to take ownership.

C. Change return type to &i32 to return a reference.

D. Change call to add_one(&x) to pass a reference.

Solution

1. Step 1: Check function parameter type

   Function expects a reference &i32, but main passes x (i32) directly.

2. Step 2: Fix function call

   Pass a reference &x to match parameter type and avoid type mismatch error.

3. Final Answer:

   Change call to add_one(&x) to pass a reference. -> Option D

4. Quick Check:

   Pass reference when function expects &i32 [OK]

Hint: Match argument type with parameter type exactly [OK]

Common Mistakes:

• Passing value instead of reference
• Changing return type unnecessarily
• Ignoring compiler type mismatch errors

5. Given this Rust code, what is the correct way to write a function that takes ownership of a vector, adds an element, and returns ownership back?

fn add_element(vec: Vec<i32>) -> Vec<i32> {
    // Add 10 to the vector
    vec.push(10);
    vec
}

hard

A. fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec }

B. fn add_element(vec: &Vec<i32>) -> Vec<i32> { vec.push(10); vec.clone() }

C. fn add_element(vec: Vec<i32>) { vec.push(10); }

D. fn add_element(vec: &mut Vec<i32>) { vec.push(10); }

Solution

1. Step 1: Understand ownership and mutability

   To modify the vector, the function must take ownership and have mutable access, so parameter must be mutable.

2. Step 2: Return ownership after modification

   Return the modified vector to give ownership back to the caller.

3. Final Answer:

   fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec } -> Option A

4. Quick Check:

   Mutable ownership + return ownership = fn add_element(mut vec: Vec<i32>) -> Vec<i32> { vec.push(10); vec } [OK]

Hint: Use mut parameter and return vector to keep ownership [OK]

Common Mistakes:

• Forgetting mut keyword for parameter
• Trying to modify borrowed vector without mut
• Not returning the vector after modification