Get Length of Linked List in DSA Rust - Time & Space Complexity

Time Complexity: Get Length of Linked List

O(n)

Understanding Time Complexity

We want to understand how long it takes to find the length of a linked list as the list grows.

How does the time needed change when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct Node {
    value: i32,
    next: Option<Box<Node>>,
}

fn get_length(head: &Option<Box<Node>>) -> usize {
    let mut count = 0;
    let mut current = head.as_ref();
    while let Some(node) = current {
        count += 1;
        current = node.next.as_ref();
    }
    count
}
    

This code counts how many nodes are in a linked list by moving from the first node to the end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that visits each node once.
• How many times: Exactly once per node in the list, from start to end.

How Execution Grows With Input

As the list gets longer, the time to count nodes grows directly with the number of nodes.

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

Pattern observation: The steps increase one-for-one with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the length grows in direct proportion to the list size.

Common Mistake

[X] Wrong: "Counting nodes is constant time because we just keep a count."

[OK] Correct: Even though counting is simple, we must visit each node to count it, so time grows with list length.

Interview Connect

Knowing how to analyze simple linked list operations like length counting builds your confidence for more complex list problems.

Self-Check

"What if the linked list was doubly linked? Would the time complexity to get length change?"

Practice

1. What does the function to get the length of a linked list typically do?

easy

A. Counts the number of nodes by traversing from head to end

B. Returns the value of the first node

C. Deletes all nodes in the list

D. Finds the maximum value in the list

Solution

1. Step 1: Understand the linked list traversal

   The function starts at the head node and moves through each node one by one.

2. Step 2: Count each node visited

   For every node visited, the function increments a counter until it reaches the end (null).

3. Final Answer:

   Counts the number of nodes by traversing from head to end -> Option A

4. Quick Check:

   Length = number of nodes visited [OK]

Hint: Traverse and count nodes until null [OK]

Common Mistakes:

• Returning the value of the first node instead of count
• Not handling empty list (null head)
• Trying to count without traversal

2. Which Rust code snippet correctly initializes a linked list node with value 5 and next pointer None?

easy

A. let node = Node::new(5);

B. let node = Node(5, None);

C. let node = Node { value: 5, next: None };

D. let node = Node { value: 5 };

Solution

1. Step 1: Check struct initialization syntax

   Rust structs are initialized with field names and values inside braces.

2. Step 2: Match field names and order

   let node = Node { value: 5, next: None }; uses correct field names and syntax. Tuple syntax requires tuple struct definition. Node::new(5) assumes an impl with new method. Omitting next field causes compile error.

3. Final Answer:

   let node = Node { value: 5, next: None }; -> Option C

4. Quick Check:

   Struct init uses field: value pairs [OK]

Hint: Use struct field names with braces for initialization [OK]

Common Mistakes:

• Using tuple syntax for struct without tuple struct definition
• Missing field names in initialization
• Omitting required fields like next

3. What is the output of this Rust code that calculates the length of a linked list?

struct Node {
    value: i32,
    next: Option,
}

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while let Some(node) = current {
        count += 1;
        current = &node.next;
    }
    count
}

fn main() {
    let node3 = Node { value: 3, next: None };
    let node2 = Node { value: 2, next: Some(Box::new(node3)) };
    let node1 = Node { value: 1, next: Some(Box::new(node2)) };
    let head = Some(Box::new(node1));
    println!("{}", length(&head));
}

medium

A. 1

B. 2

C. 0

D. 3

Solution

1. Step 1: Understand the linked list structure

   There are three nodes linked: node1 -> node2 -> node3 -> None.

2. Step 2: Trace the length function

   The function counts nodes by traversing from head. It visits node1, node2, and node3, incrementing count each time.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   3 nodes counted [OK]

Hint: Count nodes by following next pointers until None [OK]

Common Mistakes:

• Counting only head node
• Not dereferencing Option correctly
• Confusing Option with Box

4. Identify the error in this Rust function to get linked list length:

fn length(head: &Option) -> usize {
    let mut count = 0;
    let mut current = head;
    while current.is_some() {
        count += 1;
        current = current.unwrap().next;
    }
    count
}

medium

A. Using unwrap() on &Option causes a move error

B. Loop condition should be current.is_none()

C. Count should start at 1, not 0

D. Function should return i32, not usize

Solution

1. Step 1: Analyze unwrap usage on reference

   Calling unwrap() on &Option tries to move out of a borrowed reference, which is not allowed.

2. Step 2: Correct way to access next node

   Use pattern matching or while let Some(node) = current to borrow without moving.

3. Final Answer:

   Using unwrap() on &Option causes a move error -> Option A

4. Quick Check:

   unwrap() moves ownership, not allowed on references [OK]

Hint: Use pattern matching, avoid unwrap on references [OK]

Common Mistakes:

• Using unwrap() on borrowed Option causing move error
• Wrong loop condition logic
• Incorrect return type for length

5. Given a linked list where each node contains an integer, how would you modify the length function to count only nodes with even values?

fn length_even(head: &Option) -> usize {
    // Your code here
}

hard

A. Use recursion without checking node values

B. Traverse list, increment count only if node.value % 2 == 0

C. Return length() / 2 assuming half are even

D. Count all nodes, then subtract odd values count

Solution

1. Step 1: Traverse the linked list nodes

   Start from head and move through each node until None.

2. Step 2: Check each node's value for evenness

   Increment count only if node.value % 2 == 0 (even number).

3. Step 3: Return the count of even-valued nodes

   This gives the length of nodes with even values.

4. Final Answer:

   Traverse list, increment count only if node.value % 2 == 0 -> Option B

5. Quick Check:

   Count only even nodes by checking value % 2 == 0 [OK]

Hint: Count nodes while checking even condition [OK]

Common Mistakes:

• Counting all nodes without filtering
• Assuming half nodes are even without check
• Using recursion without base case