Traversal and Printing a Linked List in DSA Rust - Time & Space Complexity

Time Complexity: Traversal and Printing a Linked List

O(n)

Understanding Time Complexity

We want to understand how long it takes to go through all items in a linked list and print them.

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>,
}

fn print_list(head: &Option>) {
    let mut current = head;
    while let Some(node) = current {
        println!("{}", node.value);
        current = &node.next;
    }
}
    

This code goes through each node in the linked list and prints its value one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop that visits each node once.
• How many times: Exactly once per node, so as many times as the list length.

How Execution Grows With Input

As the list grows, the number of print steps grows the same way.

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

Pattern observation: The work grows directly with the number of nodes, so doubling nodes doubles work.

Final Time Complexity

Time Complexity: O(n)

This means the time to print grows in a straight line with the number of nodes in the list.

Common Mistake

[X] Wrong: "Since we only print values, it must be faster than O(n)."

[OK] Correct: Even printing each value requires visiting every node once, so time still grows with list size.

Interview Connect

Knowing how to analyze simple list traversal helps build confidence for more complex data structure questions.

Self-Check

"What if we changed the list to a doubly linked list? How would the time complexity of traversal and printing change?"

Practice

1. What does traversing a linked list mean in simple terms?

easy

A. Reversing the order of nodes

B. Adding a new node at the end

C. Visiting each node one by one from start to end

D. Deleting the first node

Solution

1. Step 1: Understand traversal meaning

   Traversal means going through each element in order, like reading a list from start to finish.

2. Step 2: Match with linked list action

   Visiting each node one by one from start to end matches the idea of traversal.

3. Final Answer:

   Visiting each node one by one from start to end -> Option C

4. Quick Check:

   Traversal = Visit nodes in order [OK]

Hint: Traversal means visiting nodes one by one [OK]

Common Mistakes:

• Confusing traversal with adding or deleting nodes
• Thinking traversal means changing node order

2. Which Rust code snippet correctly traverses and prints a singly linked list named list?

easy

A. let mut current = &list.head;
while let Some(node) = current {
println!("{}", node.value);
current = &node.next;
}

B. for node in list {
println!("{}", node.value);
}

C. while list.head.is_some() {
println!("{}", list.head.value);
list.head = list.head.next;
}

D. let mut current = list.head;
loop {
println!("{}", current.value);
}

Solution

1. Step 1: Check traversal logic

   let mut current = &list.head; while let Some(node) = current { println!("{}", node.value); current = &node.next; } uses a mutable reference to current node and loops while current is Some(node), printing value and moving to next node.

2. Step 2: Verify other options

   for node in list { println!("{}", node.value); } tries for loop on list directly (not valid without iterator). while list.head.is_some() { println!("{}", list.head.value); list.head = list.head.next; } mutates list head which is unsafe. let mut current = list.head; loop { println!("{}", current.value); } loops infinitely without checking for None.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Correct traversal uses while let Some(node) = current [OK]

Hint: Use while let Some(node) = current for safe traversal [OK]

Common Mistakes:

• Trying to loop directly over list without iterator
• Not checking for None to stop loop
• Mutating list head during traversal

3. What will be the output of this Rust code that prints a linked list?

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

fn print_list(head: &Option) {
    let mut current = head;
    while let Some(node) = current {
        print!("{} -> ", node.value);
        current = &node.next;
    }
    println!("null");
}

fn main() {
    let list = Some(Box::new(Node { value: 1, next: Some(Box::new(Node { value: 2, next: Some(Box::new(Node { value: 3, next: None })) })) }));
    print_list(&list);
}

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. 1 2 3 null

D. null

Solution

1. Step 1: Understand list structure

   The list nodes are 1 pointing to 2, which points to 3, which points to None.

2. Step 2: Trace print_list function

   It prints each node's value followed by ' -> ', then prints 'null' at the end.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option B

4. Quick Check:

   Print nodes in order with arrows [OK]

Hint: Follow next pointers to print nodes in order [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing arrows or null at end
• Printing values without separators

4. Identify the error in this Rust code that tries to print a linked list and fix it:

fn print_list(mut head: Option) {
    while let Some(node) = head {
        println!("{}", node.value);
        head = node.next;
    }
}

medium

A. Code uses wrong print macro println! instead of print!

B. Code does not check for None, causing infinite loop

C. Code does not update head inside the loop

D. Code moves ownership of nodes, so it cannot be reused after printing

Solution

1. Step 1: Analyze ownership in loop

   The code takes ownership of head and nodes inside the loop, consuming the list and preventing reuse.

2. Step 2: Suggest fix using references

   Use references (&Option) and iterate without consuming ownership to allow reuse.

3. Final Answer:

   Code moves ownership of nodes, so it cannot be reused after printing -> Option D

4. Quick Check:

   Ownership moved inside loop causes consumption [OK]

Hint: Use references to avoid moving ownership during traversal [OK]

Common Mistakes:

• Ignoring ownership rules causing compile errors
• Not updating head causing infinite loops
• Confusing print! and println! macros

5. Given a linked list where each node contains an integer, how would you modify the traversal code to print only even values in order?

fn print_even(head: &Option) {
    let mut current = head;
    while let Some(node) = current {
        // Your code here
        current = &node.next;
    }
}

hard

A. Add an if check: if node.value % 2 == 0 { println!("{}", node.value); }

B. Print all values and filter even ones after traversal

C. Change loop to skip nodes with odd values by modifying next pointer

D. Use recursion to print only odd values

Solution

1. Step 1: Add condition inside traversal

   Check if node.value is even using modulo operator and print only if true.

2. Step 2: Keep traversal unchanged

   Continue moving current to next node regardless of value to visit all nodes.

3. Final Answer:

   Add an if check: if node.value % 2 == 0 { println!("{}", node.value); } -> Option A

4. Quick Check:

   Filter values inside traversal with if condition [OK]

Hint: Use if condition inside loop to print only even values [OK]

Common Mistakes:

• Trying to modify next pointer during traversal
• Filtering after traversal instead of during
• Printing odd values instead of even