Traversal Forward and Backward in DSA Rust - Time & Space Complexity

Time Complexity: Traversal Forward and Backward

O(n)

Understanding Time Complexity

We want to understand how long it takes to move through a list from start to end and then back again.

How does the time needed change as the list grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn traverse_forward_backward(list: &Vec<i32>) {
    for i in 0..list.len() {
        let _ = list[i]; // forward traversal
    }
    for i in (0..list.len()).rev() {
        let _ = list[i]; // backward traversal
    }
}
    

This code goes through the list from the first item to the last, then goes back from the last to the first.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two separate loops each visiting every item once.
• How many times: Each loop runs exactly n times, where n is the list size.

How Execution Grows With Input

As the list gets bigger, the total steps grow roughly twice as fast because we go forward and backward.

Input Size (n)	Approx. Operations
10	20 (10 forward + 10 backward)
100	200 (100 forward + 100 backward)
1000	2000 (1000 forward + 1000 backward)

Pattern observation: The total steps double as input size doubles, showing a linear growth.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete grows directly in proportion to the list size.

Common Mistake

[X] Wrong: "Because there are two loops, the time complexity is O(n²)."

[OK] Correct: The loops run one after another, not nested, so their times add up, not multiply.

Interview Connect

Understanding how simple forward and backward traversals scale helps you reason about more complex data operations confidently.

Self-Check

"What if we combined the forward and backward traversal into a single loop that visits two elements per step? How would the time complexity change?"

Practice

1. What does "Traversal Forward and Backward" mean in the context of a doubly linked list?

easy

A. Skipping nodes while moving forward

B. Moving through the list from the first node to the last and then back from the last to the first

C. Only moving from the first node to the last node

D. Moving randomly between nodes

Solution

1. Step 1: Understand doubly linked list structure

   A doubly linked list has nodes connected both forward and backward, allowing movement in both directions.

2. Step 2: Define traversal forward and backward

   Traversal forward means moving from the first node to the last node, and traversal backward means moving from the last node back to the first node.

3. Final Answer:

   Moving through the list from the first node to the last and then back from the last to the first -> Option B

4. Quick Check:

   Forward and backward traversal = Moving both ways [OK]

Hint: Think of walking to the end and then returning back [OK]

Common Mistakes:

• Confusing traversal with skipping nodes
• Thinking traversal is only forward
• Assuming random movement is traversal

2. Which of the following Rust code snippets correctly traverses a doubly linked list forward using a mutable reference?

easy

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

B. while let Some(node) = current { println!("{}", node.value); current = node.prev.as_deref_mut(); }

C. while let Some(node) = current { println!("{}", node.value); current = node.next.as_deref_mut(); }

D. loop { println!("{}", current.value); current = current.prev; }

Solution

1. Step 1: Identify forward traversal in Rust

   Forward traversal moves from current node to next node using the next pointer.

2. Step 2: Check code correctness

   while let Some(node) = current { println!("{}", node.value); current = node.next.as_deref_mut(); } uses current = node.next.as_deref_mut() which correctly moves forward. while let Some(node) = current { println!("{}", node.value); current = node.prev.as_deref_mut(); } moves backward, C is invalid syntax for linked list, D loops backward without condition.

3. Final Answer:

   while let Some(node) = current { println!("{}", node.value); current = node.next.as_deref_mut(); } -> Option C

4. Quick Check:

   Forward traversal uses next pointer [OK]

Hint: Forward traversal uses next pointer, backward uses prev [OK]

Common Mistakes:

• Using prev pointer for forward traversal
• Incorrect loop syntax for linked list
• Ignoring option handling in Rust

3. Given the following Rust code for a doubly linked list traversal backward, what will be printed?

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

fn traverse_backward(mut current: Option<&Node>) {
    while let Some(node) = current {
        print!("{} ", node.value);
        current = node.prev.as_deref();
    }
}

// Assume list nodes with values 1 -> 2 -> 3, current starts at node with value 3

medium

A. 3 2 1

B. 1 2 3

C. 3 1 2

D. 1 3 2

Solution

1. Step 1: Understand backward traversal

   Starting at node with value 3, moving backward means following prev pointers to 2, then 1.

2. Step 2: Trace the print output

   Prints 3 first, then 2, then 1, each followed by a space.

3. Final Answer:

   3 2 1 -> Option A

4. Quick Check:

   Backward traversal prints nodes in reverse order [OK]

Hint: Backward traversal prints nodes from tail to head [OK]

Common Mistakes:

• Confusing forward and backward order
• Assuming traversal starts at head
• Ignoring Option handling in Rust

4. Identify the error in the following Rust code snippet that attempts to traverse a doubly linked list forward:

fn traverse(mut current: Option<&Node>) {
    while current.is_some() {
        println!("{}", current.unwrap().value);
        current = current.unwrap().next.as_deref();
    }
}

medium

A. The loop condition should be while current.is_none()

B. println! macro syntax is incorrect

C. next pointer should be accessed with prev instead

D. Using unwrap() multiple times causes panic if current is None

Solution

1. Step 1: Analyze unwrap() usage

   Calling unwrap() twice without checking can panic if current becomes None between calls.

2. Step 2: Suggest safer approach

   Use while let Some(node) = current to safely unwrap once per iteration.

3. Final Answer:

   Using unwrap() multiple times causes panic if current is None -> Option D

4. Quick Check:

   Multiple unwrap() calls risk panic [OK]

Hint: Use while let Some(node) = current to avoid unwrap panic [OK]

Common Mistakes:

• Confusing is_some() and is_none()
• Accessing wrong pointer for forward traversal
• Misusing println! macro

5. You have a doubly linked list with nodes containing values [10, 20, 30, 40]. Write a Rust function that traverses forward to collect values into a vector, then traverses backward from the last node to collect values into another vector. What will be the content of both vectors after traversal?

hard

A. Forward: [10, 20, 30, 40], Backward: [40, 30, 20, 10]

B. Forward: [40, 30, 20, 10], Backward: [10, 20, 30, 40]

C. Forward: [10, 20, 30], Backward: [40, 30, 20]

D. Forward: [], Backward: []

Solution

1. Step 1: Traverse forward collecting values

   Starting at head (10), collect values in order: 10, 20, 30, 40.

2. Step 2: Traverse backward from last node collecting values

   Starting at tail (40), collect values backward: 40, 30, 20, 10.

3. Final Answer:

   Forward: [10, 20, 30, 40], Backward: [40, 30, 20, 10] -> Option A

4. Quick Check:

   Forward and backward vectors match traversal order [OK]

Hint: Forward collects from head to tail, backward from tail to head [OK]

Common Mistakes:

• Mixing order of forward and backward vectors
• Missing last node in traversal
• Confusing vector contents with node pointers