Enqueue Using Linked List in DSA Rust - Time & Space Complexity

Time Complexity: Enqueue Using Linked List

O(1)

Understanding Time Complexity

We want to understand how the time to add an item to a linked list queue grows as the queue gets bigger.

How does the number of steps change when we add more items?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

struct Queue {
    head: Option>,
    tail: Option<*mut Node>,
}

impl Queue {
    fn enqueue(&mut self, val: i32) {
        let mut new_node = Box::new(Node { value: val, next: None });
        let raw_node: *mut _ = &mut *new_node;
        if let Some(tail) = self.tail {
            unsafe {
                (*tail).next = Some(new_node);
            }
            self.tail = Some(raw_node);
        } else {
            self.head = Some(new_node);
            self.tail = Some(raw_node);
        }
    }
}
    

This code adds a new item to the end of a linked list queue by updating the tail pointer.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding a new node by updating pointers.
• How many times: This happens once per enqueue call, no loops or traversals.

How Execution Grows With Input

Each time we add an item, the steps stay about the same because we keep track of the tail.

Input Size (n)	Approx. Operations
10	~1 pointer update
100	~1 pointer update
1000	~1 pointer update

Pattern observation: The work to add one item does not grow with the queue size.

Final Time Complexity

Time Complexity: O(1)

This means adding an item takes the same small amount of time no matter how many items are already in the queue.

Common Mistake

[X] Wrong: "Adding an item takes longer as the queue grows because we must find the end each time."

[OK] Correct: We keep a pointer to the tail, so we add directly without searching, keeping time constant.

Interview Connect

Knowing that enqueue in a linked list queue is fast and constant time shows you understand efficient data structure design, a key skill in coding interviews.

Self-Check

"What if we did not keep a tail pointer and had to find the end each time? How would the time complexity change?"

Practice

1. What does the enqueue operation do in a linked list implementation of a queue?

easy

A. Adds a new element at the beginning (head) of the linked list

B. Removes the first element from the linked list

C. Adds a new element at the end (tail) of the linked list

D. Deletes the last element from the linked list

Solution

1. Step 1: Understand enqueue operation in queue

   Enqueue means adding an element to the queue, which is done at the end (tail) in a linked list.

2. Step 2: Identify linked list position for enqueue

   In a linked list queue, new nodes are added at the tail to maintain FIFO order.

3. Final Answer:

   Adds a new element at the end (tail) of the linked list -> Option C

4. Quick Check:

   Enqueue = Add at tail [OK]

Hint: Enqueue always adds at the tail in linked list queue [OK]

Common Mistakes:

• Confusing enqueue with dequeue operation
• Adding new element at head instead of tail
• Thinking enqueue removes elements

2. Which of the following Rust code snippets correctly enqueues a new node with value val into a linked list queue?

easy

A. let new_node = Box::new(Node { data: val, next: None }); tail.next = Some(new_node); tail = tail.next.as_mut().unwrap();

B. let new_node = Box::new(Node { data: val, next: None }); head = Some(new_node);

C. let new_node = Box::new(Node { data: val, next: head }); head = Some(new_node);

D. let new_node = Box::new(Node { data: val, next: None }); tail = Some(new_node);

Solution

1. Step 1: Create new node with value and no next

   The new node is created with data = val and next = None, correct for enqueue at tail.

2. Step 2: Link new node at tail and update tail pointer

   Setting tail.next to new_node and moving tail pointer to new_node is correct enqueue logic.

3. Final Answer:

   let new_node = Box::new(Node { data: val, next: None }); tail.next = Some(new_node); tail = tail.next.as_mut().unwrap(); -> Option A

4. Quick Check:

   Enqueue = add at tail and update tail [OK]

Hint: Enqueue updates tail.next and tail pointer [OK]

Common Mistakes:

• Assigning new node to head instead of tail
• Not updating tail pointer after enqueue
• Linking new node's next to head (wrong for enqueue)

3. Given this Rust code snippet for enqueue operation on a linked list queue, what will be the printed queue after enqueuing values 10, 20, 30?

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

struct Queue {
    head: Option,
    tail: *mut Node,
}

impl Queue {
    fn new() -> Self {
        Queue { head: None, tail: std::ptr::null_mut() }
    }

    fn enqueue(&mut self, val: i32) {
        let mut new_node = Box::new(Node { data: val, next: None });
        let raw_node: *mut _ = &mut *new_node;
        if self.tail.is_null() {
            self.head = Some(new_node);
            self.tail = raw_node;
        } else {
            unsafe { (*self.tail).next = Some(new_node); }
            self.tail = raw_node;
        }
    }

    fn print_queue(&self) {
        let mut current = &self.head;
        while let Some(node) = current {
            print!("{} -> ", node.data);
            current = &node.next;
        }
        println!("null");
    }
}

fn main() {
    let mut q = Queue::new();
    q.enqueue(10);
    q.enqueue(20);
    q.enqueue(30);
    q.print_queue();
}

medium

A. 30 -> 20 -> 10 -> null

B. 10 -> 20 -> 30 -> null

C. null

D. 10 -> null

Solution

1. Step 1: Trace enqueue calls adding nodes at tail

   Enqueue adds 10, then 20, then 30 at the tail, preserving order.

2. Step 2: Print queue from head to tail

   Prints nodes in order: 10 -> 20 -> 30 -> null.

3. Final Answer:

   10 -> 20 -> 30 -> null -> Option B

4. Quick Check:

   Enqueue order preserved in print [OK]

Hint: Enqueue adds at tail, print from head shows insertion order [OK]

Common Mistakes:

• Assuming enqueue adds at head, reversing order
• Not updating tail pointer causing print errors
• Printing empty queue due to wrong head initialization

4. Identify the bug in this Rust enqueue method for a linked list queue:

fn enqueue(&mut self, val: i32) {
    let new_node = Box::new(Node { data: val, next: None });
    if self.tail.is_null() {
        self.head = Some(new_node);
        self.tail = &mut *new_node;
    } else {
        unsafe { (*self.tail).next = Some(new_node); }
        self.tail = &mut *new_node;
    }
}

medium

A. Using unsafe block unnecessarily

B. Not setting new_node.next to None

C. Not updating head pointer when queue is empty

D. Assigning tail pointer to reference of new_node which is dropped after function ends

Solution

1. Step 1: Analyze new_node lifetime

   new_node is a local variable and will be dropped after function ends, so tail pointer becomes invalid.

2. Step 2: Understand pointer assignment issue

   Assigning tail to &mut *new_node stores pointer to dropped memory, causing undefined behavior.

3. Final Answer:

   Assigning tail pointer to reference of new_node which is dropped after function ends -> Option D

4. Quick Check:

   Pointer to dropped node causes bug [OK]

Hint: Tail must point to heap node, not local variable [OK]

Common Mistakes:

• Using local variable reference for tail pointer
• Ignoring unsafe pointer lifetime issues
• Not updating tail pointer at all

5. You want to implement a queue using a singly linked list in Rust that supports enqueue and dequeue efficiently. Which approach correctly handles enqueue operation to maintain O(1) time complexity and safe memory management? A) Add new node at head and update head pointer. B) Add new node at tail, update tail pointer, and use Option for next pointers. C) Add new node at tail but do not update tail pointer. D) Add new node at head and update tail pointer.

hard

A. Add new node at tail, update tail pointer, and use Option for next pointers.

B. Add new node at head and update head pointer.

C. Add new node at tail but do not update tail pointer.

D. Add new node at head and update tail pointer.

Solution

1. Step 1: Understand enqueue efficiency in linked list queue

   Enqueue must add at tail and update tail pointer to keep O(1) time complexity.

2. Step 2: Use Option for safe memory management

   Using Option for next pointers ensures ownership and safe heap allocation in Rust.

3. Step 3: Reject other options

   Adding at head breaks FIFO order; not updating tail pointer breaks enqueue logic.

4. Final Answer:

   Add new node at tail, update tail pointer, and use Option for next pointers. -> Option A

5. Quick Check:

   Enqueue at tail + safe pointers = O(1) enqueue [OK]

Hint: Enqueue at tail with Option ensures O(1) and safety [OK]

Common Mistakes:

• Adding at head breaks queue order
• Not updating tail pointer after enqueue
• Using raw pointers without ownership