Data Structures & AlgorithmsLinked List-Based Queue Implementation

Linked List-Based Queue Implementation

LevelIntermediate

Duration55 mins

TopicLinked List-Based Queue Implementation

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Head as Front, Tail as Rear

The Two-Pointer Architecture

In the previous page, we established that a singly linked list with both head and tail pointers can efficiently support queue operations. Now we dive deeper into the specific roles of these pointers and the invariants that govern their behavior.

This isn't merely an implementation detail—the correct assignment of roles and rigorous invariant maintenance is what separates a working queue from a subtly broken one. Understanding this architecture thoroughly will enable you to:

Implement queue operations correctly the first time
Debug linked queue issues quickly by checking invariants
Extend the basic queue with additional features without introducing bugs
Reason formally about queue correctness

The head-as-front, tail-as-rear mapping is not arbitrary. It's the natural and optimal choice given the asymmetric efficiency profile of singly linked lists.

What You Will Learn

By the end of this page, you will understand exactly why head corresponds to front and tail to rear, master the invariants that govern these pointers, handle all edge cases (especially the single-element case), and be able to trace through queue state transitions with confidence.

Why Head = Front and Tail = Rear?

The assignment of head to front and tail to rear follows directly from the efficiency profile of singly linked lists. Let's trace through the logic:

Singly Linked List Efficiency (with tail pointer):

Operation	Time Complexity	Reason
Insert at head	O(1)	Create node, link to old head, update head
Remove from head	O(1)	Update head to head.next
Insert at tail	O(1)	Link current tail.next to new node, update tail
Remove from tail	O(n)	Must traverse to find second-to-last node

Queue Operation Requirements:

Operation	Location	Nature
Enqueue	Rear	Insertion
Dequeue	Front	Removal

Now, consider our choices:

Option A: Head=Front, Tail=Rear ✓

•Enqueue: Insert at tail → O(1) ✓
•Dequeue: Remove from head → O(1) ✓
•Both operations are constant time!

Option B: Head=Rear, Tail=Front ✗

•Enqueue: Insert at head → O(1) ✓
•Dequeue: Remove from tail → O(n) ✗
•Dequeue becomes linear time!

The asymmetry is critical:

Singly linked lists can insert at either end in O(1), but they can only remove from the head in O(1). Removing from the tail requires traversal because we need to find the second-to-last node to become the new tail—and singly linked lists have no backward pointers.

Since queues require removal from one end (front) and insertion at the other (rear), we MUST map:

Front (dequeue happens here) → Head (supports O(1) removal)
Rear (enqueue happens here) → Tail (supports O(1) insertion)

This is the only mapping that yields O(1) for both operations. The alternative gives O(n) dequeue, which is unacceptable.

Remember This Rule

In a singly linked list, you can efficiently INSERT at either end, but you can only efficiently REMOVE from the head. Therefore, the end where REMOVAL happens (queue front) must be the linked list head. This is not a convention—it's a consequence of the data structure's fundamental properties.

Visual Architecture of a Linked Queue

Let's build a clear mental model of how the queue looks at various states. Understanding these visuals is crucial for debugging and reasoning about your implementation.

Standard Queue State (Multiple Elements):

Converting Mermaid diagram...

Interpretation:

Elements enter from the right (enqueue at tail)
Elements exit from the left (dequeue at head)
Following the next pointers takes you from front toward rear
The tail's next is always null

This matches FIFO semantics:

Element 10 was enqueued first, it's at the front, will be dequeued first
Element 30 was enqueued last, it's at the rear, will be dequeued last

Single Element State:

Converting Mermaid diagram...

Critical observation: When there's only one element, both head and tail point to the same node. This single-element case requires special handling in both enqueue (to empty queue) and dequeue (of last element) operations.

Empty Queue State:

Converting Mermaid diagram...

When empty: Both head and tail are null. There are no nodes to reference. This state is trivially distinguishable from all non-empty states.

The Three Queue States

A linked queue exists in exactly one of three states: (1) Empty: head=null, tail=null, count=0; (2) Single element: head=tail≠null, head.next=null, count=1; (3) Multiple elements: head≠tail, both non-null, count≥2. Your invariant checks should verify mutual consistency between these pointers.

Invariant Specification: The Rules That Must Always Hold

Invariants are conditions that must be true at all observable states of the data structure—specifically, before and after every public method call. Invariants are the contract your implementation promises to maintain.

For a linked list queue with head, tail, and count, the invariants are:

Linked Queue Invariants

•Nullity Consistency: head is null if and only if tail is null. (You can't have a front without a rear or vice versa.)
•Empty Equivalence: head is null if and only if count equals 0. (Null head means empty, and empty means null head.)
•Tail Termination: If tail is not null, then tail.next is null. (The rear node has no successor.)
•Reachability: If the queue is non-empty, following next pointers from head exactly count - 1 times reaches tail.
•Single Element Consistency: If count equals 1, then head equals tail (same object reference, not just equal values).
•No Cycles: Following next pointers from head eventually reaches null (no circular references within the queue structure itself).

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// Invariant checking method (useful for debugging)
private checkInvariants(): void {
    // Invariant 1: Nullity consistency
    const headIsNull = this.head === null;
    const tailIsNull = this.tail === null;
    if (headIsNull !== tailIsNull) {
        throw new Error("Invariant violation: head/tail nullity mismatch");
    }
    
    // Invariant 2: Empty equivalence
    if (headIsNull !== (this.count === 0)) {
        throw new Error("Invariant violation: empty state inconsistency");
    }
    
    // Invariant 3: Tail termination
    if (this.tail !== null && this.tail.next !== null) {
        throw new Error("Invariant violation: tail.next must be null");
    }
    
    // Invariant 4 & 5: Reachability and single-element consistency
    if (this.count > 0) {
        let current = this.head;
        let steps = 0;
        while (current !== null && current !== this.tail) {
            current = current.next;
            steps++;
            if (steps > this.count) {
                throw new Error("Invariant violation: cycle detected or count mismatch");
            }
        }
        if (current !== this.tail) {
            throw new Error("Invariant violation: cannot reach tail from head");
        }
        if (steps !== this.count - 1) {
            throw new Error("Invariant violation: count doesn't match structure");
        }
    }
}

When to Check Invariants

In development builds, call checkInvariants() at the beginning and end of every public method. This catches bugs immediately when they occur, not later when symptoms manifest. In production, disable these checks for performance—but keep them available for debugging production issues.

The Single-Element Case: Where Most Bugs Hide

If there's one section of this module you should memorize, it's this one. The vast majority of bugs in linked queue implementations occur when transitioning through the single-element state.

Why is this case special?

In the single-element state, head and tail point to the same node. This creates coupling between operations that don't exist in other states:

Single-Element Transitions
Transition	What Changes	Common Bug
Empty → Single element (enqueue to empty queue)	Must set BOTH head AND tail to new node	Forgetting to set head, leaving it null
Single element → Empty (dequeue last item)	Must set BOTH head AND tail to null	Forgetting to set tail to null
Single element → Multiple (enqueue second item)	Only tail changes; head stays same	Accidentally modifying head
Multiple → Single element (dequeue to one remaining)	Only head changes; tail stays same	Accidentally modifying tail

Let's trace through each transition:

1. Enqueue to Empty Queue (Empty → Single):

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// Before: head = null, tail = null, count = 0
enqueue(element: T): void {
    const newNode = new QueueNode<T>(element);
    
    if (this.isEmpty()) {
        // CRITICAL: Empty queue - new node becomes BOTH head AND tail
        this.head = newNode;
        this.tail = newNode;
    } else {
        // Normal case: append to tail
        this.tail!.next = newNode;
        this.tail = newNode;
    }
    
    this.count++;
}
// After (if was empty): head = newNode, tail = newNode, count = 1
// Note: head === tail (same reference)

2. Dequeue from Single-Element Queue (Single → Empty):

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// Before: head = someNode, tail = someNode (same reference), count = 1
dequeue(): T {
    if (this.isEmpty()) {
        throw new Error("Cannot dequeue from empty queue");
    }
    
    const data = this.head!.data;
    this.head = this.head!.next;  // head becomes null (since head.next was null)
    
    if (this.head === null) {
        // CRITICAL: We just removed the last element - tail must also become null
        this.tail = null;
    }
    
    this.count--;
    return data;
}
// After: head = null, tail = null, count = 0

Converting Mermaid diagram...

The Most Common Bug

Bug #1: Dequeueing the last element and forgetting to set tail to null. Your head correctly becomes null, but tail still points to the (now orphaned) old node. Next enqueue checks isEmpty() using head, sees it's null, sets both head and tail to new node—but now the old node's data may still be accessible through stale references, or your invariant checks will fail.

Pointer Update Patterns: A Systematic Approach

To avoid bugs, follow a systematic pattern for pointer updates. Rather than thinking case-by-case, use a template that handles all cases correctly.

Enqueue Pattern:

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enqueue(element: T): void {
    const newNode = new QueueNode<T>(element);
    
    // Step 1: Link new node after current tail (if exists)
    if (this.tail !== null) {
        this.tail.next = newNode;
    }
    
    // Step 2: Update tail to new node (always)
    this.tail = newNode;
    
    // Step 3: If queue was empty, head must also point to new node
    if (this.head === null) {
        this.head = newNode;
    }
    
    // Step 4: Increment count
    this.count++;
}
 
// This pattern works for ALL cases:
// - Empty queue: tail was null (step 1 skipped), both pointers set to newNode
// - Non-empty: old tail linked to new, tail updated, head unchanged

Dequeue Pattern:

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dequeue(): T {
    // Step 1: Check for empty queue
    if (this.head === null) {
        throw new Error("Queue is empty");
    }
    
    // Step 2: Store data to return
    const data = this.head.data;
    
    // Step 3: Advance head to next node
    this.head = this.head.next;
    
    // Step 4: If head is now null, queue is empty - clear tail too
    if (this.head === null) {
        this.tail = null;
    }
    
    // Step 5: Decrement count
    this.count--;
    
    return data;
}
 
// This pattern works for ALL cases:
// - Single element: head.next was null, head becomes null, tail set to null
// - Multiple elements: head advances, tail unchanged

The Key Insight

Notice how both patterns handle the single-element case through general conditions rather than explicit if-else branches. This 'always update one pointer, conditionally update the other' approach reduces opportunities for bugs and makes the code more maintainable.

Why Not Use a Doubly Linked List?

A reasonable question arises: wouldn't a doubly linked list be even better? With prev pointers, we could remove from either end in O(1). While true, this introduces unnecessary complexity and overhead for a standard queue.

The Cost of Doubly Linked Lists:

Singly vs. Doubly Linked Lists for Queues
Aspect	Singly Linked + Tail	Doubly Linked
Memory per node	data + 1 pointer (8 bytes)	data + 2 pointers (16 bytes)
Enqueue complexity	O(1)	O(1)
Dequeue complexity	O(1)	O(1)
Pointer updates per operation	1-2	2-4
Invariants to maintain	3-4	6-8
Bug opportunity surface	Smaller	Larger

For a standard queue:

We only need:

O(1) insertion at rear ✓ (singly linked with tail pointer)
O(1) removal from front ✓ (singly linked with head pointer)

We do NOT need:

O(1) removal from rear (deques need this, queues don't)
O(1) insertion at front (deques need this, queues don't)
Bidirectional traversal (we never traverse backwards in a queue)

Doubly linked lists are overkill for queues. They add memory overhead, implementation complexity, and more invariants to maintain—all for capabilities we don't use.

When doubly linked lists ARE appropriate:

If you're implementing a deque (double-ended queue), where you need O(1) operations at BOTH ends (including removal), then doubly linked lists become necessary. But for a standard FIFO queue, a singly linked list with tail pointer is the optimal choice.

Principle of Minimal Mechanism

Use the simplest data structure that meets your requirements. Singly linked lists with tail pointers are simpler than doubly linked lists, with fewer pointers to maintain and fewer opportunities for bugs. For queues, they provide exactly the capabilities needed—no more, no less.

Complete State Trace: Seeing It All Together

Let's trace through a complete sequence of operations, showing the exact state of head, tail, and count at each step. This exercise cements understanding and serves as a debugging reference.

Operation Sequence: enqueue(10) → enqueue(20) → enqueue(30) → dequeue() → dequeue() → enqueue(40) → dequeue() → dequeue()

Complete Queue State Trace
Step	Operation	Queue Contents	head→	tail→	count
0	(initial)	[]	null	null	0
1	enqueue(10)	[10]	Node(10)	Node(10)	1
2	enqueue(20)	[10, 20]	Node(10)	Node(20)	2
3	enqueue(30)	[10, 20, 30]	Node(10)	Node(30)	3
4	dequeue() → 10	[20, 30]	Node(20)	Node(30)	2
5	dequeue() → 20	[30]	Node(30)	Node(30)	1
6	enqueue(40)	[30, 40]	Node(30)	Node(40)	2
7	dequeue() → 30	[40]	Node(40)	Node(40)	1
8	dequeue() → 40	[]	null	null	0

Key observations from the trace:

Steps 1, 5, 7: Single-element states where head === tail
Step 1: Enqueue to empty queue sets both pointers
Step 5: Dequeue to single element—only head changes
Step 8: Dequeue last element—both pointers become null
Steps 2-3, 6: Enqueue to non-empty—only tail changes
Step 4: Dequeue from multiple—only head changes

Notice how the count always matches the actual number of elements, and head/tail nullity is always consistent.

Debugging Technique

When debugging queue issues, create a trace table like this showing the state after each operation. Compare against expected values. The mismatch reveals exactly where your implementation diverges from correct behavior.

Summary: The Two-Pointer Architecture Mastered

We've thoroughly examined how head and tail pointers work together to implement an efficient queue. Let's consolidate the key insights:

Key Takeaways

•Head = Front, Tail = Rear is the only mapping that gives O(1) for both enqueue and dequeue. The reverse mapping yields O(n) dequeue.
•Invariants are crucial: Head and tail nullity must be consistent; count must match reachable nodes; tail.next must be null.
•The single-element case requires special attention—both enqueue to empty and dequeue of last element must update BOTH pointers.
•Use systematic patterns that handle all cases through conditions rather than separate code paths.
•Singly linked list is sufficient—doubly linked adds overhead for capabilities queues don't need.
•State tracing is a powerful debugging and reasoning tool.

What's next:

With the pointer architecture and invariants firmly understood, we'll now implement and analyze the core operations: enqueue and dequeue. We'll prove that both achieve O(1) time complexity and examine the exact steps involved in each operation.

Page Complete

You now have a deep understanding of why head serves as front and tail as rear in a linked queue, the invariants that must be maintained, and how to handle the critical single-element case. This foundation will make implementing and debugging linked queues straightforward.

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Data Structures & AlgorithmsLinked List-Based Queue Implementation

Linked List-Based Queue Implementation

LevelIntermediate

Duration55 mins

TopicLinked List-Based Queue Implementation

2 / 4

Head as Front, Tail as Rear

The Two-Pointer Architecture

Implement queue operations correctly the first time
Debug linked queue issues quickly by checking invariants
Extend the basic queue with additional features without introducing bugs
Reason formally about queue correctness

The head-as-front, tail-as-rear mapping is not arbitrary. It's the natural and optimal choice given the asymmetric efficiency profile of singly linked lists.

What You Will Learn

Why Head = Front and Tail = Rear?

The assignment of head to front and tail to rear follows directly from the efficiency profile of singly linked lists. Let's trace through the logic:

Singly Linked List Efficiency (with tail pointer):

Operation	Time Complexity	Reason
Insert at head	O(1)	Create node, link to old head, update head
Remove from head	O(1)	Update head to head.next
Insert at tail	O(1)	Link current tail.next to new node, update tail
Remove from tail	O(n)	Must traverse to find second-to-last node

Queue Operation Requirements:

Operation	Location	Nature
Enqueue	Rear	Insertion
Dequeue	Front	Removal

Now, consider our choices:

Option A: Head=Front, Tail=Rear ✓

•Enqueue: Insert at tail → O(1) ✓
•Dequeue: Remove from head → O(1) ✓
•Both operations are constant time!

Option B: Head=Rear, Tail=Front ✗

•Enqueue: Insert at head → O(1) ✓
•Dequeue: Remove from tail → O(n) ✗
•Dequeue becomes linear time!

The asymmetry is critical:

Since queues require removal from one end (front) and insertion at the other (rear), we MUST map:

Front (dequeue happens here) → Head (supports O(1) removal)
Rear (enqueue happens here) → Tail (supports O(1) insertion)

This is the only mapping that yields O(1) for both operations. The alternative gives O(n) dequeue, which is unacceptable.

Remember This Rule

Visual Architecture of a Linked Queue

Let's build a clear mental model of how the queue looks at various states. Understanding these visuals is crucial for debugging and reasoning about your implementation.

Standard Queue State (Multiple Elements):

Converting Mermaid diagram...

Interpretation:

Elements enter from the right (enqueue at tail)
Elements exit from the left (dequeue at head)
Following the next pointers takes you from front toward rear
The tail's next is always null

This matches FIFO semantics:

Element 10 was enqueued first, it's at the front, will be dequeued first
Element 30 was enqueued last, it's at the rear, will be dequeued last

Single Element State:

Converting Mermaid diagram...

Empty Queue State:

Converting Mermaid diagram...

When empty: Both head and tail are null. There are no nodes to reference. This state is trivially distinguishable from all non-empty states.

The Three Queue States

Invariant Specification: The Rules That Must Always Hold

For a linked list queue with head, tail, and count, the invariants are:

Linked Queue Invariants

•Nullity Consistency: head is null if and only if tail is null. (You can't have a front without a rear or vice versa.)
•Empty Equivalence: head is null if and only if count equals 0. (Null head means empty, and empty means null head.)
•Tail Termination: If tail is not null, then tail.next is null. (The rear node has no successor.)
•Reachability: If the queue is non-empty, following next pointers from head exactly count - 1 times reaches tail.
•Single Element Consistency: If count equals 1, then head equals tail (same object reference, not just equal values).
•No Cycles: Following next pointers from head eventually reaches null (no circular references within the queue structure itself).

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// Invariant checking method (useful for debugging)
private checkInvariants(): void {
    // Invariant 1: Nullity consistency
    const headIsNull = this.head === null;
    const tailIsNull = this.tail === null;
    if (headIsNull !== tailIsNull) {
        throw new Error("Invariant violation: head/tail nullity mismatch");
    }
    
    // Invariant 2: Empty equivalence
    if (headIsNull !== (this.count === 0)) {
        throw new Error("Invariant violation: empty state inconsistency");
    }
    
    // Invariant 3: Tail termination
    if (this.tail !== null && this.tail.next !== null) {
        throw new Error("Invariant violation: tail.next must be null");
    }
    
    // Invariant 4 & 5: Reachability and single-element consistency
    if (this.count > 0) {
        let current = this.head;
        let steps = 0;
        while (current !== null && current !== this.tail) {
            current = current.next;
            steps++;
            if (steps > this.count) {
                throw new Error("Invariant violation: cycle detected or count mismatch");
            }
        }
        if (current !== this.tail) {
            throw new Error("Invariant violation: cannot reach tail from head");
        }
        if (steps !== this.count - 1) {
            throw new Error("Invariant violation: count doesn't match structure");
        }
    }
}

When to Check Invariants

The Single-Element Case: Where Most Bugs Hide

If there's one section of this module you should memorize, it's this one. The vast majority of bugs in linked queue implementations occur when transitioning through the single-element state.

Why is this case special?

In the single-element state, head and tail point to the same node. This creates coupling between operations that don't exist in other states:

Single-Element Transitions
Transition	What Changes	Common Bug
Empty → Single element (enqueue to empty queue)	Must set BOTH head AND tail to new node	Forgetting to set head, leaving it null
Single element → Empty (dequeue last item)	Must set BOTH head AND tail to null	Forgetting to set tail to null
Single element → Multiple (enqueue second item)	Only tail changes; head stays same	Accidentally modifying head
Multiple → Single element (dequeue to one remaining)	Only head changes; tail stays same	Accidentally modifying tail

Let's trace through each transition:

1. Enqueue to Empty Queue (Empty → Single):

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// Before: head = null, tail = null, count = 0
enqueue(element: T): void {
    const newNode = new QueueNode<T>(element);
    
    if (this.isEmpty()) {
        // CRITICAL: Empty queue - new node becomes BOTH head AND tail
        this.head = newNode;
        this.tail = newNode;
    } else {
        // Normal case: append to tail
        this.tail!.next = newNode;
        this.tail = newNode;
    }
    
    this.count++;
}
// After (if was empty): head = newNode, tail = newNode, count = 1
// Note: head === tail (same reference)

2. Dequeue from Single-Element Queue (Single → Empty):

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// Before: head = someNode, tail = someNode (same reference), count = 1
dequeue(): T {
    if (this.isEmpty()) {
        throw new Error("Cannot dequeue from empty queue");
    }
    
    const data = this.head!.data;
    this.head = this.head!.next;  // head becomes null (since head.next was null)
    
    if (this.head === null) {
        // CRITICAL: We just removed the last element - tail must also become null
        this.tail = null;
    }
    
    this.count--;
    return data;
}
// After: head = null, tail = null, count = 0

Converting Mermaid diagram...

The Most Common Bug

Pointer Update Patterns: A Systematic Approach

To avoid bugs, follow a systematic pattern for pointer updates. Rather than thinking case-by-case, use a template that handles all cases correctly.

Enqueue Pattern:

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enqueue(element: T): void {
    const newNode = new QueueNode<T>(element);
    
    // Step 1: Link new node after current tail (if exists)
    if (this.tail !== null) {
        this.tail.next = newNode;
    }
    
    // Step 2: Update tail to new node (always)
    this.tail = newNode;
    
    // Step 3: If queue was empty, head must also point to new node
    if (this.head === null) {
        this.head = newNode;
    }
    
    // Step 4: Increment count
    this.count++;
}
 
// This pattern works for ALL cases:
// - Empty queue: tail was null (step 1 skipped), both pointers set to newNode
// - Non-empty: old tail linked to new, tail updated, head unchanged

Dequeue Pattern:

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dequeue(): T {
    // Step 1: Check for empty queue
    if (this.head === null) {
        throw new Error("Queue is empty");
    }
    
    // Step 2: Store data to return
    const data = this.head.data;
    
    // Step 3: Advance head to next node
    this.head = this.head.next;
    
    // Step 4: If head is now null, queue is empty - clear tail too
    if (this.head === null) {
        this.tail = null;
    }
    
    // Step 5: Decrement count
    this.count--;
    
    return data;
}
 
// This pattern works for ALL cases:
// - Single element: head.next was null, head becomes null, tail set to null
// - Multiple elements: head advances, tail unchanged

The Key Insight

Why Not Use a Doubly Linked List?

The Cost of Doubly Linked Lists:

Singly vs. Doubly Linked Lists for Queues
Aspect	Singly Linked + Tail	Doubly Linked
Memory per node	data + 1 pointer (8 bytes)	data + 2 pointers (16 bytes)
Enqueue complexity	O(1)	O(1)
Dequeue complexity	O(1)	O(1)
Pointer updates per operation	1-2	2-4
Invariants to maintain	3-4	6-8
Bug opportunity surface	Smaller	Larger

For a standard queue:

We only need:

O(1) insertion at rear ✓ (singly linked with tail pointer)
O(1) removal from front ✓ (singly linked with head pointer)

We do NOT need:

O(1) removal from rear (deques need this, queues don't)
O(1) insertion at front (deques need this, queues don't)
Bidirectional traversal (we never traverse backwards in a queue)

Doubly linked lists are overkill for queues. They add memory overhead, implementation complexity, and more invariants to maintain—all for capabilities we don't use.

When doubly linked lists ARE appropriate:

Principle of Minimal Mechanism

Complete State Trace: Seeing It All Together

Let's trace through a complete sequence of operations, showing the exact state of head, tail, and count at each step. This exercise cements understanding and serves as a debugging reference.

Operation Sequence: enqueue(10) → enqueue(20) → enqueue(30) → dequeue() → dequeue() → enqueue(40) → dequeue() → dequeue()

Complete Queue State Trace
Step	Operation	Queue Contents	head→	tail→	count
0	(initial)	[]	null	null	0
1	enqueue(10)	[10]	Node(10)	Node(10)	1
2	enqueue(20)	[10, 20]	Node(10)	Node(20)	2
3	enqueue(30)	[10, 20, 30]	Node(10)	Node(30)	3
4	dequeue() → 10	[20, 30]	Node(20)	Node(30)	2
5	dequeue() → 20	[30]	Node(30)	Node(30)	1
6	enqueue(40)	[30, 40]	Node(30)	Node(40)	2
7	dequeue() → 30	[40]	Node(40)	Node(40)	1
8	dequeue() → 40	[]	null	null	0

Key observations from the trace:

Steps 1, 5, 7: Single-element states where head === tail
Step 1: Enqueue to empty queue sets both pointers
Step 5: Dequeue to single element—only head changes
Step 8: Dequeue last element—both pointers become null
Steps 2-3, 6: Enqueue to non-empty—only tail changes
Step 4: Dequeue from multiple—only head changes

Notice how the count always matches the actual number of elements, and head/tail nullity is always consistent.

Debugging Technique

Summary: The Two-Pointer Architecture Mastered

We've thoroughly examined how head and tail pointers work together to implement an efficient queue. Let's consolidate the key insights:

Key Takeaways

•Head = Front, Tail = Rear is the only mapping that gives O(1) for both enqueue and dequeue. The reverse mapping yields O(n) dequeue.
•Invariants are crucial: Head and tail nullity must be consistent; count must match reachable nodes; tail.next must be null.
•The single-element case requires special attention—both enqueue to empty and dequeue of last element must update BOTH pointers.
•Use systematic patterns that handle all cases through conditions rather than separate code paths.
•Singly linked list is sufficient—doubly linked adds overhead for capabilities queues don't need.
•State tracing is a powerful debugging and reasoning tool.

What's next:

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