Queue Implementation Using Linked List in DSA C - Time & Space Complexity

Time Complexity: Queue Implementation Using Linked List

O(1)

Understanding Time Complexity

We want to understand how the time taken by queue operations changes as the number of items grows.

How does adding or removing items affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following queue operations using a linked list.

// Node structure
struct Node {
    int data;
    struct Node* next;
};

// Enqueue operation
void enqueue(struct Node** front, struct Node** rear, int value) {
    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = value;
    newNode->next = NULL;
    if (*rear == NULL) {
        *front = *rear = newNode;
        return;
    }
    (*rear)->next = newNode;
    *rear = newNode;
}

// Dequeue operation
int dequeue(struct Node** front, struct Node** rear) {
    if (*front == NULL) return -1; // Queue empty
    struct Node* temp = *front;
    int val = temp->data;
    *front = (*front)->next;
    if (*front == NULL) *rear = NULL;
    free(temp);
    return val;
}
    

This code adds items to the end and removes from the front of a linked list to implement a queue.

Identify Repeating Operations

Look for loops or repeated steps inside enqueue and dequeue.

• Primary operation: Adding or removing a single node pointer.
• How many times: Each enqueue or dequeue does a fixed number of steps, no loops.

How Execution Grows With Input

Each operation works with only the front or rear pointer, so the work stays the same no matter how many items are in the queue.

Input Size (n)	Approx. Operations
10	5-7 steps per enqueue or dequeue
100	5-7 steps per enqueue or dequeue
1000	5-7 steps per enqueue or dequeue

Pattern observation: The number of steps does not increase with more items; it stays constant.

Final Time Complexity

Time Complexity: O(1)

This means each enqueue or dequeue takes the same amount of time regardless of queue size.

Common Mistake

[X] Wrong: "Enqueue or dequeue takes longer as the queue grows because we have to move through the list."

[OK] Correct: We keep track of front and rear pointers, so we never walk through the whole list during these operations.

Interview Connect

Knowing that queue operations with linked lists run in constant time shows you understand efficient data structure design, a key skill in coding interviews.

Self-Check

"What if we did not keep a rear pointer and only had a front pointer? How would the time complexity of enqueue change?"

Practice

1. What is the main advantage of implementing a queue using a linked list instead of an array?

easy

A. Uses less memory than arrays always

B. Faster access to elements by index

C. Simpler code with fewer pointers

D. Dynamic size allows efficient memory use without fixed limits

Solution

1. Step 1: Understand queue size constraints in arrays

   Arrays have fixed size, so queues implemented with arrays have a maximum capacity.

2. Step 2: Analyze linked list flexibility

   Linked lists can grow or shrink dynamically, so queues using linked lists can handle any number of elements limited only by memory.

3. Final Answer:

   Dynamic size allows efficient memory use without fixed limits -> Option D

4. Quick Check:

   Linked list queues grow dynamically = A [OK]

Hint: Linked list queues grow dynamically, arrays don't [OK]

Common Mistakes:

• Thinking arrays always use less memory
• Confusing access speed with size flexibility
• Assuming linked lists have simpler code

2. Which of the following is the correct way to declare a node structure for a queue using a linked list in C?

easy

A. typedef struct Node { int data; Node* next; };

B. struct Node { int data; int next; };

C. typedef struct Node { int data; struct Node* next; } Node;

D. struct Node { int data; struct Node next; };

Solution

1. Step 1: Check pointer usage in struct

   The next pointer must be a pointer to struct Node, so 'struct Node* next;' is correct.

2. Step 2: Validate typedef syntax

   Using typedef with struct Node allows using 'Node' as type directly, which is correct in typedef struct Node { int data; struct Node* next; } Node;.

3. Final Answer:

   typedef struct Node { int data; struct Node* next; } Node; -> Option C

4. Quick Check:

   Pointer to struct Node with typedef = A [OK]

Hint: Use 'struct Node* next;' for pointer inside struct [OK]

Common Mistakes:

• Using 'int next;' instead of pointer
• Omitting 'struct' keyword before Node*
• Declaring next as struct Node instead of pointer

3. Consider the following C code snippet implementing enqueue operation for a queue using linked list:

void enqueue(Node** front, Node** rear, int value) {
    Node* newNode = (Node*)malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = NULL;
    if (*rear == NULL) {
        *front = *rear = newNode;
    } else {
        (*rear)->next = newNode;
        *rear = newNode;
    }
}

What will be the state of the queue after these operations? enqueue(&front, &rear, 10); enqueue(&front, &rear, 20); enqueue(&front, &rear, 30);

medium

A. 10 -> 20 -> 30 -> NULL

B. 30 -> 20 -> 10 -> NULL

C. 10 -> 30 -> 20 -> NULL

D. 20 -> 10 -> 30 -> NULL

Solution

1. Step 1: Trace first enqueue call

   Queue is empty, so front and rear point to new node with data 10.

2. Step 2: Trace second and third enqueue calls

   New nodes with 20 and 30 are added at rear, linked in order: 10 -> 20 -> 30 -> NULL.

3. Final Answer:

   10 -> 20 -> 30 -> NULL -> Option A

4. Quick Check:

   Enqueue adds at rear, order preserved = C [OK]

Hint: Enqueue adds at rear, so order is insertion order [OK]

Common Mistakes:

• Confusing enqueue with push (stack behavior)
• Reversing order of nodes
• Not updating rear pointer correctly

4. The following dequeue function for a queue using linked list has a bug. Identify the bug:

int dequeue(Node** front, Node** rear) {
    if (*front == NULL) {
        printf("Queue is empty\n");
        return -1;
    }
    Node* temp = *front;
    int value = temp->data;
    *front = (*front)->next;
    free(temp);
    return value;
}

medium

A. Does not free memory of dequeued node

B. Does not update rear when queue becomes empty

C. Returns wrong value after dequeue

D. Does not check if front is NULL before dequeue

Solution

1. Step 1: Analyze dequeue steps

   Function removes front node and updates front pointer correctly.

2. Step 2: Check rear pointer update

   If after dequeue front becomes NULL (queue empty), rear should also be set to NULL but this is missing.

3. Final Answer:

   Does not update rear when queue becomes empty -> Option B

4. Quick Check:

   Rear must be NULL if queue empty = B [OK]

Hint: Set rear to NULL when front becomes NULL after dequeue [OK]

Common Mistakes:

• Forgetting to update rear on empty queue
• Not freeing dequeued node memory
• Returning wrong data value

5. You want to implement a function reverseQueue that reverses the elements of a queue implemented using a linked list. Which approach is the most efficient and correct?

hard

A. Use a stack to push all elements, then enqueue them back to queue

B. Traverse the linked list and swap data of front and rear nodes repeatedly

C. Dequeue all elements and enqueue them back in the same order

D. Create a new linked list and copy elements in reverse order without modifying original

Solution

1. Step 1: Understand reversing queue with linked list

   Reversing requires changing order of elements, which is not straightforward by swapping nodes in singly linked list.

2. Step 2: Evaluate stack approach

   Using a stack to push all elements then enqueue back reverses order efficiently and correctly.

3. Step 3: Check other options

   Swapping data repeatedly is complex and error-prone; copying to new list wastes memory; re-enqueueing same order does nothing.

4. Final Answer:

   Use a stack to push all elements, then enqueue them back to queue -> Option A

5. Quick Check:

   Stack reverses order efficiently = D [OK]

Hint: Use stack to reverse queue elements easily [OK]

Common Mistakes:

• Trying to swap nodes in singly linked list
• Re-enqueueing without changing order
• Copying to new list wastes memory