Enqueue Operation in DSA C - Time & Space Complexity

Time Complexity: Enqueue Operation

O(1)

Understanding Time Complexity

We want to understand how the time to add an item to a queue changes as the queue grows.

How does the enqueue operation's time depend on the number of items already in the queue?

Scenario Under Consideration

Analyze the time complexity of the following enqueue code for a queue implemented with a linked list.

// Add a new node at the end of the queue
void enqueue(Queue* q, int value) {
    Node* newNode = malloc(sizeof(Node));
    newNode->data = value;
    newNode->next = NULL;
    if (q->rear == NULL) {
        q->front = newNode;
        q->rear = newNode;
    } else {
        q->rear->next = newNode;
        q->rear = newNode;
    }
}
    

This code adds a new element to the end of a queue using a linked list with front and rear pointers.

Identify Repeating Operations

Look for loops or repeated steps that depend on queue size.

• Primary operation: Creating a new node and updating pointers.
• How many times: This happens once per enqueue call, no loops involved.

How Execution Grows With Input

Adding one item always takes the same steps regardless of queue size.

Input Size (n)	Approx. Operations
10	5-6 steps
100	5-6 steps
1000	5-6 steps

Pattern observation: The number of steps stays about the same no matter how big the queue is.

Final Time Complexity

Time Complexity: O(1)

This means adding an item to the queue takes a constant amount of time, no matter how many items are already there.

Common Mistake

[X] Wrong: "Enqueue takes longer as the queue grows because it has to find the end each time."

[OK] Correct: The queue keeps a pointer to the last item, so it does not search; it directly adds the new item, keeping time constant.

Interview Connect

Understanding constant time operations like enqueue helps you design efficient data structures and shows you know how to keep operations fast as data grows.

Self-Check

"What if the queue was implemented using an array without a rear pointer? How would the enqueue time complexity change?"

Practice

1. What does the enqueue operation do in a queue data structure?

easy

A. Adds an element to the rear of the queue

B. Removes an element from the front of the queue

C. Checks if the queue is empty

D. Returns the size of the queue

Solution

1. Step 1: Understand the queue operations

   In a queue, enqueue means adding an element, while dequeue means removing one.

2. Step 2: Identify the position of addition

   Elements are always added at the rear (end) of the queue during enqueue.

3. Final Answer:

   Adds an element to the rear of the queue -> Option A

4. Quick Check:

   Enqueue = Add rear [OK]

Hint: Enqueue always adds at the back of the queue [OK]

Common Mistakes:

• Confusing enqueue with dequeue operation
• Thinking enqueue removes elements
• Assuming enqueue adds at the front

2. Which of the following is the correct syntax to enqueue an element 10 in a queue implemented using an array named queue with rear index rear in C?

easy

A. queue[rear] = ++10;

B. queue[rear++] = 10;

C. queue[rear] = 10++;

D. queue[++rear] = 10;

Solution

1. Step 1: Understand post-increment vs pre-increment

   Post-increment (rear++) uses rear as index then increments it. Pre-increment (++rear) increments rear first then uses it as index.

2. Step 2: Choose correct index update for enqueue

   To add at current rear position and then move rear forward, use post-increment: queue[rear++] = 10;.

3. Final Answer:

   queue[rear++] = 10; -> Option B

4. Quick Check:

   Post-increment rear after assignment [OK]

Hint: Use rear++ to assign then move rear forward [OK]

Common Mistakes:

• Using pre-increment causing skip of first index
• Trying to increment the value instead of index
• Syntax errors with invalid increments

3. What will be the state of the queue array after executing the following code?

int queue[5];
int rear = -1;
queue[++rear] = 5;
queue[++rear] = 10;
queue[++rear] = 15;

medium

A. [5, 10, 0, 15, 0]

B. [0, 5, 10, 15, 0]

C. [5, 10, 15, 0, 0]

D. [15, 10, 5, 0, 0]

Solution

1. Step 1: Trace each enqueue operation

   Initially, rear = -1.
   First enqueue: ++rear = 0, queue[0] = 5.
   Second enqueue: ++rear = 1, queue[1] = 10.
   Third enqueue: ++rear = 2, queue[2] = 15.

2. Step 2: Determine final queue array state

   Positions 0,1,2 have 5,10,15 respectively. Others remain 0 (default uninitialized).

3. Final Answer:

   [5, 10, 15, 0, 0] -> Option C

4. Quick Check:

   Enqueue adds sequentially at indices 0,1,2 [OK]

Hint: Pre-increment rear before assignment fills array from index 0 [OK]

Common Mistakes:

• Confusing post-increment with pre-increment
• Assuming queue starts at index 1
• Ignoring initial rear value -1

4. Identify the error in the following enqueue function for a queue implemented using an array and fix it:

void enqueue(int queue[], int *rear, int value) {
    queue[*rear] = value;
    (*rear)++;
}

medium

A. The value should be assigned after incrementing rear twice

B. The rear should not be incremented at all

C. The function should decrement rear instead

D. The rear should be incremented before assignment

Solution

1. Step 1: Analyze the enqueue logic

   The function assigns value at current rear index, then increments rear.
   This causes the first element to be placed at rear=0 if rear was 0, which is correct only if rear points to next free slot.

2. Step 2: Correct the order for standard enqueue

   Usually, rear points to last filled index, so increment rear first, then assign:
   queue[++(*rear)] = value;.

3. Final Answer:

   The rear should be incremented before assignment -> Option D

4. Quick Check:

   Increment rear first, then assign value [OK]

Hint: Increment rear before assignment to avoid overwrite [OK]

Common Mistakes:

• Assigning value before incrementing rear
• Incrementing rear twice accidentally
• Decrementing rear instead of incrementing

5. Given a circular queue of size 5 with front = 3 and rear = 1, which of the following enqueue operations is valid and will not cause overflow?

hard

A. Enqueue at rear = 2

B. Enqueue at rear = 3

C. Enqueue at rear = 4

D. Enqueue at rear = 0

Solution

1. Step 1: Understand circular queue indexing

   Queue size = 5, indices 0 to 4.
   front = 3, rear = 1 means elements occupy positions from front to rear wrapping around.

2. Step 2: Check next rear position for enqueue

   Next rear = (rear + 1) % size = (1 + 1) % 5 = 2.
   If next rear equals front, queue is full (overflow). Here, next rear 2 != front 3, so enqueue at 2 is valid.
   But options show enqueue at rear = 2, 3, 4, 0. We must check which is correct next rear.

3. Step 3: Validate each option

   Enqueue at rear = 2: rear=2 is next rear, valid.
   Enqueue at rear = 3: rear=3 equals front, overflow.
   Enqueue at rear = 4: rear=4 is not next rear, invalid.
   Enqueue at rear = 0: rear=0 is not next rear, invalid.

4. Final Answer:

   Enqueue at rear = 2 -> Option A

5. Quick Check:

   Next rear = (rear+1)%size = 2, enqueue valid [OK]

Hint: Next rear = (rear + 1) mod size; must not equal front [OK]

Common Mistakes:

• Ignoring circular wrap-around logic
• Enqueueing at front index causing overflow
• Assuming rear can be any index