Queue Implementation Using Array in DSA C - Time & Space Complexity

Time Complexity: Queue Implementation Using Array

O(1)

Understanding Time Complexity

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

Specifically, how fast can we add or remove items in a queue implemented with an array?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#define MAX 100
int queue[MAX];
int front = -1, rear = -1;

void enqueue(int x) {
  if (rear == MAX - 1) return; // queue full
  if (front == -1) front = 0;
  queue[++rear] = x;
}

int dequeue() {
  if (front == -1 || front > rear) return -1; // queue empty
  return queue[front++];
}
    

This code adds (enqueue) and removes (dequeue) elements from a queue using a fixed-size array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single-step insert or remove at array index.
• How many times: Each enqueue or dequeue does one operation regardless of queue size.

How Execution Grows With Input

Each operation takes the same small number of steps no matter how many items are in the queue.

Input Size (n)	Approx. Operations
10	1 operation per enqueue or dequeue
100	1 operation per enqueue or dequeue
1000	1 operation per enqueue or dequeue

Pattern observation: The time does not increase with more elements; it stays constant.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing an item takes the same short time no matter how big the queue is.

Common Mistake

[X] Wrong: "Removing an item from the queue takes longer as the queue grows because we have to shift elements."

[OK] Correct: In this implementation, we do not shift elements; we just move the front index forward, so removal stays fast.

Interview Connect

Understanding how queue operations stay fast helps you explain efficient data handling in real systems.

Self-Check

"What if we changed the queue to shift all elements left on dequeue? How would the time complexity change?"

Practice

1. What is the main characteristic of a queue implemented using an array?

easy

A. It follows First-In-First-Out (FIFO) order

B. It follows Last-In-First-Out (LIFO) order

C. It allows random access to elements

D. It stores elements in sorted order

Solution

1. Step 1: Understand queue behavior

   A queue is a data structure where the first element added is the first one to be removed, which is FIFO.

2. Step 2: Compare with options

   It follows First-In-First-Out (FIFO) order correctly states FIFO order. It follows Last-In-First-Out (LIFO) order is LIFO (stack), C allows random access (array), D stores sorted (not queue).

3. Final Answer:

   It follows First-In-First-Out (FIFO) order -> Option A

4. Quick Check:

   Queue = FIFO [OK]

Hint: Queue always removes the oldest element first [OK]

Common Mistakes:

• Confusing queue with stack (LIFO)
• Thinking queue allows random access
• Assuming queue sorts elements automatically

2. Which of the following is the correct way to declare a queue array of size 5 in C?

easy

A. queue int[5];

B. int queue[5];

C. int queue();

D. int queue = new int[5];

Solution

1. Step 1: Recall C array declaration syntax

   In C, arrays are declared as type name[size], e.g., int queue[5];

2. Step 2: Check each option

   int queue[5]; matches correct syntax. queue int[5]; is invalid syntax. int queue(); declares a function, not an array. int queue = new int[5]; is Java-style syntax.

3. Final Answer:

   int queue[5]; -> Option B

4. Quick Check:

   C array declaration = int name[size] [OK]

Hint: C arrays use type name[size] syntax [OK]

Common Mistakes:

• Using Java or other language syntax
• Confusing function declaration with array
• Missing semicolon or brackets

3. Consider this C code snippet for a queue implemented using an array of size 3:

int queue[3];
int front = -1, rear = -1;

// Enqueue 10
rear++;
queue[rear] = 10;
if(front == -1) front = 0;

// Enqueue 20
rear++;
queue[rear] = 20;

// Dequeue
int val = queue[front];
front++;

// Enqueue 30
rear++;
queue[rear] = 30;

// Current queue elements from front to rear?

medium

A. 30 -> null

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

C. 20 -> 30 -> null

D. 10 -> 30 -> null

Solution

1. Step 1: Trace enqueue and dequeue operations

   Initially front = -1, rear = -1. Enqueue 10: rear=0, queue[0]=10, front=0. Enqueue 20: rear=1, queue[1]=20. Dequeue: val=queue[front]=queue[0]=10, front=1. Enqueue 30: rear=2, queue[2]=30.

2. Step 2: Identify current queue elements

   front=1, rear=2, so elements are queue[1]=20 and queue[2]=30. Hence queue is 20 -> 30 -> null.

3. Final Answer:

   20 -> 30 -> null -> Option C

4. Quick Check:

   After dequeue, front moves forward; queue = 20, 30 [OK]

Hint: After dequeue, front moves forward; queue shows elements from front to rear [OK]

Common Mistakes:

• Not updating front after dequeue
• Including dequeued elements in output
• Confusing front and rear indices

4. Identify the error in this queue enqueue function using array in C:

void enqueue(int queue[], int *rear, int size, int value) {
    if (*rear == size) {
        printf("Queue is full\n");
        return;
    }
    (*rear)++;
    queue[*rear] = value;
}

medium

A. rear should be compared with size - 1, not size

B. rear should be incremented after inserting value

C. Function should return int instead of void

D. queue array should be dynamically allocated

Solution

1. Step 1: Analyze the full condition

   The check if(*rear == size) is wrong. Array indices are 0 to size-1, so full when *rear == size-1 before incrementing. Current check allows overflow.

2. Step 2: Compare options

   rear should be compared with size - 1, not size identifies the correct fix: compare with size-1, not size. Others are incorrect.

3. Final Answer:

   rear should be compared with size - 1, not size -> Option A

4. Quick Check:

   Full: *rear == size-1 [OK]

Hint: Full check: rear == size - 1 [OK]

Common Mistakes:

• Comparing rear == size instead of size-1
• Incorrect overflow condition
• Ignoring zero-based indexing

5. You have a queue implemented using an array of size 4 with front = 1 and rear = 3. After dequeuing two elements and enqueuing one element, what will be the correct front and rear values if you implement a circular queue?

hard

A. front = 2, rear = 0

B. front = 3, rear = 1

C. front = 0, rear = 1

D. front = 3, rear = 0

Solution

1. Step 1: Understand initial state and operations

   Initial: front=1, rear=3, size=4. Dequeue twice: front moves forward twice. Since circular, front = (1 + 2) % 4 = 3.

2. Step 2: Enqueue one element

   rear moves forward by one circularly: rear = (3 + 1) % 4 = 0.

3. Final Answer:

   front = 3, rear = 0 -> Option D

4. Quick Check:

   Circular queue uses modulo for front and rear [OK]

Hint: Use modulo to wrap front and rear in circular queue [OK]

Common Mistakes:

• Not using modulo for circular increment
• Updating front or rear incorrectly
• Confusing front and rear positions