Double Ended Queue Deque in DSA C - Time & Space Complexity

Time Complexity: Double Ended Queue Deque

O(1)

Understanding Time Complexity

We want to understand how fast operations on a double ended queue (deque) run as the number of items grows.

Specifically, how does adding or removing items from either end affect the time it takes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Simple deque operations
void pushFront(Deque* dq, int val) {
    // Add val at front
    dq->front = (dq->front - 1 + dq->capacity) % dq->capacity;
    dq->arr[dq->front] = val;
    dq->size++;
}

int popBack(Deque* dq) {
    // Remove from back
    int val = dq->arr[(dq->front + dq->size - 1) % dq->capacity];
    dq->size--;
    return val;
}
    

This code adds an item at the front and removes an item from the back of a deque implemented with a circular array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Simple index calculations and direct array access.
• How many times: Each operation runs once per call, no loops or recursion inside.

How Execution Grows With Input

Each push or pop does a fixed number of steps regardless of how many items are in the deque.

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

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

Final Time Complexity

Time Complexity: O(1)

This means adding or removing items from either end takes the same short time no matter how many items are stored.

Common Mistake

[X] Wrong: "Removing from the back of a deque takes longer as it gets bigger because you have to shift items."

[OK] Correct: The deque uses a circular array with index math, so no shifting is needed. Each remove or add is a quick step.

Interview Connect

Knowing that deque operations run in constant time shows you understand efficient data structures, a key skill in coding interviews and real projects.

Self-Check

"What if the deque was implemented using a linked list instead of an array? How would the time complexity change?"

Practice

1. What is a Double Ended Queue (Deque)?

easy

A. A queue where insertion happens only at the rear and deletion at the front

B. A data structure where insertion and deletion can happen at both ends

C. A stack where insertion and deletion happen only at one end

D. A list that allows random access to elements

Solution

1. Step 1: Understand the definition of Deque

   A Deque allows adding and removing elements from both front and rear ends.

2. Step 2: Compare with other data structures

   Unlike a queue or stack, Deque supports operations at both ends.

3. Final Answer:

   A data structure where insertion and deletion can happen at both ends -> Option B

4. Quick Check:

   Deque = Both ends insertion/deletion [OK]

Hint: Deque means double-ended queue, both ends allowed [OK]

Common Mistakes:

• Confusing deque with stack or queue
• Thinking insertion is only at one end
• Assuming random access like arrays

2. Which of the following is the correct syntax to initialize an empty deque in C using an array implementation?

easy

A. int deque[] = {}; int front = 0, rear = 0;

B. int deque[SIZE] = NULL; int front = 0, rear = 0;

C. int deque[SIZE]; int front = -1, rear = -1;

D. int deque[SIZE]; int front = 0, rear = -1;

Solution

1. Step 1: Understand initialization of deque with front and rear

   Front and rear are set to -1 to indicate empty deque.

2. Step 2: Check array declaration and initial values

   int deque[SIZE]; int front = -1, rear = -1; correctly declares array and sets front and rear to -1.

3. Final Answer:

   int deque[SIZE]; int front = -1, rear = -1; -> Option C

4. Quick Check:

   Empty deque front and rear = -1 [OK]

Hint: Empty deque front and rear start at -1 [OK]

Common Mistakes:

• Setting front or rear to 0 initially
• Using NULL for array initialization
• Leaving array size unspecified

3. Consider the following operations on an empty deque of size 5 implemented using an array:
insertRear(10); insertRear(20); insertFront(5); deleteRear(); insertFront(15);
What is the content of the deque from front to rear after these operations?

medium

A. 15 -> 5 -> 10 -> null

B. 5 -> 15 -> 10 -> null

C. 15 -> 5 -> null

D. 10 -> 5 -> 15 -> null

Solution

1. Step 1: Perform insertRear(10) and insertRear(20)

   Deque: 10 -> 20

2. Step 2: Perform insertFront(5)

   Deque: 5 -> 10 -> 20

3. Step 3: Perform deleteRear()

   Removes 20, Deque: 5 -> 10

4. Step 4: Perform insertFront(15)

   Deque: 15 -> 5 -> 10

5. Final Answer:

   15 -> 5 -> 10 -> null -> Option A

6. Quick Check:

   Operations result in 15,5,10 order [OK]

Hint: Track front and rear insertions stepwise [OK]

Common Mistakes:

• Forgetting deleteRear removes last element
• Mixing order of insertFront and insertRear
• Assuming insertFront adds at rear

4. Given the following code snippet for deque insertion at front, identify the error:

void insertFront(int deque[], int *front, int *rear, int value) {
  if (*front == 0) {
    printf("Deque is full at front\n");
    return;
  }
  (*front)--;
  deque[*front] = value;
}

medium

A. It incorrectly checks full condition only at front == 0

B. It decrements front before insertion causing invalid index

C. It does not check if deque is empty before insertion

D. It should increment front instead of decrement

Solution

1. Step 1: Analyze the full condition check

   Checking only if front == 0 is not enough to detect full deque.

2. Step 2: Understand circular deque logic

   Deque can be full even if front != 0, so this check is incomplete.

3. Final Answer:

   It incorrectly checks full condition only at front == 0 -> Option A

4. Quick Check:

   Full check must consider rear position too [OK]

Hint: Full deque check needs both front and rear positions [OK]

Common Mistakes:

• Assuming front == 0 means full always
• Not handling circular wrap-around
• Decrementing front without boundary check

5. You have a deque of size 4 implemented using a circular array. Initially empty, you perform:
insertRear(1); insertRear(2); insertFront(3); insertFront(4); deleteRear(); insertRear(5);
What is the final state of the deque from front to rear?

hard

A. 4 -> 3 -> 1 -> null

B. 4 -> 3 -> 5 -> null

C. 3 -> 4 -> 1 -> 5 -> null

D. 4 -> 3 -> 1 -> 5 -> null

Solution

1. Step 1: InsertRear(1), InsertRear(2)

   Deque: 1 -> 2

2. Step 2: InsertFront(3), InsertFront(4)

   Deque: 4 -> 3 -> 1 -> 2 (deque full)

3. Step 3: DeleteRear()

   Removes 2, Deque: 4 -> 3 -> 1

4. Step 4: InsertRear(5)

   Deque: 4 -> 3 -> 1 -> 5 (full again)

5. Final Answer:

   4 -> 3 -> 1 -> 5 -> null -> Option D

6. Quick Check:

   Circular deque final state matches 4 -> 3 -> 1 -> 5 -> null [OK]

Hint: Track circular wrap and full capacity carefully [OK]

Common Mistakes:

• Ignoring circular wrap-around
• Misordering front and rear elements
• Forgetting deleteRear removes last element