Double Ended Queue Deque in DSA Java - 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, we ask: how does the time to add or remove items from either end change with size?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public class SimpleDeque {
    private int[] data = new int[1000];
    private int front = 0, rear = 0;

    public void addFront(int val) {
        front = (front - 1 + data.length) % data.length;
        data[front] = val;
    }

    public void addRear(int val) {
        data[rear] = val;
        rear = (rear + 1) % data.length;
    }

    public int removeFront() {
        int val = data[front];
        front = (front + 1) % data.length;
        return val;
    }

    public int removeRear() {
        rear = (rear - 1 + data.length) % data.length;
        return data[rear];
    }
}
    

This code shows adding and removing items from both ends of a fixed-size deque using a circular array.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Single-step index updates and array assignments or retrievals.
• How many times: Each add or remove operation runs once per call, no loops inside.

How Execution Grows With Input

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

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

Pattern observation: The time stays the same no matter how many items are stored.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing from either end takes the same short time no matter the deque size.

Common Mistake

[X] Wrong: "Removing from the rear takes longer because you have to move all items."

[OK] Correct: The code uses index math to jump directly to the rear item, so no moving of other items happens.

Interview Connect

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

Self-Check

"What if the deque used a linked list instead of an array? How would the time complexity of add and remove operations change?"

Practice

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

easy

A. A queue where elements can only be added at the rear

B. A data structure where elements can be added or removed from both ends

C. A stack where elements can only be added or removed from the top

D. A list that allows random access to elements

Solution

1. Step 1: Understand the definition of Deque

   A Deque allows insertion and deletion from both front and rear ends.

2. Step 2: Compare with other data structures

   Unlike queues or stacks, Deque supports operations at both ends.

3. Final Answer:

   A data structure where elements can be added or removed from both ends -> Option B

4. Quick Check:

   Deque = double-ended queue [OK]

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

Common Mistakes:

• Confusing Deque with stack or simple queue
• Thinking Deque allows random access like arrays
• Assuming Deque only works from one end

2. Which of the following is the correct way to create a Deque in Java using the standard library?

easy

A. Deque<Integer> dq = new ArrayDeque<>();

B. Deque<Integer> dq = new Deque<>();

C. Deque<Integer> dq = new Stack<>();

D. Deque<Integer> dq = new Queue<>();

Solution

1. Step 1: Identify valid Deque implementations in Java

   ArrayDeque implements the Deque interface and is the preferred implementation.

2. Step 2: Check syntax correctness

   ArrayDeque is instantiated with new ArrayDeque<>(), which is correct syntax.

3. Final Answer:

   Deque<Integer> dq = new ArrayDeque<>(); -> Option A

4. Quick Check:

   ArrayDeque is standard Deque implementation [OK]

Hint: Use ArrayDeque for Deque in Java [OK]

Common Mistakes:

• Using Stack which is not a Deque
• Trying to instantiate interface Queue directly
• Trying to instantiate the Deque interface directly

3. What will be the output of the following Java code?

Deque<String> dq = new ArrayDeque<>();
dq.addFirst("A");
dq.addLast("B");
dq.addFirst("C");
System.out.println(dq);

medium

A. [A, C, B]

B. [A, B, C]

C. [C, A, B]

D. [B, A, C]

Solution

1. Step 1: Trace addFirst and addLast operations

   addFirst("A") adds A at front: [A]
   addLast("B") adds B at rear: [A, B]
   addFirst("C") adds C at front: [C, A, B]

2. Step 2: Print the Deque

   Printing dq shows elements in order: [C, A, B]

3. Final Answer:

   [C, A, B] -> Option C

4. Quick Check:

   addFirst adds front, addLast adds rear [OK]

Hint: addFirst puts element at front, addLast at rear [OK]

Common Mistakes:

• Mixing order of addFirst and addLast
• Assuming addFirst adds at rear
• Confusing print output format

4. Identify the error in the following Java code using Deque:

Deque<Integer> dq = new ArrayDeque<>();
dq.addFirst(10);
dq.addLast(20);
dq.removeLast();
dq.removeFirst();
dq.removeFirst();

medium

A. No error, code runs fine

B. Compilation error due to wrong method names

C. Throws NullPointerException on removeLast()

D. Throws NoSuchElementException on last removeFirst()

Solution

1. Step 1: Trace the Deque operations

   Start: empty
   addFirst(10): [10]
   addLast(20): [10, 20]
   removeLast(): removes 20 -> [10]
   removeFirst(): removes 10 -> []
   removeFirst(): tries to remove from empty Deque

2. Step 2: Identify exception on removing from empty Deque

   removeFirst() on empty Deque throws NoSuchElementException

3. Final Answer:

   Throws NoSuchElementException on last removeFirst() -> Option D

4. Quick Check:

   Removing from empty Deque throws NoSuchElementException [OK]

Hint: Removing from empty Deque throws NoSuchElementException [OK]

Common Mistakes:

• Assuming removeFirst returns null instead of exception
• Confusing method names causing compile error
• Thinking removeLast throws NullPointerException

5. You want to implement a task scheduler that processes tasks in the order they arrive but sometimes needs to prioritize urgent tasks by adding them to the front. Which Deque operation combination is best suited for this scenario?

hard

A. Use addLast() for normal tasks and addFirst() for urgent tasks

B. Use addFirst() for normal tasks and addLast() for urgent tasks

C. Use only addLast() for all tasks

D. Use only addFirst() for all tasks

Solution

1. Step 1: Understand task order requirements

   Normal tasks should be processed in arrival order (queue behavior). Urgent tasks should be processed before normal tasks (front insertion).

2. Step 2: Match Deque operations to requirements

   addLast() adds normal tasks at rear (end), preserving arrival order.
   addFirst() adds urgent tasks at front, prioritizing them.

3. Final Answer:

   Use addLast() for normal tasks and addFirst() for urgent tasks -> Option A

4. Quick Check:

   Normal at rear, urgent at front [OK]

Hint: Normal tasks rear, urgent tasks front with addLast/addFirst [OK]

Common Mistakes:

• Reversing addFirst and addLast roles
• Using only one end for all tasks losing priority
• Confusing queue and stack behavior