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

Time Complexity: Double Ended Queue Deque

O(n) for front operations, O(1) for back operations

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 adding or removing items from either end affects the time taken.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

class Deque {
  private items: any[] = [];

  addFront(element: any) {
    this.items.unshift(element);
  }

  addBack(element: any) {
    this.items.push(element);
  }

  removeFront() {
    return this.items.shift();
  }

  removeBack() {
    return this.items.pop();
  }
}
    

This code shows a deque with methods to add or remove items from both front and back using an array.

Identify Repeating Operations

Look at the main operations that repeat when the deque grows.

• Primary operation: Array methods like unshift and shift that move elements.
• How many times: These methods may move many elements depending on the deque size n.

How Execution Grows With Input

Adding or removing at the back uses push or pop, which do not move other elements.

Adding or removing at the front uses unshift or shift, which move all existing elements one step.

Input Size (n)	Approx. Operations for addFront/removeFront
10	About 10 moves
100	About 100 moves
1000	About 1000 moves

Pattern observation: Operations at the front grow linearly with the number of items, while back operations stay constant.

Final Time Complexity

Time Complexity: O(n) for front operations, O(1) for back operations

This means adding or removing at the back is fast no matter the size, but at the front it takes longer as the deque grows.

Common Mistake

[X] Wrong: "All deque operations take the same constant time because it's a queue."

[OK] Correct: Using an array, front operations move many elements, so they take longer as size grows, unlike back operations.

Interview Connect

Understanding these differences helps you choose the right data structure and explain your choices clearly in interviews.

Self-Check

What if we implemented the deque using a linked list instead of an array? How would the time complexity of front operations change?

Practice

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

easy

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

B. A queue where elements can only be added at the rear and removed from the front

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 a deque

   A deque allows insertion and deletion from both front and rear ends, unlike a queue or stack.

2. Step 2: Compare with other data structures

   Queue allows operations only at one end for insertion and the other for deletion; stack only at one end.

3. Final Answer:

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

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 queue
• Thinking deque allows random access
• Assuming insertion only at one end

2. Which of the following is the correct TypeScript method to add an element at the front of a deque implemented as an array named deque?

easy

A. deque.shift(element);

B. deque.push(element);

C. deque.pop(element);

D. deque.unshift(element);

Solution

1. Step 1: Recall array methods for front insertion

   In TypeScript/JavaScript, unshift() adds an element at the start of an array.

2. Step 2: Verify other methods

   push() adds at the end, pop() removes from end, shift() removes from front.

3. Final Answer:

   deque.unshift(element); -> Option D

4. Quick Check:

   unshift() adds front [OK]

Hint: Use unshift() to add front in arrays [OK]

Common Mistakes:

• Using push() instead of unshift() for front
• Confusing pop() and shift() as insertion
• Passing arguments to pop() or shift()

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

const deque: number[] = [];
deque.push(10);
deque.unshift(20);
deque.push(30);
deque.shift();
console.log(deque);

medium

A. [20, 10, 30]

B. [20, 30]

C. [10, 30]

D. [30, 20]

Solution

1. Step 1: Trace each operation on deque

   Start: []
   push(10) -> [10]
   unshift(20) -> [20, 10]
   push(30) -> [20, 10, 30]
   shift() removes front -> removes 20 -> [10, 30]

2. Step 2: Final state of deque

   After all operations, deque is [10, 30].

3. Final Answer:

   [10, 30] -> Option C

4. Quick Check:

   After shift(), front removed -> [10, 30] [OK]

Hint: Remember shift() removes front element [OK]

Common Mistakes:

• Forgetting shift() removes front
• Mixing order of push and unshift
• Assuming shift() removes last element

4. Identify the error in the following TypeScript code that implements a deque using an array:

const deque: number[] = [];
deque.push(5);
deque.unshift(10);
deque.pop(5);
console.log(deque);

medium

A. push() cannot be used on arrays

B. pop() should not have any arguments

C. unshift() removes elements, not adds

D. console.log syntax is incorrect

Solution

1. Step 1: Check usage of pop() method

   pop() removes the last element and does not take any arguments. Passing 5 is incorrect.

2. Step 2: Verify other methods and syntax

   push() adds element at end, unshift() adds at front, console.log syntax is correct.

3. Final Answer:

   pop() should not have any arguments -> Option B

4. Quick Check:

   pop() takes no arguments [OK]

Hint: pop() never takes arguments [OK]

Common Mistakes:

• Passing argument to pop()
• Confusing unshift() as removal
• Thinking push() is invalid on arrays

5. You want to implement a deque that supports adding and removing elements from both ends efficiently. Which data structure is best suited for this in TypeScript, and why?

hard

A. Use a linked list because it allows O(1) insertion and deletion at both ends

B. Use an array because it provides O(1) insertion and deletion at both ends

C. Use a binary tree because it supports fast search and insertion

D. Use a stack because it supports adding/removing from both ends

Solution

1. Step 1: Analyze efficiency of array for deque operations

   Arrays have O(1) insertion/deletion at the end but O(n) at the front due to shifting elements.

2. Step 2: Consider linked list advantages

   Linked lists allow O(1) insertion and deletion at both front and rear by adjusting pointers.

3. Step 3: Evaluate other options

   Binary trees are for search, not efficient deque operations; stacks only allow one end operations.

4. Final Answer:

   Use a linked list because it allows O(1) insertion and deletion at both ends -> Option A

5. Quick Check:

   Linked list = efficient deque [OK]

Hint: Linked list supports fast front and rear operations [OK]

Common Mistakes:

• Assuming arrays are efficient at front insertion
• Confusing stack with deque
• Choosing binary tree for deque operations