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.

Specifically, how the time changes when we add or remove items from either end.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <deque>
#include <iostream>

int main() {
    std::deque<int> dq;
    dq.push_back(10);    // add to back
    dq.push_front(20);   // add to front
    dq.pop_back();       // remove from back
    dq.pop_front();      // remove from front
    return 0;
}
    

This code adds and removes elements from both ends of a deque.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Adding or removing elements at front or back.
• How many times: Each operation happens once here, but in general can happen many times.

How Execution Grows With Input

Each add or remove operation takes about the same time no matter how many items are in the deque.

Input Size (n)	Approx. Operations per add/remove
10	1
100	1
1000	1

Pattern observation: The time stays constant even as the deque grows larger.

Final Time Complexity

Time Complexity: O(1)

This means adding or removing from either end takes the same small amount of time no matter how big the deque is.

Common Mistake

[X] Wrong: "Adding to the front of a deque takes longer as it grows because it has to move all elements."

[OK] Correct: A deque is designed to add or remove at both ends quickly without moving all elements, so time stays constant.

Interview Connect

Knowing that deque operations at both ends are fast helps you choose the right data structure in real problems.

Self-Check

"What if we used a vector instead of a deque? How would the time complexity for adding to the front change?"

Practice

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

easy

A. A stack where elements are added and removed only from the top

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

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

D. A list where elements can be accessed randomly by index

Solution

1. Step 1: Understand the definition of Deque

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

2. Step 2: Compare with other data structures

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

3. Final Answer:

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

4. Quick Check:

   Deque = double-ended queue allowing both ends operations [OK]

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

Common Mistakes:

• Confusing deque with stack or queue
• Thinking deque allows random access like arrays
• Assuming deque only works like a queue

2. Which of the following is the correct syntax to declare a deque of integers in C++?

easy

A. deque<int> d = new deque();

B. deque<int> d;

C. deque<int> d[];

D. deque d<int>;

Solution

1. Step 1: Recall C++ STL deque declaration syntax

   The correct syntax uses template syntax: deque<int> d; declares a deque of integers named d.

2. Step 2: Check other options for syntax errors

   'deque<int> d = new deque();' uses invalid 'new' syntax for STL containers, 'deque d<int>;' has wrong template placement, 'deque<int> d[];' declares an array of deques incorrectly.

3. Final Answer:

   deque<int> d; -> Option B

4. Quick Check:

   Correct C++ deque declaration = deque<int> d; [OK]

Hint: Use template syntax: deque<Type> name; [OK]

Common Mistakes:

• Placing template brackets incorrectly
• Using Java or other language syntax
• Declaring arrays of deques without proper syntax

3. What will be the output of the following C++ code?

#include <iostream>
#include <deque>
using namespace std;
int main() {
  deque<int> d;
  d.push_back(10);
  d.push_front(20);
  d.push_back(30);
  d.pop_front();
  for (auto x : d) cout << x << " ";
  return 0;
}

medium

A. 10 30

B. 30 10

C. 20 10 30

D. 10 20 30

Solution

1. Step 1: Trace deque operations step-by-step

   Initially empty deque. push_back(10) -> [10]. push_front(20) -> [20, 10]. push_back(30) -> [20, 10, 30]. pop_front() removes 20 -> [10, 30].

2. Step 2: Print elements in deque

   Loop prints elements in order: 10 30

3. Final Answer:

   10 30 -> Option A

4. Quick Check:

   After pop_front, deque = 10 30 [OK]

Hint: Track push_front/back and pop_front carefully [OK]

Common Mistakes:

• Forgetting pop_front removes front element
• Mixing order of push_front and push_back
• Assuming pop_front removes from back

4. Identify the error in the following C++ code using deque:

#include <deque>
using namespace std;
int main() {
  deque<int> d;
  d.push_front(5);
  d.pop_back();
  d.pop_back();
  return 0;
}

medium

A. deque<int> d; is invalid declaration

B. push_front() is not a valid deque method

C. Missing #include <iostream> for cout

D. Calling pop_back() on empty deque causes undefined behavior

Solution

1. Step 1: Analyze deque operations

   After push_front(5), deque has one element [5]. First pop_back() removes 5, deque becomes empty []. Second pop_back() is called on empty deque.

2. Step 2: Understand behavior of pop_back on empty deque

   Calling pop_back() on empty deque causes undefined behavior or runtime error.

3. Final Answer:

   Calling pop_back() on empty deque causes undefined behavior -> Option D

4. Quick Check:

   pop_back on empty deque = error [OK]

Hint: Never pop from empty deque to avoid errors [OK]

Common Mistakes:

• Assuming pop_back on empty deque is safe
• Confusing push_front with invalid method
• Ignoring missing iostream is irrelevant here

5. You want to implement a sliding window maximum using a deque in C++. Which deque operation is essential to maintain the window elements in decreasing order?

hard

A. Remove elements from the back while they are smaller than the new element

B. Remove elements from the front while they are smaller than the new element

C. Add new elements only at the front

D. Add new elements only at the back without removal

Solution

1. Step 1: Understand sliding window maximum logic

   To keep maximum at front, remove all smaller elements from back before adding new element at back.

2. Step 2: Identify correct deque operation

   Removing smaller elements from back maintains decreasing order, ensuring front is max.

3. Final Answer:

   Remove elements from the back while they are smaller than the new element -> Option A

4. Quick Check:

   Sliding window max uses back removals for order [OK]

Hint: Remove smaller from back to keep max at front [OK]

Common Mistakes:

• Removing from front instead of back
• Adding elements without removing smaller ones
• Adding only at front breaks order