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

Time Complexity: Double Ended Queue Deque

O(n) for front insert/remove, O(1) for back insert/remove

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.

struct Deque {
    data: Vec<i32>,
}

impl Deque {
    fn push_front(&mut self, val: i32) {
        self.data.insert(0, val);
    }

    fn push_back(&mut self, val: i32) {
        self.data.push(val);
    }

    fn pop_front(&mut self) -> Option<i32> {
        if self.data.is_empty() { None } else { Some(self.data.remove(0)) }
    }

    fn pop_back(&mut self) -> Option<i32> {
        self.data.pop()
    }
}
    

This code defines a deque using a vector. It adds or removes items from front or back.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Inserting or removing at the front uses insert(0, val) or remove(0), which shifts all elements.
• How many times: Each front operation shifts all existing elements, so it repeats for all elements after the first.
• Back operations: Adding or removing at the back uses push or pop, which are fast and do not shift elements.

How Execution Grows With Input

Adding or removing at the back stays fast no matter how many items there are.

Adding or removing at the front takes longer as the number of items grows, because all elements must move.

Input Size (n)	Approx. Operations for front insert/remove
10	About 10 shifts
100	About 100 shifts
1000	About 1000 shifts

Pattern observation: Front operations grow linearly with the number of items; back operations stay constant time.

Final Time Complexity

Time Complexity: O(n) for front insert/remove, O(1) for back insert/remove

This means adding or removing at the front takes longer as the deque grows, but back operations stay quick.

Common Mistake

[X] Wrong: "All deque operations are always fast and take the same time."

[OK] Correct: Using a vector, front operations shift many elements, making them slower as size grows, unlike back operations.

Interview Connect

Understanding how different operations scale helps you choose the right data structure and explain your choices clearly in interviews.

Self-Check

"What if we used a linked list instead of a vector? How would the time complexity of front insert and remove change?"

Practice

1. What is a deque in data structures?

easy

A. A double-ended queue allowing insertion and removal from both ends

B. A queue that only allows insertion at the front

C. A stack that only allows removal from the top

D. A list that only supports random access

Solution

1. Step 1: Understand the deque concept

   A deque is a data structure where elements can be added or removed from both the front and the back.

2. Step 2: Compare options with deque definition

   Only double-ended queue allowing insertion and removal from both ends correctly describes this behavior; others describe different structures.

3. Final Answer:

   A double-ended queue allowing insertion and removal 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 regular queue
• Thinking deque only allows insertion at one end
• Assuming deque is a fixed-size array

2. Which of the following is the correct way to create an empty VecDeque in Rust?

easy

A. let dq = VecDeque::empty();

B. let dq = VecDeque();

C. let dq = new VecDeque();

D. let dq = VecDeque::new();

Solution

1. Step 1: Recall Rust VecDeque creation syntax

   In Rust, the standard way to create an empty VecDeque is by calling VecDeque::new().

2. Step 2: Check each option for Rust syntax correctness

   let dq = VecDeque::new(); uses correct Rust syntax. The other options are invalid Rust syntax for VecDeque creation.

3. Final Answer:

   let dq = VecDeque::new(); -> Option D

4. Quick Check:

   VecDeque::new() creates empty deque [OK]

Hint: Use VecDeque::new() to create empty deque in Rust [OK]

Common Mistakes:

• Using constructor syntax like VecDeque() which is invalid in Rust
• Trying to call new as a method without ::
• Using non-existent empty() method

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

use std::collections::VecDeque;

fn main() {
    let mut dq = VecDeque::new();
    dq.push_back(10);
    dq.push_front(20);
    dq.push_back(30);
    dq.pop_front();
    println!("{:?}", dq);
}

medium

A. [10, 30]

B. [20, 10, 30]

C. [30, 10]

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, resulting in [10, 30]

2. Step 2: Confirm final deque state

   After all operations, the deque contains [10, 30].

3. Final Answer:

   [10, 30] -> Option A

4. Quick Check:

   After pop_front, deque = [10, 30] [OK]

Hint: Track push/pop order carefully front and back [OK]

Common Mistakes:

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

4. Identify the error in the following Rust code using VecDeque:

use std::collections::VecDeque;

fn main() {
    let mut dq = VecDeque::new();
    dq.push_front(5);
    dq.push_back(10);
    let x = dq.pop_back();
    println!("{}", x);
}

medium

A. pop_back() does not return a value

B. Cannot print Option<i32> directly with {}

C. VecDeque does not have push_front method

D. dq must be immutable to pop elements

Solution

1. Step 1: Understand pop_back return type

   pop_back() returns an Option containing the popped value or None if empty.

2. Step 2: Check println! usage

   Using {} tries to print the Option<i32> directly, which causes a type mismatch error. {:?} or unwrap() is needed.

3. Final Answer:

   Cannot print Option<i32> directly with {} -> Option B

4. Quick Check:

   Option<i32> needs {:?} or unwrap() to print [OK]

Hint: Use {:?} or unwrap() to print Option values [OK]

Common Mistakes:

• Trying to print Option with {} instead of {:?}
• Assuming pop_back returns value directly, not Option
• Thinking VecDeque lacks push_front method

5. You want to implement a sliding window maximum using a VecDeque in Rust. Which approach correctly maintains the deque to always have the maximum at the front while sliding the window?

hard

A. Remove elements from front smaller than new element before pushing it to front

B. Always push new element to front without removing any elements

C. Remove elements from back smaller than new element before pushing it to back

D. Remove elements from back larger than new element before pushing it to back

Solution

1. Step 1: Understand sliding window max logic with deque

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

2. Step 2: Analyze options for correct deque maintenance

   Remove elements from back smaller than new element before pushing it to back correctly describes removing smaller elements from back before pushing new element to back. Others do not maintain max at front properly.

3. Final Answer:

   Remove elements from back smaller than new element before pushing it to back -> Option C

4. Quick Check:

   Sliding window max uses back removals of smaller elements [OK]

Hint: Pop smaller from back before push_back new max candidate [OK]

Common Mistakes:

• Removing from front instead of back
• Pushing new elements to front incorrectly
• Removing larger elements instead of smaller