Double Ended Queue Deque in DSA Rust - Execution Trace

Concept Flow - Double Ended Queue Deque

Start

↓

Check Operation Type

↓

Insert Front

↓

Update Front Pointer

↓

Add Element at Front

↓

Check if Empty

↓

Remove Front

↓

Update Pointers

↓

End

This flow shows how a deque allows adding or removing elements from both front and rear ends by updating pointers accordingly.

Execution Sample

DSA Rust

deque.push_back(10);
deque.push_front(20);
deque.push_back(30);
deque.pop_front();
deque.pop_back();

This code adds 10 at rear, 20 at front, 30 at rear, then removes one element from front and one from rear.

Execution Table

Step	Operation	Nodes in List	Pointer Changes	Visual State
1	push_back(10)	10	rear -> 10, front -> 10	10 -> null
2	push_front(20)	20 -> 10	front -> 20, rear -> 10	20 -> 10 -> null
3	push_back(30)	20 -> 10 -> 30	rear -> 30	20 -> 10 -> 30 -> null
4	pop_front()	10 -> 30	front -> 10	10 -> 30 -> null
5	pop_back()	10	rear -> 10	10 -> null
6	End	10	No change	10 -> null

💡 No more operations; deque has one element left.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	Final
front	null	10	20	20	10	10	10
rear	null	10	10	30	30	10	10
size	0	1	2	3	2	1	1

Key Moments - 3 Insights

Why does the front pointer move forward after pop_front but rear pointer stays the same?

Because pop_front removes the first element, front must point to the next node (see step 4 in execution_table). Rear stays as last element until pop_back changes it (step 5).

What happens if we push_front on an empty deque?

Both front and rear pointers point to the new single node (see step 2 where front moves to 20 and rear moves to 10 from previous step). For an empty deque, both would point to the new node.

Why is the visual state shown with arrows ending in null?

The arrows represent links between nodes; null means the end of the deque (no next node). This helps visualize the chain of elements.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, what is the visual state of the deque?

A 10 -> 20 -> 30 -> null

B 20 -> 10 -> 30 -> null

C 30 -> 20 -> 10 -> null

D 20 -> 30 -> 10 -> null

💡 Hint

Check the 'Visual State' column at step 3 in execution_table.

At which step does the size of the deque become 2?

A After step 1

B After step 4

C After step 2

D After step 5

💡 Hint

Look at the 'size' row in variable_tracker after each step.

If we add push_front(40) after step 5, what will be the new front pointer?

A 40

B 10

C 30

D null

💡 Hint

Adding at front updates the front pointer to the new node (see step 2 for similar behavior).

Concept Snapshot

Double Ended Queue (Deque):
- Allows insertion/removal at both front and rear.
- Maintains front and rear pointers.
- push_front adds at front; push_back adds at rear.
- pop_front removes from front; pop_back removes from rear.
- Visualize as linked nodes ending with null.
- Size updates with each operation.

Full Transcript

A double ended queue, or deque, lets us add or remove items from both ends. We keep track of two pointers: front and rear. When we add at the front, we update the front pointer to the new node. When we add at the rear, we update the rear pointer. Removing from front moves the front pointer forward, removing from rear moves the rear pointer backward. The deque looks like a chain of nodes linked by arrows ending in null. Size changes as we add or remove elements. This example showed adding 10 at rear, 20 at front, 30 at rear, then removing one from front and one from rear, leaving one element in the deque.

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