Double Ended Queue Deque in DSA Go - Build from Scratch

Double Ended Queue (Deque) in Go

📖 Scenario: You are building a simple task manager that can add and remove tasks from both the front and back of the list. This helps you quickly manage urgent tasks or finish tasks in order.

🎯 Goal: Build a double ended queue (deque) using a slice in Go. You will add tasks to the front and back, remove tasks from both ends, and finally print the current list of tasks.

📋 What You'll Learn

Create a slice of strings called tasks with initial tasks

Create a variable newTask to hold a task to add

Add newTask to the front of tasks

Add newTask to the back of tasks

Remove one task from the front and one from the back of tasks

Print the final state of tasks showing all tasks in order

💡 Why This Matters

🌍 Real World

Double ended queues are useful in task scheduling, undo-redo operations, and managing buffers where you need to add or remove items from both ends efficiently.

💼 Career

Understanding deque operations helps in software development roles involving data processing, real-time systems, and UI state management.

1

Create initial tasks slice

Create a slice of strings called tasks with these exact tasks in order: "email client", "write report", "call supplier"

DSA Go

# Your code here

Hint

Use a slice literal to create tasks with the exact strings in order.

2

Add new task variable

Create a string variable called newTask and set it to "schedule meeting"

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"email client", "write report", "call supplier"}
    // Your code here
}

Hint

Declare newTask as a string and assign the exact value.

3

Add and remove tasks from deque

Add newTask to the front of tasks, then add newTask to the back of tasks. Next, remove one task from the front and one from the back of tasks.

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"email client", "write report", "call supplier"}
    newTask := "schedule meeting"
    // Your code here
}

Hint

Use append to add to front and back. Use slicing to remove from front and back.

4

Print final tasks list

Print the tasks slice to show the final list of tasks in order.

DSA Go

package main

import "fmt"

func main() {
    tasks := []string{"email client", "write report", "call supplier"}
    newTask := "schedule meeting"

    // Add newTask to the front
    tasks = append([]string{newTask}, tasks...)

    // Add newTask to the back
    tasks = append(tasks, newTask)

    // Remove one task from the front
    tasks = tasks[1:]

    // Remove one task from the back
    tasks = tasks[:len(tasks)-1]

    // Your code here
}

Hint

Use fmt.Println(tasks) to print the slice.

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

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, unlike a queue or stack.

2. Step 2: Compare with other data structures

   Queue allows insertion at rear and deletion at front; stack allows insertion and deletion 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 way to declare a deque using Go's container/list package?

easy

A. var d []int

B. d := list.New()

C. d := make(deque, 0)

D. var d map[int]int

Solution

1. Step 1: Recall Go's deque implementation

   Go does not have a built-in deque type but container/list can be used as a double-ended queue.

2. Step 2: Check syntax for creating a new list

   list.New() returns a pointer to a new doubly linked list, suitable for deque operations.

3. Final Answer:

   d := list.New() -> Option B

4. Quick Check:

   Use list.New() for deque in Go [OK]

Hint: Use list.New() to create a deque in Go [OK]

Common Mistakes:

• Using slice instead of list for deque
• Trying to make a deque type directly
• Confusing map with deque

3. What will be the output of the following Go code using container/list as a deque?

package main
import (
  "container/list"
  "fmt"
)
func main() {
  d := list.New()
  d.PushBack(10)
  d.PushFront(20)
  d.PushBack(30)
  for e := d.Front(); e != nil; e = e.Next() {
    fmt.Print(e.Value, " ")
  }
}

medium

A. 30 10 20

B. 10 20 30

C. 20 10 30

D. 20 30 10

Solution

1. Step 1: Trace the Push operations on deque

   PushBack(10) adds 10 at rear; PushFront(20) adds 20 at front; PushBack(30) adds 30 at rear.

2. Step 2: Traverse from front to back

   Elements order: 20 (front), 10 (middle), 30 (rear).

3. Final Answer:

   20 10 30 -> Option C

4. Quick Check:

   Front to back traversal = 20 10 30 [OK]

Hint: PushFront adds front, PushBack adds rear [OK]

Common Mistakes:

• Mixing order of PushFront and PushBack
• Assuming PushBack adds to front
• Confusing traversal direction

4. Identify the error in this Go code snippet using container/list as a deque:

package main
import (
  "container/list"
  "fmt"
)
func main() {
  d := list.New()
  d.PushFront(1)
  d.Remove(d.Back())
  fmt.Println(d.Back().Value)
}

medium

A. Using PushFront instead of PushBack

B. Not initializing the list with make()

C. Printing d.Back().Value without converting to string

D. Calling Remove on d.Back() before checking if d.Back() is nil

Solution

1. Step 1: Understand Remove and Back usage

   d.Remove(d.Back()) removes the last element; after removal, d.Back() may be nil if list is empty.

2. Step 2: Check the print statement

   fmt.Println(d.Back().Value) accesses Value of d.Back() without checking if d.Back() is nil, causing runtime panic if nil.

3. Final Answer:

   Calling Remove on d.Back() before checking if d.Back() is nil -> Option D

4. Quick Check:

   Check nil before accessing Back() after Remove [OK]

Hint: Check nil after Remove before accessing Back() [OK]

Common Mistakes:

• Ignoring nil check after removal
• Confusing PushFront and PushBack
• Thinking fmt.Println needs string conversion

5. You want to implement a sliding window maximum using a deque in Go. Which deque operation sequence correctly maintains the window elements in decreasing order from front to rear?

// Assume d is a deque of indices
for i := 0; i < len(nums); i++ {
  for d.Len() > 0 && nums[d.Back().Value.(int)] <= nums[i] {
    d.Remove(d.Back())
  }
  d.PushBack(i)
  if d.Front().Value.(int) <= i - k {
    d.Remove(d.Front())
  }
}

hard

A. Remove from back while current element is greater or equal, then add current index at back

B. Remove from front while current element is smaller, then add current index at front

C. Add current index at front without removals

D. Remove from back while current element is smaller, then add current index at front

Solution

1. Step 1: Understand sliding window max logic

   We remove smaller or equal elements from back to keep deque decreasing, then add current index at back.

2. Step 2: Remove indices out of window from front

   If front index is out of current window (i - k), remove it from front to maintain window size.

3. Final Answer:

   Remove from back while current element is greater or equal, then add current index at back -> Option A

4. Quick Check:

   Sliding window max uses back removals and back insertions [OK]

Hint: Remove smaller from back, add new at back for max window [OK]

Common Mistakes:

• Removing from front instead of back for smaller elements
• Adding new elements at front incorrectly
• Not removing out-of-window indices