Queue Using Two Stacks in DSA Go - Build from Scratch

Queue Using Two Stacks

📖 Scenario: Imagine you are managing a line of customers waiting for service. You want to keep track of who comes first and who comes last. A queue is perfect for this because it works like a real line: first in, first out.But you only have two stacks, which work like piles of plates: last in, first out. Your task is to use these two stacks to build a queue that behaves correctly.

🎯 Goal: Build a queue using two stacks in Go. You will create two stacks, then write methods to add items to the queue and remove items from the queue, making sure the order is correct.

📋 What You'll Learn

Create two stacks using slices of integers

Write a method to add (enqueue) an integer to the queue

Write a method to remove (dequeue) an integer from the queue

Print the queue state after operations

💡 Why This Matters

🌍 Real World

Queues are used in many places like customer service lines, task scheduling, and printing jobs. Using stacks to build a queue helps understand how data structures can work together.

💼 Career

Understanding how to implement a queue with stacks shows problem-solving skills and knowledge of data structures, useful for software development and technical interviews.

1

Create two stacks

Create two variables called stack1 and stack2 as slices of integers. Both should start empty.

DSA Go

# Your code here

Hint

Use var stack1 []int and var stack2 []int to create empty stacks.

2

Add an enqueue function

Add a function called enqueue that takes a pointer to stack1 and an integer value. It should add value to the end of stack1.

DSA Go

package main

import "fmt"

func main() {
    var stack1 []int
    var stack2 []int
}

// Write enqueue function here
# Your code here

Hint

Use append to add value to *stack1.

3

Add a dequeue function

Add a function called dequeue that takes pointers to stack1 and stack2. It should remove and return the first inserted integer from the queue using the two stacks. If stack2 is empty, move all elements from stack1 to stack2 in reverse order before popping from stack2.

DSA Go

package main

import "fmt"

func enqueue(stack1 *[]int, value int) {
    *stack1 = append(*stack1, value)
}

// Write dequeue function here
# Your code here

func main() {
    var stack1 []int
    var stack2 []int
}

Hint

Move all elements from stack1 to stack2 if stack2 is empty, then pop from stack2.

4

Test the queue and print results

In main, enqueue the integers 10, 20, and 30 using enqueue. Then dequeue two integers using dequeue. Finally, print the remaining elements in stack1 and stack2 to show the queue state.

DSA Go

package main

import "fmt"

func enqueue(stack1 *[]int, value int) {
    *stack1 = append(*stack1, value)
}

func dequeue(stack1 *[]int, stack2 *[]int) int {
    if len(*stack2) == 0 {
        for len(*stack1) > 0 {
            lastIndex := len(*stack1) - 1
            *stack2 = append(*stack2, (*stack1)[lastIndex])
            *stack1 = (*stack1)[:lastIndex]
        }
    }
    if len(*stack2) == 0 {
        return -1 // Queue is empty
    }
    lastIndex := len(*stack2) - 1
    val := (*stack2)[lastIndex]
    *stack2 = (*stack2)[:lastIndex]
    return val
}

func main() {
    var stack1 []int
    var stack2 []int

    // Enqueue 10, 20, 30
    // Dequeue twice
    // Print stacks
    # Your code here
}

Hint

Call enqueue three times, then call dequeue twice, printing the returned values and the stacks.

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use two stacks to store elements in sorted order

B. Use one stack for enqueue and another for dequeue to reverse order

C. Use two stacks to store elements in a tree structure

D. Use one stack to store even elements and another for odd elements

Solution

1. Step 1: Understand queue and stack behavior

   A queue follows FIFO (first in, first out), while a stack follows LIFO (last in, first out).

2. Step 2: Use two stacks to reverse order

   By pushing elements into one stack and popping them into another, the order reverses, simulating queue behavior.

3. Final Answer:

   Use one stack for enqueue and another for dequeue to reverse order -> Option B

4. Quick Check:

   Two stacks reverse order to mimic queue = B [OK]

Hint: Remember: two stacks reverse order to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Confusing stack behavior with queue behavior
• Using stacks to separate even and odd elements

2. Which of the following is the correct syntax to push an element x into a stack named stack1 in Go?

easy

A. stack1 = append(stack1, x)

B. stack1.Push(x)

C. stack1.add(x)

D. stack1.insert(x)

Solution

1. Step 1: Understand Go slice push operation

   In Go, stacks are often implemented using slices, and pushing means appending an element.

2. Step 2: Use append to add element

   The correct way to push is using stack1 = append(stack1, x) to add x at the end.

3. Final Answer:

   stack1 = append(stack1, x) -> Option A

4. Quick Check:

   Go slice push uses append = A [OK]

Hint: Use append(slice, x) to push in Go stack [OK]

Common Mistakes:

• Using non-existent Push() method on slice
• Using add() or insert() which are not Go slice methods
• Trying to push without reassigning slice

3. Given the following Go code implementing a queue using two stacks, what will be printed?

type Queue struct {
  stack1 []int
  stack2 []int
}

func (q *Queue) Enqueue(x int) {
  q.stack1 = append(q.stack1, x)
}

func (q *Queue) Dequeue() int {
  if len(q.stack2) == 0 {
    for len(q.stack1) > 0 {
      n := len(q.stack1) - 1
      q.stack2 = append(q.stack2, q.stack1[n])
      q.stack1 = q.stack1[:n]
    }
  }
  n := len(q.stack2) - 1
  val := q.stack2[n]
  q.stack2 = q.stack2[:n]
  return val
}

func main() {
  q := Queue{}
  q.Enqueue(10)
  q.Enqueue(20)
  q.Enqueue(30)
  fmt.Println(q.Dequeue())
  fmt.Println(q.Dequeue())
  q.Enqueue(40)
  fmt.Println(q.Dequeue())
  fmt.Println(q.Dequeue())
}

medium

A. 20
10
30
40

B. 30
20
10
40

C. 10
20
40
30

D. 10
20
30
40

Solution

1. Step 1: Trace enqueue operations

   Enqueue 10, 20, 30 adds to stack1: [10, 20, 30]

2. Step 2: Dequeue operations transfer stack1 to stack2 and pop

   First dequeue moves all from stack1 to stack2 reversing order: stack2 = [30, 20, 10]. Pop from stack2 returns 10, then 20. Enqueue 40 adds to stack1: [40]. Next dequeue pops 30 from stack2. Last dequeue pops 40 from stack1 after transfer.

3. Final Answer:

   10 20 30 40 -> Option D

4. Quick Check:

   Queue order preserved by two stacks = A [OK]

Hint: Remember stack2 reverses stack1 to dequeue in FIFO order [OK]

Common Mistakes:

• Assuming stack1 pops directly for dequeue
• Not transferring all elements from stack1 to stack2
• Mixing order of elements after transfer

4. Identify the bug in this Go code snippet for dequeue operation in a queue using two stacks:

func (q *Queue) Dequeue() int {
  if len(q.stack2) == 0 {
    for len(q.stack1) > 0 {
      n := len(q.stack1) - 1
      q.stack2 = append(q.stack2, q.stack1[n])
      q.stack1 = q.stack1[:n]
    }
  }
  val := q.stack2[0]
  q.stack2 = q.stack2[1:]
  return val
}

medium

A. Appending elements to stack2 instead of pushing

B. Not checking if both stacks are empty before dequeue

C. Accessing q.stack2[0] instead of last element causes wrong dequeue order

D. Using len(q.stack1) - 1 incorrectly in loop

Solution

1. Step 1: Understand stack2 usage in dequeue

   stack2 stores elements in reversed order; popping should remove last element (end of slice).

2. Step 2: Identify incorrect element access

   Accessing q.stack2[0] removes first element, breaking LIFO order of stack2 and queue FIFO behavior.

3. Final Answer:

   Accessing q.stack2[0] instead of last element causes wrong dequeue order -> Option C

4. Quick Check:

   Dequeue must pop last stack2 element = C [OK]

Hint: Pop from end of stack slice, not from start [OK]

Common Mistakes:

• Removing first element of stack2 instead of last
• Not checking empty stacks before dequeue
• Misusing slice indices in loop

5. You want to implement a Peek() function for the queue using two stacks in Go that returns the front element without removing it. Which approach correctly returns the front element?

hard

A. If stack2 is empty, transfer all elements from stack1 to stack2, then return last element of stack2

B. Return first element of stack1 if not empty, else last element of stack2

C. Return last element of stack1 always

D. Transfer elements from stack2 to stack1 and return first element of stack1

Solution

1. Step 1: Understand peek logic in two-stack queue

   Peek must return the front element, which is at the top of stack2 if not empty.

2. Step 2: Transfer elements if stack2 empty and return last element

   If stack2 is empty, transfer all from stack1 to stack2 to reverse order, then peek last element of stack2 without removing.

3. Final Answer:

   If stack2 is empty, transfer all elements from stack1 to stack2, then return last element of stack2 -> Option A

4. Quick Check:

   Peek front = top of stack2 after transfer = D [OK]

Hint: Peek top of stack2 after transfer if empty [OK]

Common Mistakes:

• Peeking first element of stack1 which is wrong order
• Always returning last element of stack1 ignoring stack2
• Transferring elements in wrong direction