Queue Using Two Stacks in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of enqueue and dequeue operations

What is the printed state of the queue after these operations using two stacks?

DSA Go

package main
import "fmt"

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)
            q.stack2 = append(q.stack2, q.stack1[n-1])
            q.stack1 = q.stack1[:n-1]
        }
    }
    n := len(q.stack2)
    x := q.stack2[n-1]
    q.stack2 = q.stack2[:n-1]
    return x
}

func (q *Queue) Print() string {
    result := []int{}
    for i := len(q.stack2) - 1; i >= 0; i-- {
        result = append(result, q.stack2[i])
    }
    for _, v := range q.stack1 {
        result = append(result, v)
    }
    s := ""
    for i, v := range result {
        if i > 0 {
            s += " -> "
        }
        s += fmt.Sprint(v)
    }
    s += " -> null"
    return s
}

func main() {
    q := Queue{}
    q.Enqueue(1)
    q.Enqueue(2)
    q.Enqueue(3)
    q.Dequeue()
    q.Enqueue(4)
    fmt.Println(q.Print())
}

A 4 -> null

B 2 -> 3 -> 4 -> null

C 1 -> 2 -> 3 -> 4 -> null

D 3 -> 4 -> null

💡 Hint

Remember that dequeue moves elements from stack1 to stack2 only when stack2 is empty.

After enqueueing 1,2,3 and dequeuing once, 1 is removed. Then enqueue 4. The queue order is 2,3,4.

❓ Predict Output

intermediate

Result of multiple dequeues

What is the output of the following code snippet using queue with two stacks?

DSA Go

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())

A

10
20
30
40

B

10
20
40
30

C

10
30
20
40

D

20
10
30
40

💡 Hint

Dequeue removes elements in the order they were enqueued.

Dequeue returns 10, then 20, then after enqueue 40, returns 30 and 40 in order.

🔧 Debug

advanced

Identify the error in dequeue implementation

Which option shows the code that will cause a runtime error when calling Dequeue on an empty queue?

DSA Go

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

A

func (q *Queue) Dequeue() int {
    n := len(q.stack2)
    x := q.stack2[n-1]
    q.stack2 = q.stack2[:n-1]
    return x
}

B

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

C

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

D

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

💡 Hint

Check what happens if both stacks are empty before accessing elements.

Option A does not check if stacks are empty before accessing, causing runtime panic on empty queue.

🧠 Conceptual

advanced

Time complexity of dequeue operation

What is the amortized time complexity of the Dequeue operation in a queue implemented using two stacks?

A O(log n) amortized time

B O(n) worst case for every dequeue

C O(1) amortized time

D O(n^2) worst case

💡 Hint

Consider how often elements are moved between stacks.

Each element is moved at most twice, so average cost per dequeue is constant time.

🚀 Application

expert

Final queue state after mixed operations

Given the following sequence of operations on a queue implemented with two stacks, what is the final printed queue state?

DSA Go

q := Queue{}
q.Enqueue(5)
q.Enqueue(10)
q.Dequeue()
q.Enqueue(15)
q.Enqueue(20)
q.Dequeue()
q.Enqueue(25)
fmt.Println(q.Print())

A 5 -> 10 -> 15 -> 20 -> 25 -> null

B 10 -> 15 -> 20 -> 25 -> null

C 20 -> 25 -> null

D 15 -> 20 -> 25 -> null

💡 Hint

Track each enqueue and dequeue carefully to see which elements remain.

Dequeues remove 5 and 10, remaining elements are 15, 20, 25 in order.

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