Queue Using Two Stacks in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind implementing a queue using two stacks?

Use one stack to handle incoming elements (enqueue) and another stack to handle outgoing elements (dequeue). When the dequeue stack is empty, transfer all elements from the enqueue stack to reverse their order, simulating queue behavior.

beginner

Why do we transfer elements from the first stack to the second stack in a queue implemented with two stacks?

Transferring elements reverses their order, so the oldest element (which was pushed first) ends up on top of the second stack, allowing us to dequeue in FIFO order.

intermediate

What is the time complexity of enqueue and dequeue operations in a queue implemented using two stacks?

Enqueue is O(1) because we just push onto the first stack. Dequeue is amortized O(1) because transferring elements happens only when the second stack is empty, spreading the cost over multiple operations.

intermediate

Show the state of two stacks after enqueueing 1, 2, 3 and then dequeuing one element in a queue using two stacks.

After enqueue 1,2,3: stack1 = [1,2,3], stack2 = []. After dequeue: transfer all from stack1 to stack2, stack1 = [], stack2 = [3,2,1]. Dequeue pops 1 from stack2, so stack2 = [3,2].

beginner

What happens if you try to dequeue from an empty queue implemented with two stacks?

Both stacks are empty, so there are no elements to dequeue. This should be handled gracefully, usually by returning an error or a special value indicating the queue is empty.

In a queue implemented with two stacks, where do new elements get added?

A First stack (enqueue stack)

B Second stack (dequeue stack)

C Both stacks simultaneously

D Neither stack

New elements are always pushed onto the first stack to keep track of incoming elements.

What triggers the transfer of elements from the first stack to the second stack?

A When the second stack is empty and a dequeue is requested

B After every enqueue operation

C When the first stack is full

D When both stacks are empty

Elements are transferred only when the second stack is empty and we need to dequeue, to maintain correct order.

What is the order of elements in the second stack after transferring from the first stack?

A Random order

B Same order as first stack

C Reversed order of first stack

D Sorted order

Transferring elements reverses their order, so the oldest element is on top for dequeue.

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

A O(log n)

B O(1)

C O(n)

D O(n^2)

Dequeue is amortized O(1) because transfers happen only when needed, spreading the cost.

If both stacks are empty, what does dequeue operation return?

A The last element enqueued

B The first element enqueued

C Zero

D An error or indication that queue is empty

Dequeue from an empty queue should return an error or special value indicating no elements.

Explain how a queue can be implemented using two stacks and why this works.

Think about how stacks reverse order and how that helps simulate a queue.

Describe the steps and stack states when enqueueing three elements and then dequeuing one element in a queue using two stacks.

Visualize the stacks as piles of plates and how moving plates changes their 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