Implement Stack Using Queue in DSA Java - Execution Trace

Concept Flow - Implement Stack Using Queue

Push operation

↓

Enqueue new element

↓

Rotate queue to put new element at front

↓

Top element is at front

↓

Pop operation

↓

Dequeue front element

↓

Stack top removed

Push adds element by enqueue then rotates queue to front; Pop removes front element to simulate stack behavior.

Execution Sample

DSA Java

Queue<Integer> q = new LinkedList<>();
// Push 1
q.add(1);
// Rotate queue
for (int i = 0; i < q.size() - 1; i++) q.add(q.remove());
// Pop
int top = q.remove();

This code pushes 1 onto the stack using queue rotation, then pops the top element.

Execution Table

Step	Operation	Queue State Before	Pointer Changes	Queue State After	Visual State
1	Push 1	[]	Enqueue 1	[1]	1 -> null
2	Rotate queue	[1]	Remove 1, Enqueue 1	[1]	1 -> null
3	Push 2	[1]	Enqueue 2	[1, 2]	1 -> 2 -> null
4	Rotate queue	[1, 2]	Remove 1, Enqueue 1	[2, 1]	2 -> 1 -> null
5	Push 3	[2, 1]	Enqueue 3	[2, 1, 3]	2 -> 1 -> 3 -> null
6	Rotate queue	[2, 1, 3]	Remove 2, Enqueue 2; Remove 1, Enqueue 1	[3, 2, 1]	3 -> 2 -> 1 -> null
7	Pop	[3, 2, 1]	Dequeue 3	[2, 1]	2 -> 1 -> null
8	Pop	[2, 1]	Dequeue 2	[1]	1 -> null
9	Pop	[1]	Dequeue 1	[]	null
10	Pop	[]	Queue empty, cannot dequeue	[]	null

💡 Step 10 stops because queue is empty, no elements to pop.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	After Step 10
Queue q	[]	[1]	[1]	[1, 2]	[2, 1]	[2, 1, 3]	[3, 2, 1]	[2, 1]	[1]	[]	[]

Key Moments - 3 Insights

Why do we rotate the queue after each push?

Rotating the queue moves the newly added element to the front, making it behave like the top of a stack. See execution_table rows 2, 4, and 6 where rotation changes the queue order.

Why does pop remove from the front of the queue?

Because after rotation, the front of the queue is the last pushed element, simulating stack's top. Refer to execution_table rows 7, 8, and 9 where dequeue removes the top element.

What happens if we pop when the queue is empty?

No element can be removed; the queue remains empty and pop operation cannot proceed. See execution_table row 10 where pop fails due to empty queue.

Visual Quiz - 3 Questions

Look at the execution_table at step 6, what is the queue state after rotation?

A [1, 2, 3]

B [3, 2, 1]

C [2, 1, 3]

D [1, 3, 2]

💡 Hint

Check the 'Queue State After' column in row 6 of execution_table.

At which step does the queue first contain the element '3' at the front?

A Step 6

B Step 5

C Step 3

D Step 7

💡 Hint

Look at the 'Visual State' column and see when '3' is at the front.

If we skip the rotation after push, what will happen to the pop operation?

A Pop will remove nothing

B Pop will remove the last pushed element correctly

C Pop will remove the oldest element, not the last pushed

D Queue will be empty immediately

💡 Hint

Rotation moves the new element to front; without it, front is oldest element (see key_moments about rotation).

Concept Snapshot

Implement Stack Using Queue:
- Use one queue
- Push: enqueue element, then rotate queue to front
- Pop: dequeue front element
- Top is always at front after rotation
- Rotation simulates LIFO behavior using FIFO queue

Full Transcript

This visualization shows how to implement a stack using a single queue. When pushing, we add the new element to the queue and then rotate the queue so that this new element moves to the front. This makes the front of the queue behave like the top of the stack. When popping, we simply remove the front element from the queue. The execution table traces each step, showing queue states and pointer changes. Key moments clarify why rotation is necessary and what happens when popping from an empty queue. The visual quiz tests understanding of queue state changes and the importance of rotation. The concept snapshot summarizes the approach in simple steps.

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use one queue and reorder elements to simulate stack behavior

B. Use two stacks to simulate a queue

C. Use a linked list to implement both stack and queue

D. Use an array to simulate queue operations

Solution

1. Step 1: Understand stack and queue behaviors

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue

   By reordering elements in a single queue after each push, we can simulate LIFO behavior of a stack.

3. Final Answer:

   Use one queue and reorder elements to simulate stack behavior -> Option A

4. Quick Check:

   Stack via queue = reorder in one queue [OK]

Hint: Remember stack is LIFO; reorder queue to mimic it [OK]

Common Mistakes:

• Confusing stack with queue behavior
• Using two stacks instead of queues
• Ignoring element order after push

2. Which of the following is the correct Java syntax to declare a queue of integers using LinkedList?

easy

A. LinkedList<Integer> queue = new Queue<>();

B. Queue<int> queue = new LinkedList<>();

C. Queue<Integer> queue = new LinkedList<>();

D. Queue<Integer> queue = new Queue<>();

Solution

1. Step 1: Use interface Queue with LinkedList implementation

   In Java, Queue is an interface and LinkedList implements it. Use generics with wrapper class Integer.

2. Step 2: Check syntax correctness

   Queue<Integer> queue = new LinkedList<>(); is correct syntax for declaring a queue of integers.

3. Final Answer:

   Queue<Integer> queue = new LinkedList<>(); -> Option C

4. Quick Check:

   Queue<Integer> with LinkedList = correct syntax [OK]

Hint: Use wrapper classes and interface for queue declaration [OK]

Common Mistakes:

• Using primitive int instead of Integer
• Trying to instantiate interface Queue directly
• Confusing LinkedList and Queue syntax

3. What will be the output of the following code snippet implementing stack push using one queue?

Queue<Integer> q = new LinkedList<>();
// Push 1
q.add(1);
// Push 2
q.add(2);
// Reorder queue to simulate stack push
for (int i = 0; i < q.size() - 1; i++) {
  q.add(q.remove());
}
System.out.println(q.remove());

medium

A. 1

B. 2

C. Queue is empty

D. Runtime error

Solution

1. Step 1: Add elements 1 and 2 to queue

   Queue after adds: front -> 1, 2 -> rear

2. Step 2: Reorder queue to simulate stack push

   Loop runs q.size()-1 = 1 time: remove 1, add 1 back. Queue becomes front -> 2, 1 -> rear

3. Step 3: Remove and print element

   q.remove() removes front element 2, which is printed.

4. Final Answer:

   2 -> Option B

5. Quick Check:

   Reordered queue front = last pushed = 2 [OK]

Hint: After push reorder, front is last pushed element [OK]

Common Mistakes:

• Not reordering queue after push
• Printing wrong element without remove
• Confusing queue front with rear

4. Identify the error in this stack pop implementation using one queue:

public int pop() {
  if (queue.isEmpty()) return -1;
  return queue.remove();
}

medium

A. No error, code works correctly

B. Queue should be cleared after pop

C. Should use queue.peek() instead of remove()

D. Pop removes the oldest element, not the last pushed

Solution

1. Step 1: Understand queue remove() behavior

   queue.remove() removes the front element, which is the oldest inserted, not the last pushed.

2. Step 2: Check stack pop requirement

   Stack pop must remove the last pushed element (LIFO), but this code removes FIFO element.

3. Final Answer:

   Pop removes the oldest element, not the last pushed -> Option D

4. Quick Check:

   Queue remove() = FIFO, stack pop = LIFO [OK]

Hint: Pop must remove last pushed, not queue front [OK]

Common Mistakes:

• Using queue.remove() directly for pop
• Ignoring element order after push
• Not reordering queue to simulate stack

5. You want to implement a stack using two queues. Which sequence of operations correctly simulates the pop() method?

hard

A. Move all but last element from queue1 to queue2, remove last from queue1, swap queues

B. Remove front element from queue1 directly

C. Add element to queue2, then remove from queue1

D. Swap queues without removing any element

Solution

1. Step 1: Understand two-queue stack pop logic

   To pop last pushed element, move all except last from queue1 to queue2, then remove last from queue1.

2. Step 2: Swap queues for next operations

   After removal, swap queue1 and queue2 references to maintain stack state.

3. Final Answer:

   Move all but last element from queue1 to queue2, remove last from queue1, swap queues -> Option A

4. Quick Check:

   Pop with two queues = move all but last, remove last, swap [OK]

Hint: Pop = move all but last, remove last, then swap queues [OK]

Common Mistakes:

• Removing front element directly
• Not swapping queues after pop
• Adding elements to wrong queue during pop