Implement Stack Using Queue in DSA C++ - Execution Trace

Concept Flow - Implement Stack Using Queue

Push operation

↓

Enqueue element in queue

↓

Rotate queue to put new element at front

↓

Top element is front of queue

↓

Pop operation

↓

Dequeue front element

↓

Stack size decreases

Push adds element by enqueue and rotates queue to keep stack order; pop removes front element simulating stack top.

Execution Sample

DSA C++

queue<int> q;
// Push 1
q.push(1);
// Rotate queue to put new element at front
int sz = q.size();
for (int i = 0; i < sz - 1; ++i) {
    int val = q.front();
    q.pop();
    q.push(val);
}
// Push 2
q.push(2);
// Rotate queue to put new element at front
sz = q.size();
for (int i = 0; i < sz - 1; ++i) {
    int val = q.front();
    q.pop();
    q.push(val);
}
// Pop
q.pop();

This code demonstrates queue operations (push/enqueue and pop/dequeue) used in the stack implementation. Full stack push requires additional rotation after enqueue, as shown in the execution_table.

Execution Table

Step	Operation	Queue State Before	Action	Queue State After	Visual State
1	Push 1	[]	Enqueue 1	[1]	1 -> null
2	Rotate after push 1	[1]	Rotate 0 times (size-1=0)	[1]	1 -> null
3	Push 2	[1]	Enqueue 2	[1, 2]	1 -> 2 -> null
4	Rotate after push 2	[1, 2]	Rotate 1 time: dequeue 1, enqueue 1	[2, 1]	2 -> 1 -> null
5	Pop	[2, 1]	Dequeue front (2)	[1]	1 -> null
6	Pop	[1]	Dequeue front (1)	[]	empty

💡 Queue empty after last pop, stack is empty.

Variable Tracker

Variable	Start	After Step 1	After Step 3	After Step 4	After Step 5	After Step 6
queue	[]	[1]	[1, 2]	[2, 1]	[1]	[]
size	0	1	2	2	1	0

Key Moments - 3 Insights

Why do we rotate the queue after pushing an element?

Rotating the queue moves the newly added element to the front, simulating stack's top. See step 4 in execution_table where rotation changes [1, 2] to [2, 1].

Why does pop remove the front element of the queue?

Because after rotation, the front of the queue is the last pushed element, matching stack's top. Step 5 shows dequeue of front element 2, which is the stack top.

What happens if we pop when the queue is empty?

The queue is empty, so no element to remove; stack is empty. Step 6 shows queue becomes empty after last pop.

Visual Quiz - 3 Questions

Look at the execution_table, what is the queue state after step 4?

A [1, 2]

B [1]

C [2, 1]

D []

💡 Hint

Check the 'Queue State After' column for step 4 in execution_table.

At which step does the queue first contain two elements?

A Step 3

B Step 4

C Step 1

D Step 5

💡 Hint

Look at 'Queue State After' column to find when queue size becomes 2.

If we skip the rotation after push, what would be the queue state after pushing 2?

A [2, 1]

B [1, 2]

C [2]

D [1]

💡 Hint

Rotation moves last pushed to front; skipping it keeps enqueue order.

Concept Snapshot

Implement stack using a single queue:
- Push: enqueue element, then rotate queue to front
- Pop: dequeue front element
- Top: front of queue
- Rotation keeps LIFO order using FIFO queue
- Queue size tracks stack size

Full Transcript

This concept shows how to build a stack using one queue. When pushing, we add the element to the queue and then rotate the queue so the new element moves to the front. This simulates the stack's top. When popping, we remove the front element of the queue, which is the last pushed element. The execution table traces each step showing queue states and rotations. Key moments clarify why rotation is needed and how pop works. The visual quiz tests understanding of queue states and rotation effects. The snapshot summarizes the approach to implement stack operations using queue operations.

Practice

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

easy

A. Use two stacks to simulate a queue

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

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

D. Use recursion to reverse the queue elements

Solution

1. Step 1: Understand stack and queue behavior

   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, the newest element is always at the front, simulating stack pop behavior.

3. Final Answer:

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

4. Quick Check:

   Stack via queue = reorder elements [OK]

Hint: Rearrange queue after push to mimic stack top [OK]

Common Mistakes:

• Confusing stack with queue behavior
• Using two stacks instead of queues
• Not reordering queue after push

2. Which of the following is the correct syntax to declare a queue of integers in C++ STL?

easy

A. queue q<int>;

B. queue<int> q;

C. queue<int> q()

D. queue<int> q[];

Solution

1. Step 1: Recall C++ STL queue declaration

   The correct syntax uses template syntax: queue<int> q; declares a queue of integers.

2. Step 2: Check syntax correctness

   queue<int> q; matches the correct syntax; others have syntax errors or invalid declarations.

3. Final Answer:

   queue<int> q; -> Option B

4. Quick Check:

   Correct STL queue syntax = queue<int> q; [OK]

Hint: Use template syntax queue<int> q; for integer queue [OK]

Common Mistakes:

• Misplacing template brackets
• Using parentheses instead of semicolon
• Declaring queue as array

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

class Stack {
  queue<int> q;
public:
  void push(int x) {
    q.push(x);
    int n = q.size();
    for (int i = 0; i < n - 1; i++) {
      q.push(q.front());
      q.pop();
    }
  }
  int pop() {
    int val = q.front();
    q.pop();
    return val;
  }
};

int main() {
  Stack s;
  s.push(10);
  s.push(20);
  s.push(30);
  cout << s.pop() << " ";
  s.push(40);
  cout << s.pop() << " ";
  cout << s.pop() << " ";
  return 0;
}

medium

A. 30 40 20

B. 10 20 30

C. 30 20 10

D. 20 30 40

Solution

1. Step 1: Trace push operations

   Push 10: queue = [10]
   Push 20: queue = [20, 10] after rotation
   Push 30: queue = [30, 20, 10] after rotation

2. Step 2: Trace pop and push

   Pop: returns 30, queue = [20, 10]
   Push 40: queue = [40, 20, 10]
   Pop: returns 40, queue = [20, 10]
   Pop: returns 20, queue = [10]

3. Final Answer:

   30 40 20 -> Option A

4. Quick Check:

   Stack pop order = 30, 40, 20 [OK]

Hint: After push, rotate queue to front for stack order [OK]

Common Mistakes:

• Not rotating queue after push
• Confusing queue front with stack top
• Incorrect pop order

4. Identify the error in the following stack implementation using two queues:

class Stack {
  queue<int> q1, q2;
public:
  void push(int x) {
    q1.push(x);
  }
  int pop() {
    if (q1.empty()) return -1;
    while (q1.size() > 1) {
      q2.push(q1.front());
      q1.pop();
    }
    int val = q1.front();
    q1.pop();
    swap(q1, q2);
    return val;
  }
};

medium

A. Pop operation does not handle empty stack correctly

B. Push operation should push to q2 instead of q1

C. Swap should be done before popping from q1

D. Pop operation is correct; no error

Solution

1. Step 1: Analyze pop operation logic

   Pop moves all but last element from q1 to q2, then pops last element from q1, then swaps q1 and q2.

2. Step 2: Check correctness of push and pop

   Push adds to q1; pop removes last pushed element correctly by transferring elements and swapping queues.

3. Final Answer:

   Pop operation is correct; no error -> Option D

4. Quick Check:

   Two-queue stack pop logic = correct [OK]

Hint: Pop moves elements and swaps queues to simulate stack pop [OK]

Common Mistakes:

• Swapping queues too early
• Pushing to wrong queue
• Not checking empty queue before pop

5. You want to implement a stack using a single queue but optimize the pop operation to O(1) time. Which approach is best?

hard

A. Push elements by enqueueing and dequeueing all elements except the new one

B. Use a queue and a stack together for hybrid implementation

C. Use two queues to separate push and pop operations

D. Push elements normally and rotate queue on each pop

Solution

1. Step 1: Understand time complexity trade-offs

   Push elements normally and rotate queue on each pop (push normally, rotate on pop) makes push O(1), pop O(n). Use a queue and a stack together for hybrid implementation is not pure queue. Use two queues to separate push and pop operations uses two queues, not single queue.

2. Step 2: Optimize pop to O(1) with single queue

   Enqueue new element, then dequeue and enqueue all previous elements to move new element to front. This makes pop O(1) as front is always top (matches Push elements by enqueueing and dequeueing all elements except the new one).

3. Final Answer:

   Push elements by enqueueing and dequeueing all elements except the new one -> Option A

4. Quick Check:

   Rotate queue after push for O(1) pop [OK]

Hint: Rotate queue after push to keep top at front [OK]

Common Mistakes:

• Trying to optimize pop without reordering queue
• Using two queues instead of one
• Mixing stack and queue data structures