Implement Stack Using Queue in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after these stack operations?

Consider a stack implemented using two queues. The operations are:

• push(1)
• push(2)
• pop()
• push(3)
• pop()

What is the printed output of the popped elements?

DSA C

/* Using two queues q1 and q2 to implement stack */
// push(x): enqueue x to q2, then enqueue all elements of q1 to q2, swap q1 and q2
// pop(): dequeue from q1

// Operations:
// push(1)
// push(2)
// pop() -> prints popped element
// push(3)
// pop() -> prints popped element

A [1, 2]

B [1, 3]

C [2, 1]

D [2, 3]

💡 Hint

Remember that the last pushed element is popped first in a stack.

After push(1) and push(2), the stack top is 2. pop() removes 2. Then push(3) adds 3 on top. pop() removes 3.

🧠 Conceptual

intermediate

Which queue operation is used to simulate stack's pop?

In a stack implemented using queues, which queue operation corresponds to the stack's pop operation?

A Enqueue (adding element to the back)

B Dequeue (removing element from the front)

C Peek (viewing front element without removing)

D Size (counting elements in queue)

💡 Hint

Pop removes the top element from the stack.

Pop removes the last inserted element, which is simulated by dequeueing from the front of the queue after reordering.

🔧 Debug

advanced

Find the error in this stack push implementation using one queue

Given this push function for stack using a single queue, what error will occur?

void push(int x) {
    q.push(x);
    for (int i = 0; i < q.size() - 1; i++) {
        int val = q.front();
        q.pop();
        q.push(val);
    }
}

A No error, works correctly

B Compilation error due to missing queue declaration

C Infinite loop because q.size() changes inside loop

D Runtime error due to modifying queue size during loop

💡 Hint

The queue size remains constant during the loop because each pop is followed by a push.

No error occurs. This is the standard O(n) implementation for push in a stack using one queue. The newly pushed element is rotated to the front by moving the previous elements to the back. The queue size stays constant, so the loop condition works correctly.

❓ Predict Output

advanced

What is the stack content after these operations using one queue?

Using a stack implemented with one queue, perform these operations:

• push(10)
• push(20)
• push(30)
• pop()

What is the state of the queue representing the stack after these operations?

DSA C

/* push(x): enqueue x, then rotate queue to put x at front
   pop(): dequeue front element */

A 30 -> 20 -> 10 -> null

B 10 -> 20 -> null

C 20 -> 10 -> null

D 20 -> 30 -> 10 -> null

💡 Hint

Pop removes the last pushed element, so after pop(), the top is the previous element.

After pushing 10,20,30, the queue order is 30(front),20,10. Pop removes 30, leaving 20(front),10.

🧠 Conceptual

expert

What is the time complexity of push operation in stack implemented using two queues?

In a stack implemented using two queues, the push operation involves moving elements between queues. What is its time complexity?

A O(n) linear time

B O(log n) logarithmic time

C O(1) constant time

D O(n^2) quadratic time

💡 Hint

Consider how many elements are moved during each push.

Push requires moving all existing elements to the other queue, so it takes linear time proportional to the number of elements.