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.

Practice

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

easy

A. Use one or two queues to simulate stack operations like push and pop

B. Use a queue to sort elements in ascending order

C. Use a stack to implement queue operations

D. Use two stacks to implement a queue

Solution

1. Step 1: Understand stack and queue differences

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

2. Step 2: Simulate stack using queue(s)

   By using one or two queues cleverly, we can reorder elements to behave like a stack for push and pop operations.

3. Final Answer:

   Use one or two queues to simulate stack operations like push and pop -> Option A

4. Quick Check:

   Stack via queue = simulate push/pop with queues [OK]

Hint: Remember stack is LIFO, queue is FIFO; simulate LIFO with queue(s) [OK]

Common Mistakes:

• Confusing stack and queue order
• Trying to sort elements instead of simulating stack
• Using stack to implement queue instead

2. Which of the following is the correct syntax to declare a queue in C using the standard library?

easy

A. queue<int> q;

B. struct queue q;

C. #include \nqueue q;

D. No standard queue in C; must implement manually

Solution

1. Step 1: Check C standard library support

   C language does not have a built-in queue data structure in its standard library.

2. Step 2: Understand queue implementation in C

   Queues must be implemented manually using arrays or linked lists in C.

3. Final Answer:

   No standard queue in C; must implement manually -> Option D

4. Quick Check:

   C lacks built-in queue; manual implementation needed [OK]

Hint: C has no built-in queue; implement yourself [OK]

Common Mistakes:

• Assuming C has STL-like queue
• Using C++ syntax in C code
• Including non-existent headers

3. Consider the following code snippet implementing push operation of a stack using one queue in C-like pseudocode:

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

What will be the order of elements in the queue after pushing 1, then 2, then 3?

medium

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. 2 -> 3 -> 1 -> null

D. 3 -> 1 -> 2 -> null

Solution

1. Step 1: Push 1

   Queue after push: [1]

2. Step 2: Push 2

   Push 2: queue becomes [1,2], then rotate first element: pop 1, push 1 -> queue is [2,1]

3. Step 3: Push 3

   Push 3: queue is [2,1,3], rotate first two elements: pop 2 push 2, pop 1 push 1 -> queue is [3,2,1]

4. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

5. Quick Check:

   Newest element at front after push = 3 -> 2 -> 1 [OK]

Hint: After push, rotate queue to put new element front [OK]

Common Mistakes:

• Not rotating queue after push
• Assuming queue order is same as push order
• Confusing front and back of queue

4. The following code attempts to implement pop operation for a stack using two queues. Identify the error:

int pop() {
    if (q1.empty()) return -1;
    while (q1.size() > 1) {
        q2.push(q1.front());
        q1.pop();
    }
    int top = q1.front();
    q1.pop();
    swap(q1, q2);
    return top;
}

medium

A. The queues q1 and q2 are not initialized properly

B. The pop operation does not handle empty stack correctly

C. No error; the code correctly implements pop

D. The swap(q1, q2) should be before popping the last element

Solution

1. Step 1: Analyze pop logic

   Elements except last are moved from q1 to q2, last element is popped from q1.

2. Step 2: Swap queues

   Swapping q1 and q2 makes q1 the queue with remaining elements for next operations.

3. Final Answer:

   No error; the code correctly implements pop -> Option C

4. Quick Check:

   Pop moves all but last, returns last, swaps queues [OK]

Hint: Pop moves all but last to second queue, then swap [OK]

Common Mistakes:

• Swapping queues too early
• Not checking empty queue before pop
• Forgetting to swap queues after pop

5. You want to implement a stack using two queues. Which approach ensures that the pop() operation is O(1) time complexity, and push() is O(n)?

hard

A. Make pop costly by moving all but last from q1 to q2, then popping last from q1

B. Make push costly by pushing element to empty q2, then moving all from q1 to q2, swap q1 and q2

C. Use a single queue and rotate elements after each push

D. Use two stacks instead of queues

Solution

1. Step 1: Understand push costly approach

   Push inserts element into empty q2, moves all q1 elements to q2, then swaps q1 and q2. This makes newest element front in q1.

2. Step 2: Pop becomes O(1)

   Pop just dequeues from q1 front, which is the top of stack, so pop is O(1).

3. Final Answer:

   Make push costly by pushing element to empty q2, then moving all from q1 to q2, swap q1 and q2 -> Option B

4. Quick Check:

   Push costly + pop O(1) = push rearranges queues [OK]

Hint: Push costly rearranges queues; pop just dequeues front [OK]

Common Mistakes:

• Confusing which operation is costly
• Using single queue for O(1) pop incorrectly
• Choosing two stacks instead of queues