Implement Stack Using Queue in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the queue size in the stack structure.

DSA C

typedef struct {
    int arr[100];
    int front, rear;
} Queue;

typedef struct {
    Queue q;
    int [1];
} Stack;

void initStack(Stack* s) {
    s->[1] = 0;
    s->q.front = 0;
    s->q.rear = -1;
}

Drag options to blanks, or click blank then click option'

Asize

Blength

Ccount

Dtop

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'size' or 'length' which are not updated properly.

Confusing queue indices with stack top.

The 'top' variable keeps track of the number of elements in the stack.

2fill in blank

medium

Complete the code to enqueue an element into the queue.

DSA C

void enqueue(Queue* q, int val) {
    q->rear++;
    q->arr[q->[1]] = val;
}

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Arear

Bfront

Ctop

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'front' instead of 'rear' causes overwriting the wrong position.

Using 'top' which is unrelated to queue indices.

We insert the new value at the 'rear' index of the queue.

3fill in blank

hard

Fix the error in the pop function to correctly remove the top element from the stack using the queue.

DSA C

int pop(Stack* s) {
    if (s->top == 0) return -1; // empty
    int i, val;
    for (i = 0; i < s->top - 1; i++) {
        val = s->q.arr[s->q.front];
        s->q.front++;
        s->q.arr[++s->q.rear] = val;
    }
    val = s->q.arr[s->q.front];
    s->q.front++;
    s->top--;
    [1];
    return val;
}

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As->q.front = 0

Bs->q.rear = s->q.front - 1

Cs->q.rear = s->q.front

Ds->q.front = s->q.rear

Attempts:

3 left

💡 Hint

Common Mistakes

Not updating rear causes incorrect queue state.

Setting rear equal to front causes empty queue misinterpretation.

After removing elements, rear should be reset to front - 1 to maintain queue state.

4fill in blank

hard

Fill both blanks to implement the push function that adds an element to the stack using the queue.

DSA C

void push(Stack* s, int val) {
    s->q.arr[++s->q.[1]] = val;
    s->top++;
    int i, size = s->top;
    for (i = 0; i < size - 1; i++) {
        int temp = s->q.arr[s->q.[2]];
        s->q.front++;
        s->q.arr[++s->q.rear] = temp;
    }
}

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Arear

Bfront

Ctop

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Using front instead of rear to insert new element.

Not rotating elements correctly after insertion.

We add the new element at rear and then rotate elements from front to rear to simulate stack push.

5fill in blank

hard

Fill all three blanks to implement the isEmpty function that checks if the stack is empty.

DSA C

int isEmpty(Stack* s) {
    if (s->[1] == 0 && s->q.[2] > s->q.[3]) {
        return 1;
    }
    return 0;
}

Drag options to blanks, or click blank then click option'

Atop

Bfront

Crear

Dsize

Attempts:

3 left

💡 Hint

Common Mistakes

Checking only top or only queue indices.

Using size which is not maintained.

Stack is empty if top is zero and queue front is greater than rear (empty queue).

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