Queue Practice Questions

To reverse the elements of Q1 into Q2, we need to process them in a Last-In, First-Out (LIFO) order. The element '4' must be the first to enter Q2. A simple way to conceptualize this reversal is with recursion.

Consider a recursive function Reverse(Q1, Q2). In each call, we would $Dequeue$ an element from Q1, make a recursive call for the rest of Q1, and upon return, $Enqueue$ the dequeued element into Q2. This process naturally reverses the sequence. For example, '1' is dequeued first, but enqueued into Q2 last, after '2', '3', and '4' have been processed.

This recursive algorithm consists solely of $Dequeue(Q1)$ and $Enqueue(Q2)$ operations. It performs zero $Enqueue$ operations on Q1. While this approach uses the function call stack as implicit storage, it is the standard method to achieve queue reversal, and it demonstrates that the reversal can be achieved with 0 $Enqueue$ operations on Q1. Therefore, the minimum number is 0.

The enqueue operation is implemented by inserting a new node at the head of the singly linked list. This involves creating a new node, setting its next pointer to the current head, and then updating the head pointer to the new node. These are all constant-time operations, so the complexity is $\Theta(1)$.

The dequeue operation is implemented by deleting a node from the tail. In a singly linked list, to remove the tail node, one must update the next pointer of the second-to-last node to NULL. Finding this predecessor node requires traversing the entire list from the head, which takes time proportional to the number of nodes, $n$. Therefore, the complexity is $\Theta(n)$.

Breadth-First Search ($BFS$) traverses a graph by visiting all neighbor nodes at the current depth level before proceeding to the next depth. To manage this process, the algorithm requires a First-In, First-Out ($FIFO$) ordering mechanism to ensure nodes are visited in the order they are discovered. A Queue data structure natively supports this $FIFO$ operation, making it ideal for storing vertices during $BFS$ traversal. In contrast, a Stack utilizes Last-In, First-Out ($LIFO$) logic, which is suitable for Depth-First Search ($DFS$), not $BFS$. Therefore, a Queue is the correct data structure for this task.

For a circular queue implemented with a singly linked list to support O(1) insertion and deletion, an efficient structure is required. Deletion occurs at the front and insertion at the rear.
If the next pointer of the rear node points to the front node, we can access the front from the rear in O(1) time. This allows for efficient deletion of the front node and insertion after the rear node.
Statement II, "Next pointer of rear node points to the front node," describes this correct and efficient implementation.
Statement I is incorrect; having the front node's pointer point to the rear node does not facilitate standard O(1) queue operations.

A queue implemented using a circular array maintains two pointers, $front$ and $rear$, to track the current positions of the elements. For an $ENQUEUE$ operation, the element is inserted at the index $rear$, and the pointer is updated as $rear = (rear + 1) \pmod N$, taking $O(1)$ time. For a $DEQUEUE$ operation, the element is retrieved from $front$, and the pointer is updated as $front = (front + 1) \pmod N$, also taking $O(1)$ time. Since circular indexing avoids the need to shift array elements, both operations achieve $O(1)$ worst-case time complexity. Thus, both operations are performed efficiently in constant time.

The provided algorithm describes a sorting process that utilizes a queue $Q$ and a stack $S$ to order elements. In the worst-case scenario, such as when the input queue is sorted in reverse order (descending), the algorithm forces elements to be moved back and forth between the queue and the stack repeatedly. This mechanism mimics the behavior of quadratic sorting algorithms like bubble sort or insertion sort, where the total number of operations is $O(N^2)$. For a queue containing $N = 16$ elements, the maximum number of iterations is $N^2$, which yields $16^2 = 256$.

[CORRECT_OPTION: N/A]

A queue operates on the First-In-First-Out (FIFO) principle, while a stack follows Last-In-First-Out (LIFO). To implement a queue using stacks, we utilize two stacks, $S_1$ and $S_2$. The enqueue operation pushes elements into $S_1$. For the dequeue operation, if $S_2$ is empty, we transfer all elements from $S_1$ into $S_2$ by popping them one by one, which reverses their order. Popping an element from $S_2$ then retrieves the element that was inserted first, satisfying the FIFO requirement. Therefore, two stacks are necessary and sufficient to implement a queue.

In a standard circular queue, the insertion position is calculated by incrementing the rear pointer as $rear = (rear + 1) \pmod{size}$. With the initial $rear = 2$ and $size = 11$, adding the $k$-th element updates the rear pointer to $(initial\_rear + k) \pmod{size}$. For the 9th element ($k = 9$):
$rear_{9} = (2 + 9) \pmod{11} = 11 \pmod{11} = 0$
Since the result is $0$, the 9th element is stored at position $q[0]$.

Let's analyze the time complexity of a sequence of $n$ queue operations, including Enqueue, Dequeue, and MultiDequeue.

1. Enqueue/Dequeue: A single Enqueue or Dequeue operation on a queue takes $O(1)$ time.
2. MultiDequeue: The MultiDequeue function dequeues elements one by one until either the queue is empty or $m$ (initialized to $k$) elements have been removed. This means MultiDequeue takes $O(k)$ time in the worst case if the queue has at least $k$ elements, or $O(Q_{size})$ if the queue has fewer than $k$ elements.
3. Amortized Analysis: Consider $n$ operations in total. Each Enqueue operation adds one element. Each Dequeue operation removes one element. The total number of Dequeue operations across all MultiDequeue calls and explicit Dequeue calls cannot exceed the total number of Enqueue operations, which is at most $n$.
4. Overall Complexity: Since each element is enqueued once and dequeued at most once, the total cost for all Enqueue and Dequeue operations (including those within MultiDequeue) is $O(n)$. The MultiDequeue function's loop runs based on $m$ and queue size, but the actual work done is limited by dequeuing elements, which is bounded by $n$.
   Therefore, the total worst-case time complexity for $n$ operations is $\Theta(n)$.

A circular queue of capacity $(n-1)$ elements uses an array of $n$ elements.
Initially, both REAR and FRONT pointers are $0$.
A queue is empty when REAR == FRONT. This is because no elements have been added or all elements have been removed, bringing the pointers to the same position.
A queue is full when (REAR + 1) % n == FRONT. This condition reserves one slot in the array to distinguish between a full and an empty queue. If REAR advances one position and meets FRONT, it means the queue is full.

The function $f(Q)$ utilizes the system call stack to reverse the order of elements in the queue $Q$. For a queue containing elements $[q_1, q_2, \dots, q_n]$, each recursive call removes the front element $q_i$ and stores it in the local stack frame. This continues until the queue is empty, at which point the recursion unwinds. As the recursion returns, each stored element is inserted at the rear of the queue in the reverse order of removal ($q_n, q_{n-1}, \dots, q_1$). This process effectively reverses the queue $Q$.

To implement the queue, $n$ insertions perform $n$ pushes onto $S_1$. When $m$ deletions occur, elements are transferred from $S_1$ to $S_2$ only when $S_2$ is empty. Let $k$ be the total number of elements transferred from $S_1$ to $S_2$.

1. The total number of push operations $x = n + k$ (initial $n$ pushes plus $k$ pushes to $S_2$).
2. The total number of pop operations $y = m + k$ ($m$ pops from $S_2$ for the deletions plus $k$ pops from $S_1$ during transfers).
3. To perform $m$ delete operations, we must transfer at least $m$ elements ($k \geq m$) and can transfer at most all $n$ elements ($k \leq n$).
4. Substituting the bounds of $k$ into the expressions:
   For $x$: $n + m \leq x \leq n + n \implies n + m \leq x \leq 2n$.
   For $y$: $m + m \leq y \leq m + n \implies 2m \leq y \leq n + m$.

A queue operates on First-In-First-Out (FIFO) logic, while a stack follows Last-In-First-Out (LIFO) logic. To bridge this difference, we use two stacks: $S_{\in}$ for enqueueing and $S_{out}$ for dequeueing. When a dequeue operation is requested and $S_{out}$ is empty, we pop all elements from $S_{\in}$ and push them into $S_{out}$, effectively reversing the order of arrival. This inversion ensures the first element pushed to $S_{\in}$ becomes the first to be popped from $S_{out}$, satisfying FIFO requirements. Therefore, two stacks are the minimum required.

A stack follows LIFO (Last-In, First-Out) behavior, meaning the most recently added element must be the first one removed. The DELETEMIN(Q) operation removes the element with the smallest key. To ensure the last pushed item is retrieved first, the priority queue must assign the smallest key to the most recent PUSH operation. If keys $K_1, K_2, \dots, K_n$ are assigned to sequential PUSH operations, we require $K_1 > K_2 > \dots > K_n$ so that $K_n$ is the minimum. Therefore, the keys must be in strictly decreasing order.

(i) False: First-in-first-out (FIFO) operations are the defining characteristic of QUEUES, not STACKS (which are LIFO).
(ii) True: Linked lists allow $O(1)$ insertion and deletion given a reference, whereas arrays require $O(n)$ time to shift elements for these operations.
(iii) True: Circular arrays allow $O(1)$ enqueue/dequeue operations using modulo arithmetic, preventing the wasted space or the $O(n)$ shifting required by linear arrays with fixed indices.
(iv) False: Last-in-first-out (LIFO) operations are the defining characteristic of STACKS, not QUEUES.
Since statements (ii) and (iii) are true, Option A is the correct answer.