Consider the queues Q1 containing four elements and Q2 containing none (shown as the Initial State in the figure). The only operations allowed on these two queues are Enqueue(Q, element) and Dequeue(Q). The minimum number of Enqueue operations on Q1 required to place the elements of Q1 in Q2 in reverse order (shown as the Final State in the figure) without using any additional storage is___________.

Correct Answer:

0

Solution:

Enqueue(Q, element): Insert the given element at the end of the given Queue.

Dequeue(Q): Remove and return the element from the start of the Queue

Importantly, these operations don't consume additional space. It's crucial to note that the return value and arguments of a function are stored in the function's stack frame, ensuring that no extra space is required beyond what is inherent to the function call or operation itself.

OperationsState of Queues
Initial StateQ1:1, 2, 3, 4
Q2: 
Enqueue(Q2, Dequeue(Q1))Q1:2, 3, 4
Q2: 1
Enqueue(Q2, Dequeue(Q1))Q1:3, 4
Q2: 1, 2
Enqueue(Q2, Dequeue(Q2))Q1:3, 4
Q2: 2, 1
Enqueue(Q2, Dequeue(Q1))Q1:4
Q2: 2, 1, 3
Enqueue(Q2, Dequeue(Q2))Q1:4
Q2: 1, 3, 2
Enqueue(Q2, Dequeue(Q2)) Q1:4
Q2: 3, 2, 1
Enqueue(Q2, Dequeue(Q1)) Q1: 
Q2: 3, 2, 1, 4
Enqueue(Q2, Dequeue(Q2)) Q1:
Q2: 2, 1, 4, 3
Enqueue(Q2, Dequeue(Q2))Q1:
Q2: 1, 4, 3, 2
Enqueue(Q2, Dequeue(Q2)) 
Final State

Q1:
Q2: 4, 3, 2, 1

The minimum number of Enqueue operations on Q1 required to place the elements of Q1 in Q2 in reverse order without using any additional storage is 0.