Correct option is (B) ⊖(1), ⊖(n)
For Enqueue operation, performs in constant amount of time (i.e., Θ(1)), because it modifies only two pointers, i.e.,
Create a Node P.
P-->Data = Data
P-->Next = Head
Head = P
For Dequeue operation, we need address of second last node of single linked list to make NULL of its next pointer. Since we can not access its previous node in singly linked list, so need to traverse entire linked list to get second last node of linked list.