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Data Structures | Linked List | Question 12

  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021

A circularly linked list is used to represent a Queue. A single variable p is used to access the Queue. To which node should p point such that both the operations enQueue and deQueue can be performed in constant time? (GATE 2004)
circularLinkedList

(A) rear node
(B) front node
(C) not possible with a single pointer
(D) node next to front


Answer: (A)

Explanation: Answer is not “(b) front node”, as we can not get rear from front in O(1), but if p is rear we can implement both enQueue and deQueue in O(1) because from rear we can get front in O(1). Below are sample functions. Note that these functions are just sample are not working. Code to handle base cases is missing.




/* p is pointer to address of rear (double pointer).  This function adds new 
   node after rear and updates rear which is *p to point to new node  */
void  enQueue(struct node **p, struct node *new_node)
{
    /* Missing code to handle base cases like *p is NULL */
       
     new_node->next = (*p)->next;
     (*p)->next = new_node;
     (*p) = new_node /* new is now rear */
     /* Note that p->next is again front and  p is rear */
  
 }
  
/* p is pointer to rear. This function removes the front element and 
    returns the dequeued element from the queue */
struct node *deQueue(struct node *p)
{
    /* Missing code to handle base cases like p is NULL,
        p->next is NULL,...  etc */
  
    struct node *temp = p->next;
    p->next = p->next->next;
    return temp;
    /* Note that p->next is again front and  p is rear */
}


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