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A circularly linked list is used to represent a Queue. A single variable p is used to access the Queue. To which node should point such that both the operations enqueue and dequeue can be performed in constant time?
  • a)
    rear node
  • b)
    front node
  • c)
    not possible with a single pointer
  • d)
    node next to front
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
A circularly linked list is used to represent a Queue. A single variab...
The pointer points to the Rear node.
EnQueue: Insert newNode after Rear, and make Rear point to the newly inserted node:
//struct node *newNode;
newNode->next = rear->next;
rear->next = newNode;
rear=newNode;
DeQueue: Delete the Front node, and make the second node the front node.
//rear->next points to the front node.
//front->next points to the second node.
struct node* front;
front = rear->next;
rear->next = front->next;
free(front);
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Most Upvoted Answer
A circularly linked list is used to represent a Queue. A single variab...
 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.
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Community Answer
A circularly linked list is used to represent a Queue. A single variab...
The pointer points to the Rear node.
EnQueue: Insert newNode after Rear, and make Rear point to the newly inserted node:
//struct node *newNode;
newNode->next = rear->next;
rear->next = newNode;
rear=newNode;
DeQueue: Delete the Front node, and make the second node the front node.
//rear->next points to the front node.
//front->next points to the second node.
struct node* front;
front = rear->next;
rear->next = front->next;
free(front);
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