Given a queue implemented using a linked list, the rear pointer points to the rear node, and front pointer points to the front node of the queue, which of the following statement is not possible in O(1) complexity*2 pointsinsertion at rearinsertion at frontdeletion at frontdeletion at rear

The operation that is not possible in O(1) complexity is "deletion at rear".

Here's why:

1. Insertion at rear: This operation is possible in O(1) time complexity because we have a pointer to the rear of the queue. We can simply add a new node at the location the rear pointer is pointing to and then update the rear pointer to the new node.

2. Insertion at front: This operation is also possible in O(1) time complexity because we have a pointer to the front of the queue. We can simply add a new node at the location the front pointer is pointing to and then update the front pointer to the new node.

3. Deletion at front: This operation is possible in O(1) time complexity because we have a pointer to the front of the queue. We can simply remove the node at the location the front pointer is pointing to and then update the front pointer to the next node.

4. Deletion at rear: This operation is not possible in O(1) time complexity. To delete a node at the rear of the queue, we would need to have a pointer to the node before the rear node, but we only have a pointer to the rear node itself. Therefore, we would need to traverse the queue from the front to find the node before the rear node, which takes O(n) time complexity where n is the number of nodes in the queue.