#Enqueue time complexity update#
#Enqueue time complexity code#

Let’s imagine that we started with the brute force solution described above and came to this situation: Implement Queue Using Two Stacks Solution 2: Optimal Tail = insert_node(head, tail, dequeue(s1, s2)) LinkedListNode* implement_queue(LinkedListNode* operations) Shovel all elements back to the first stack. Head of the second stack is the element to dequeue. Pop all elements from first to second stack. The only dequeue operation results in the first enqueued element, 0, to be appended to the result list. Here is how we would execute the operations and build the result list: Operation

If the queue is empty, append -1 to the result.

If the queue is not empty, dequeue current head and append it to the result.

A non-negative integer means 'enqueue me'.

Operations are given in the form of a linked list, and you need to return the result as a linked list, too. Given a sequence of enqueue and dequeue operations, return a result of their execution without using a queue implementation from a library. You feedback is highly appreciated.Implement Queue Using Two Stacks Problem Statement

#Enqueue time complexity code#

Link to the above code on my github profile, Queue using linked list. Space complexity = O(1) since no extra or any auxiliary space is used for the above operations. Time complexity of display operation is O(N) where N = number of elements in the queue, since all the elements of the queue are iterated. O(1), since only the rear and front pointers need to be updated and no iteration is performed over the elements. Time complexity of enqueue and dequeue operations is constant i.e. QueueUsingLinkedList q = new QueueUsingLinkedList() * method to display all elements of the queue */

* method to remove an element from the queue */

#Enqueue time complexity update#

attach the new node at the rear end and update rear pointer * method to add a new element to the queue */ QueueNode class representing a node in the linked list constructor to initialize front and rear pointers

To dequeue an element, a node is removed from the front end of the queue.ĭiagrammatic representation: Code implementation:.

To enqueue an element, a new node is created and attached to the rear end of the queue.Declare a QueueNode class with data as an integer variable which will hold the data of a queue node and a next pointer of type QueueNode which will point to the next node in the queue.display(): This method is used to display all the queue elements by iterating over the elements.dequeue(): This operation is used to dequeue en element from the front of the linked list.enqueue(int data): This operation is used to enqueue an element to the rear end of the linked list.Two node pointers front and rear is maintained at the head and at the tail end of the linked list. The enqueue operation is performed at the tail end of the linked list and dequeue is performed at the head end of the linked list.

To solve the above problem, queue can be implemented using linked list. But when the number of elements stored in the array becomes large it can give poor performance. This works well when the number of elements is small. In the queue implementation using array, a major drawback is when an element is dequeued from the front of the array, all the elements have to be shifted to the left by 1 position. Knowledge of Java, basic data structures, working of queue, linked list and understanding of time and space complexity. This post discusses how queue implementation using linked list gives better performance and reduces the overhead of shifting the elements to the left, every time an element is dequeued from the queue. Queue can also be implemented using linked list to overcome the disadvantages of the queue implementation using array. In the previous post I discussed about queue implementation using circular array.