Hello Everyone,
I got this question:
What's the Time Complexity of finding the 2nd smallest Element in a [sorted] Linked List?
in my exam and till this moment I don't know what's the correct answer,
But I said O[1] because Since it's sorted we already 2nd smallest Element will be always in the 2nd position no matter how big the list is, and find operation it will always be the same number of steps no matter how.
But My professor said it's O[n] without explaining it to me.
Can someone help me to determine whether it's O[n] or O[1]
Since we’re talking about a Sorted Linked List, and you’re inserting without knowing where an element is supposed to go, it will take you O[n] time [since you have to search the whole list to find the place the element goes].
What is the time complexity of inserting a node into a linked list?
Simply inserting a node is O[1] for 2 operations. The pointer to next of the previous node is set to point to this node. The next of this current node is set to the next of the previous node. You can work out insertions at the head of the linked list.
What is the time complexity of insertion sort if it is implemented on a linked list with n elements?
The time complexity of Insertion Sort is O[N^2] and works faster than other algorithms when the data set is almost sorted. Inserting an element in a sorted Linked List is a simple task which has been explored in this article in depth. Do check it out for better understanding.
What is time complexity of insertion sort?
Insertion Sort is an easy-to-implement, stable sorting algorithm with time complexity of O[n²] in the average and worst case, and O[n] in the best case. For very small n, Insertion Sort is faster than more efficient algorithms such as Quicksort or Merge Sort.
What is the time complexity for finding the middle node of a linked list?
The running time of finding the middle element this way with two pointers is O[n] because once we pass through the entire linked list of n elements, the slower pointer is at the middle node already.
What is the complexity of inserting into a sorted link list?
Let say I have 5 elements and what is the complexity to insert all of them. Think about what a single insertion into a sorted link list means. You have a new element that has to go somewhere in the list. 1] You have to find the right place in the list. This is a linear search. O [n]
How does insertion sort work in linked list?
The time complexity of Insertion Sort is O [N^2] and works faster than other algorithms when the data set is almost sorted. Inserting an element in a sorted Linked List is a simple task which has been explored in this article in depth. Do check it out for better understanding. Pseudocode of Insertion Sorting a Linked List:
What is the complexity of the insertion sort algorithm?
The overall complexity of the above algorithm will depend on the complexity of the sorting algorithm. Here we are using the insertion sort algorithm for sorting [M+N] size linked list. Time Complexity = Combine both linked lists + Sort the final list. Space Complexity = O [1] [Using Insertion Sort will take O [1] space]
Is the same complexity true for removing nodes from a linked list?
The same time complexity is also true for removing nodes from a linked list. The graph shown above represents the different time complexities of performing different actions on arrays and linked lists.
import java.util.*; class LinkedList implements java.io.Serializable { private static class Node { E item; Node next; Node prev; Node[Node prev, E element, Node next] { this.item = element; this.prev = prev; this.next = next; } } transient int size = 0; transient Node first; transient Node last; // Creates an empty list public LinkedList[] {} void insertionSort[] { // Initialize sorted linked list final Node sorted = null; // Traverse the given linked list and insert every // node to sorted Node current = first; while [current != NULL] { // Store next for next iteration Node next = current.next; // insert current in sorted linked list Unlink[current]; sortedInsert[current.data]; // Update current current = next; } } void sortedInsert[E e] { Node current = first; /* Special case for head node */ if [current == null || current.data >= e] { LinkFirst[e]; } else { while [current.next != null && current.next.data < new_node.data] current = current.next; final Node newNode = new Node[current, e, current.next]; } } //Checks whether the value x is present in linked list public boolean search[int x] { Node current = first; //Initialize current while [current != null] { if [current.data == x] return true; //data found current = current.next; } return false; //data not found } // Return Node at index "index" O[N] time public Node node[ int index] { if[index < [size >> 1]] { Node x = first; for[int i=0;iindex; --i] x = x.prev; return x; } } // Print all elements in the LinkedList public void printList[] { Node x = first; for[int i=0;i