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.
Get this book -> Problems on Array: For Interviews and Competitive Programming Problem Statement: You are given a sorted singly linked list and a key (element to be searched), find the key in the linked list using binary search algorithm. The challenge is to find the middle element as Linked List does not support random access. Binary search is used because it has a time complexity of O(N) for a sorted array.
If we follow sequential access of Linked List, then it will take O(N) time to find the middle element. With this, the overall time complexity of Binary Search on Linked List will become O(N * logN).
This will slower than linear search.
The trick is to find the middle element using two pointer approach using which the time complexity will be O(N) for Binary Search on Linked List. With Skip List which is a modification of Linked List, we can achieve O(logN) time complexity for Binary Search.
Hence, the main points are as follows:
- Binary Search: O(logN) for array
- Binary Search: O(N) for Singly Linked List
- Binary Search: O(logN) for Skip List
Sample Testcase to understand the problem:
Linked List : 1->3->5->8->9->10->11->13->NULL Input : Enter value to search : 10
Output : Found
Input : Enter value to search : 12
Output : Not Found
To perform a Binary Search Algorithm on Singly Linked Lists, determination of the middle element is important. Binary Search is fast and efficient because accessing the middle elemen is fast. In arrays, binary search takes O(1) time to access middle element. But memory allocation for the singly linked list is non-contiguous, which makes finding the middle element difficult and time consuming.
The node structure for Linked List is as follows:
// Node structure struct Node { int data; Node* next; };Finding middle element: Two pointer approach
- Traverse the singly linked list using two pointers.
- Move one pointer by one step ahead and the other pointer by two steps.
- When the fast pointer reaches the end of the singly linked list, the slow pointer will reach the middle of the singly linked list.
- Return slow pointer address.
Code
// function to find out middle element Node* middle(Node* start, Node* last) { // if start points to NULL and is empty if (start == NULL) return NULL; Node* slow = start; Node* fast = start -> next; while (fast != last) { // fast is moved one step ahead fast = fast -> next; // if fast is not the last element if (fast != last) { // slow pointer is moved one step ahead slow = slow -> next; // fast pointer is moved second step ahead fast = fast -> next; } } return slow; }BINARY SEARCH ALGORITHM
Pseudo Code Algorithm
- Start node is set to head of the list and the last node is set to NULL.
- Middle element is calculated using the two pointers approach discussed above.
- If the middle element is same as the key to be searched, we return it.
- Else if middle element is less than the key to be searched, we have to search is the right side of the singly linked list. So, we set start pointer to the next of middle element.
- Else if middle element is greater than the key to be searched, we have to search is the left side of the singly linked list. So, we set last pointer to the middle element.
- If the key is found or the entire linked list gets traversed once, we break the loop.
Code
// Node structure struct Node { int data; Node* next; }; // function to find out middle element Node* middle(Node* start, Node* last) { // if start points to NULL and is empty if (start == NULL) return NULL; Node* slow = start; Node* fast = start -> next; while (fast != last) { // fast is moved one step ahead fast = fast -> next; // if fast is not the last element if (fast != last) { // slow pointer is moved one step ahead slow = slow -> next; // fast pointer is moved second step ahead fast = fast -> next; } } return slow; } // Function for implementing the Binary Search on Singly Linked List Node* binarySearch(Node *head, int key) { Node* start = head; Node* last = NULL; do { // calling middle function to find middle element Node* mid = middle(start, last); // If middle element is empty if (mid == NULL) return NULL; // If value is present at middle, we return it if (mid -> data == key) return mid; // If value is more than mid else if (mid -> data < key) start = mid -> next; // If the value is less than mid. else last = mid; } while (last == NULL || last != start); // value not present, so we return NULL return NULL; }Complexity Analysis
We'll look at the asymptotic complexity of our solution which comes to be O(N) as we are traversing the singly linked list at least once either to find the middle element or to move the pointers for binary search.