Asking for help, clarification, or responding to other answers. It only takes a minute to sign up. 2) If the value of the node to be inserted is smaller A practical reason to do this, rather than insert the elements then sort, would be if the linked list object is shared with another thread that requires it to always be sorted. Web1) If Linked list is empty then make the node as head and return it. Linked list: advantages of preventing movement of nodes and invalidating iterators on add/remove, Average Case Analysis of Insertion Sort as dealt in Kenneth Rosen's "Discrete Mathemathematics and its Application", Complexity of insertion into a linked list, single vs double. Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. We use balanced BST augmented with pointer to slot of linked list which corresponds to key stored in node. A binary search tree would also allow enumerating the elements in sorted order in $O(n \log n)$ time. Connect and share knowledge within a single location that is structured and easy to search. You can sort linked lists in $O(n \log n)$ time (assuming a two-element comparison), for example with merge sort. appropriate node, 4) Insert the node after the appropriate node If we cannot make any assumption then you are right. From the given wording of the question, which solution is more apt? To learn more, see our tips on writing great answers. Use MathJax to format equations. keep moving until you reach a node who's value is greater than Delete - O(1). Red-Black trees: I think @VimalPatel has a better solution than sorting before insertion. @VimalPatel I think the question doesn't imply anywhere that we are allowed to use auxiliary data structures because honestly, it seems overkill to me. at the start and make it head. If you do not, you have to iterate over all elements until The proposed solution first does some preprocessing of the arguments to insert, then does the insertion proper. The node just before that is the sorting - Time complexity of insertion in linked list - Computer This assumes that the insertion process creates the list nodes as it goes (as opposed to filling existing blank nodes). @JhonRayo99 My qualm with that approach is that the question mentions "maintained in sorted order". Retrieve - O(log n). Nothing in the problem statement forbids using auxiliary data structures. So this question isn't just making strange requirements for the sake of being strange. Nothing as useful as this: Common Data Structure Operations: Time Complexity Analysis of Array - OpenGenus IQ: Insert - O(log n). However, the solution that I have says that we can first sort the elements in $O(n \log n)$ and then, we can insert them one by one in $O(n)$, giving us an overall complexity of $O(n \log n)$. the input node. This algorithm takes $\Theta(n^2)$ time in the worst case. The time complexity to insert into a doubly linked list is O (1) if you know the index you need to insert at. But then, I am not very sure either. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. found in step 3. The inner loop at step 3 takes $\Omega(k)$ time in the worst case where $k$ is the number of elements that have already been inserted. If you are only allowed to use linked lists and nothing more (no indexing of any kind), then the complexity is O(n^2) (bubble sort). Or sorting a list. At least that's how I interpret the question and hence my doubt. Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. If you happened to know that the elements are given in the correct order, you could maintain a pointer to the tail of the list, and keep inserting there, which would take $O(n)$. Information on this topic is now available on Wikipedia at: Search data structure. $ \ O(nlogn) $. A simple way to forbid auxiliary data structures would be to require $O(1)$ memory overhead. I suppose the second approach you propose implies the use of a secondary data structure like a dynamic array. I guess I will start you off with the time complexity of a linked list: 1) If Linked list is empty then make the node as Another solution with the same complexity would be to insert the elements into the target list as they come, and maintain a parallel data structure mapping element values to node pointers in the target list. Examples : Input : arr [] = {10, 20, 80, 30, 60, 50, It really is a tricky question. Check the element x at front and rear index. If element x is found return true. Else increment front and decrement rear and go to step 2. The worst case complexity is O (n/2) (equivalent to O (n)) when element is in the middle or not present in the array. The best case complexity is O (1) when element is first or last element in the array. Making statements based on opinion; back them up with references or personal experience. In my opinion, since the question mentions "linked list needs to be maintained in sorted order", I am inclined to say that we cannot sort the elements beforehand and then insert them in the sorted order. Then whenever we have to insert a new element we insert it first into BST. Front and Back Search in unsorted array - GeeksforGeeks