To find the appropriate node start from the head, A simple way to forbid auxiliary data structures would be to require $O(1)$ memory overhead. Note that there is a constant factor for the hashing algorithm, $ \ O(n) $ Second, sort the elements using merge sort. I guess I will start you off with the time complexity of a linked list: To insert each element, find the preceding element in the mapping, and insert the new element after this node. It's somewhat poorly worded because it relies on precise reading, but fails to state some key assumptions, such as the fact that obtaining the elements to insert costs $O(n)$, comparing two elements can be done in $O(1)$, and the input domain is effectively unbounded (exercise: come up with an $O(n)$ algorithm if the inputs are integers in the range $[1,42]$). There are also algorithms which are non-comparative such as Radix sort which their complexity depends on the size in bits which the numbers need to be stored in memory. Note that even under this assumption, your reasoning is wrong, or at least imprecise. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Making statements based on opinion; back them up with references or personal experience. From the given wording of the question, which solution is more apt? What is the run-time complexity of inserting an integer into an unsorted array? If its unsorted, you dont have to insert the integer in any specific place, so you can just insert it at the end. That means the time is O (1), unless you need to reallocate memory for the array. 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. Then whenever we have to insert a new element we insert it first into BST. Indexing---->O(n). I suppose the second approach you propose implies the use of a secondary data structure like a dynamic array. Nothing as useful as this: Common Data Structure Operations: What is the time complexity of indexing, inserting and So when you insert all the elements at the tail, they are not necessarily in sorted order. But the given answer is correct. How to force Unity Editor/TestRunner to run at full speed when in background? By clicking Post Your Answer, you agree to our terms of service, privacy policy and cookie policy. I know this is a general question but I really do need to clear my doubt as I am studying WebWe would like to show you a description here but the site wont allow us. 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 $O(n \log n)$. MathJax reference. We have presented the Time Complexity analysis of different operations in Array. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. So this question isn't just making strange requirements for the sake of being strange. The Time complexity of insertion sort depends on the number of inversions in the input array. In a given array, if (i < j) and (A [i] > A [j]) then the pair (i, j) is called an inversion of an array A, note that i and j are the array indexes. In my opinion, the answer should be $O(n^2)$ because in every insertion, we will have to insert the element in the right place and it is possible that every element has to be inserted at the last place, giving me a time complexity of $1 + 2 + (n-1) + n = O(n^2)$. Which was the first Sci-Fi story to predict obnoxious "robo calls"? 2) If the value of the node to be inserted is smaller than the value of the head node, then insert the node at the 2) If the value of the node to be inserted is smaller best case and worst case time complexity for insertion in Assume the array has unused slots and the elements are packed from the Insert - O(log n). First of all, the complexity of O(nlogn) applies only for the algorithms which use comparison between their elements (comparative algorithm). This assumes that the insertion process creates the list nodes as it goes (as opposed to filling existing blank nodes). Examples : Input : arr [] = {10, 20, 80, 30, 60, 50, which the input node is to be inserted. The worst case is indeed $\Theta(n^2)$, but to prove this, you have to prove that finding the insertion point in the list takes $\Theta(n)$ time, and this requires proving that the distance from any pointer you have into the list is bounded below by $\Omega(n)$. Given an unsorted array of integers and an element x, find if x is present in array using Front and Back search. The question only says that the target list needs to be maintained in sorted order. 3) In a loop, find the appropriate node after