What risks are you taking when "signing in with Google"? 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. Nothing in the problem statement forbids using auxiliary data structures. Was Aristarchus the first to propose heliocentrism? 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 I guess I will start you off with the time complexity of a linked list: The proposed solution first does some preprocessing of the arguments to insert, then does the insertion proper. Delete - O(1). 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. It only takes a minute to sign up. A simple way to forbid auxiliary data structures would be to require $O(1)$ memory overhead. Examples : Input : arr [] = {10, 20, 80, 30, 60, 50, Follow the algorithm as -. which the input node is to be inserted. So would we say that the best case complexity of insertion in an array is O (1) and worst case is O (n) or should we say both best and worst case are both O (n)? Indeed worst case insertion is O (n) if you have to copy the whole array into a larger array. But you must remember, it is the amortize cost we care about. $ \ O(nlogn) $. 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 $[1,42]$). keep moving until you reach a node who's value is greater than Assume the array has unused slots and the elements are packed from the 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 $1 + 2 + (n-1) + n = O(n^2)$. 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. In both examples, the So if we assume that we can sort the numbers beforehand with any algorithm, then we can also assume that the numbers are naturals and the maximum element is M < 10, so with radix sort you would get worst case O(10n) = O(n). 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.
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