In a singly linked list, the time taken to insert an element at an arbitrary…

Concept

Inserting into a linked list has two parts: first locate the position where the new node must go, then splice it in by rewiring a fixed number of pointers. The splice is always constant time, so the overall cost is decided entirely by how the position is reached.

Applying it to a general insertion

1. To insert at an arbitrary position, you start at the head and follow next-pointers until you reach the target spot. A linked list has no random access, so this traversal can visit up to all n nodes — that step is O(n).

2. Once the predecessor node is reached, the actual link update (set the new node's next, then point the predecessor at the new node) touches a fixed number of pointers — O(1).

3. Total general-case cost = O(n) traversal + O(1) splice = O(n). With no extra information about where to insert, O(n) is the cost we must report.

Cross-check against the other costs

• O(1): true only for the special case where the insertion position is already known — inserting at the head, or when a pointer to the predecessor node is handed to you in advance. It counts the splice but ignores the lookup that the general case still requires.

• O(n log n): this is the signature of comparison-based sorting / divide-and-conquer; a single insertion performs no sorting, so it does not arise here.

• O(n^2): this would need a nested traversal (a loop inside a loop). A single insertion makes one pass at most, so quadratic cost is far too high.

Hence, in the general case the time to insert an element into a linked list is O(n).