Unit 5 Session 2 (Click for link to problem statements)
Understand what the interviewer is asking for by using test cases and questions about the problem.
i is out of the list's current bounds? i is beyond the current end of the list, the new node should be inserted at the end of the list.Plan the solution with appropriate visualizations and pseudocode.
General Idea: Insert a new node at the specified index in the linked list, handling edge cases where i is 0 or beyond the end of the list.
1) If the list is empty (`head` is `None`), create a new node with the value and return it as the new head. 2) If the index `i` is 0, insert the new node before the current head and return the new node as the new head. 3) Use a loop to find the correct position in the list where the new node should be inserted: a) Traverse the list until reaching the position just before the index `i`. b) Insert the new node at index `i` by adjusting the `next` pointers. 4) If `i` exceeds the length of the list, append the new node at the end. 5) Return the head of the list.⚠️ Common Mistakes
next pointers correctly, which could lead to losing part of the list.i is greater than the number of nodes, which should not throw an error but append the node at the end.def ll_insert(head, val, i): # Handle the case where the list is initially empty or `i` is 0 if not head or i == 0: return Node(val, head) # Traverse to find the insertion point current = head count = 1 # Start count at 1 since we've already handled `i == 0` while current: # Insert at index `i` by adjusting the `next` pointers if count == i: new_node = Node(val, current.next) current.next = new_node return head current = current.next count += 1 # If `i` was beyond the end, append to the last if count < i: current.next = Node(val) return head