Use the two-pointer technique: advance a fast pointer n+1 steps ahead, then move both pointers until fast reaches the end. The slow pointer will be just before the node to remove, allowing a single-pass deletion with constant space. Handle edge cases like removing the head by using a dummy node.
Pro tip: Explicitly discuss edge cases (e.g., n equals list length, single-node list) and how the dummy node simplifies head removal. Also, mention that this approach is optimal for time and space, which interviewers at TikTok appreciate.
Restate the problem to ensure understanding: remove the nth node from the end of a singly linked list, return the new head, in one pass with O(1) space. Ask about constraints (e.g., n validity, list length).
Explain that you'll use two pointers, fast and slow, initially at a dummy node. Advance fast n+1 steps ahead, then move both until fast reaches null. Slow will point to the node before the one to remove.
Discuss using a dummy node to simplify removal of the head (when n equals list length). Also consider n=1 (remove last node) and single-node list.
Write clean code with clear variable names. Walk through an example (e.g., 1->2->3->4->5, n=2) to verify correctness. Mention time complexity O(L) and space O(1).
Double-check pointer manipulations for off-by-one errors. Confirm that the solution meets the single-pass and constant-space requirements. Discuss potential follow-ups (e.g., what if n is invalid).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use level-order traversal (BFS) to locate the target node and simultaneously track the deepest rightmost node and its parent. Then copy the deepest rightmost node's value into the target node and delete the deepest rightmost node by updating its parent's child pointer.
Pro tip: Clarify edge cases upfront: if the tree is empty, the key is not found, or the target node is the deepest rightmost node itself. Also mention that this approach modifies the tree in-place and runs in O(n) time with O(n) space for the queue.
Check if the root is null or if the key is not present. Confirm whether the tree can have duplicate keys and whether the deepest rightmost node is defined as the last node in level-order traversal.
Perform a BFS using a queue, storing each node along with its parent. Track the target node and its parent, and update the deepest rightmost node and its parent as you process each level.
If the target node is found, replace its value with the value of the deepest rightmost node. If the target is the root and the tree has only one node, handle removal appropriately.
Update the parent of the deepest rightmost node to remove the reference to it (set the appropriate child pointer to null). If the deepest rightmost node is the root (only one node), set root to null.
After deletion, return the original root (unless it was the only node and removed). Discuss time and space complexity: O(n) time, O(n) space due to queue.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.