Unit 6 Session 2 (Click for link to problem statements)
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
m? n nodes to remove after skipping m nodes.HAPPY CASE Input: 1 -> 2 -> 3 -> 4 -> 5 -> 6, m = 2, n = 2 Output: 1 -> 2 -> 5 -> 6 Explanation: First 2 nodes are kept, next 2 nodes are removed, and the remaining nodes are kept. EDGE CASE Input: 1 -> 2 -> 3, m = 4, n = 1 Output: 1 -> 2 -> 3 Explanation: As `m` is larger than the total number of nodes, no nodes are deleted.Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
This problem involves modifying a linked list by skipping certain nodes and removing others, using pointer manipulation without extra space.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Traverse the list, skip m nodes, then remove n nodes. Repeat the pattern until the end of the list.
1) Use a pointer to traverse the list. 2) Skip `m` nodes, ensuring the pointer does not reach the end of the list prematurely. 3) Remove `n` nodes, adjusting pointers to bypass them. 4) Repeat until the end of the list is reached. 5) Return the head of the modified list.⚠️ Common Mistakes
m or n exceeds the number of remaining nodes.Implement the code to solve the algorithm.
def skip_and_remove(head, m, n): current = head while current: # Skip m nodes for i in range(1, m): if current is None: return head current = current.next # Now current is the last node before the n nodes to delete if current is None: return head # Start deletion of next n nodes temp = current for j in range(n): if temp.next is None: break temp = temp.next # Connect the current node to the node after the N deleted nodes if current and temp.next: current.next = temp.next current = temp.next return headReview the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
m and n to ensure nodes are correctly skipped and removed.m to see that no changes are made.Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(n) where n is the total number of nodes in the list. Each node is processed up to once.O(1) because no additional space is used; modifications are done in-place.