Unit 6 Session 1 (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?
HAPPY CASE
Input: Breaking apart the assignment statement as described.
Output: Mario -> Luigi -> Wario
Explanation: The list is correctly assembled from individual nodes connected in sequence.
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 is a basic problem of constructing and manually linking a singly linked list node-by-node. It tests basic understanding of node manipulation and pointer assignment.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: Create each node independently and then connect them sequentially to form a complete linked list.
1) Create the head node 'Mario' with its next initially set to None.
2) Create the 'Luigi' node with its next also set to None.
3) Create the 'Wario' node with its next set to None.
4) Link 'Mario' to 'Luigi' and 'Luigi' to 'Wario' by setting their next properties.
⚠️ Common Mistakes
Implement the code to solve the algorithm.
class Node:
def __init__(self, value=None, next=None):
self.value = value
self.next = next
# Breaking the assignment into multiple lines
head = Node("Mario")
luigi = Node("Luigi")
wario = Node("Wario")
head.next = luigi
luigi.next = wario
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
O(1)
as the operations to create and link nodes involve a constant amount of work.O(1)
in terms of additional space as no extra data structures are used; the space used is proportional to the number of nodes created, which is fixed.