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TIP102 Unit 5 Session 2 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10-15 mins
• 🛠️ Topics: Linked Lists, Insertion

1: U-nderstand

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?

• What happens if the task list is empty?
  • If the task list is empty, the new task will be the only node in the list.

HAPPY CASE
Input: head = Node("shake tree") -> Node("dig fossils") -> Node("catch bugs"), task = "check turnip prices"
Output: Node("check turnip prices") -> Node("shake tree") -> Node("dig fossils") -> Node("catch bugs")
Explanation: The new task is added to the front of the task list.

EDGE CASE
Input: head = None, task = "start new task"
Output: Node("start new task")
Explanation: When the initial task list is empty, the new task becomes the head.

2: M-atch

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.

For Linked List problems, we want to consider the following approaches:

• Insertion in a linked list
• Updating the head of a linked list

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Insert a new node at the beginning of the linked list and make it the new head.

1) Create a new node with the given task.
2) Set the new node's next to the current head.
3) Return the new node as the new head.

⚠️ Common Mistakes

• Forgetting to update the next pointer of the new node to point to the current head.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value, next=None):
        self.value = value
        self.next = next

# For testing
def print_linked_list(head):
    current = head
    while current:
        print(current.value, end=" -> " if current.next else "\n")
        current = current.next

def add_task(head, task):
    new_node = Node(task)
    new_node.next = head
    return new_node

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Trace through your code with an input to check for the expected output
• Catch possible edge cases and off-by-one errors

Example:

task_1 = Node("shake tree")
task_2 = Node("dig fossils")
task_3 = Node("catch bugs")
task_1.next = task_2
task_2.next = task_3

# Linked List: shake tree -> dig fossils -> catch bugs
print_linked_list(add_task(task_1, "check turnip prices"))

Expected Output: check turnip prices -> shake tree -> dig fossils -> catch bugs

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work. Assume N represents the number of nodes in the linked list.

• Time Complexity: O(1) because we are simply inserting a new node at the beginning.
• Space Complexity: O(1) because we are only using a constant amount of extra space for the new node.