Pokemon Party

JCSU Unit 9 Problem Set 1 (Click for link to problem statements)

Problem Highlights

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

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 is the goal of the problem?
  • Create a linked list with nodes containing the values "Squirtle", "Pidgey", "Rattata", and "Gengar" linked sequentially.
• Are there constraints on input?
  • The Node class is provided, and the problem assumes valid input.

HAPPY CASE Input: node_1 = Node("Squirtle") node_2 = Node("Pidgey") node_3 = Node("Rattata") node_4 = Node("Gengar") Output: node_1.value = "Squirtle" node_1.next.value = "Pidgey" node_2.value = "Pidgey" node_2.next.value = "Rattata" node_3.value = "Rattata" node_3.next.value = "Gengar" node_4.value = "Gengar" node_4.next = None

EDGE CASE Input: Only one node created: Node("Squirtle") Output: node.value = "Squirtle" node.next = None Explanation: Without additional nodes, the linked list contains a single node with no next reference.

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 node creation and linking problems, we want to consider the following approaches:

• Node Creation and Assignment: Instantiate Node objects and use their next property to link them.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea:
Use the Node class to create four nodes and set the next property of each node to point to the subsequent node in the sequence.

Steps:

1. Instantiate node_1 with the value "Squirtle".
2. Instantiate node_2 with the value "Pidgey".
3. Instantiate node_3 with the value "Rattata".
4. Instantiate node_4 with the value "Gengar".
5. Link the nodes sequentially by setting the next property of each node:
   • node_1.next points to node_2.
   • node_2.next points to node_3.
   • node_3.next points to node_4.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value, next=None):
        self.value = value  # Store the value of the node
        self.next = next    # Reference to the next node (defaults to None)

# Create individual nodes
node_1 = Node("Squirtle")  # First node
node_2 = Node("Pidgey")    # Second node
node_3 = Node("Rattata")   # Third node
node_4 = Node("Gengar")    # Fourth node

# Link the nodes to form a linked list
node_1.next = node_2  # Squirtle -> Pidgey
node_2.next = node_3  # Pidgey -> Rattata
node_3.next = node_4  # Rattata -> Gengar
# Gengar -> None (default)

5: R-eview

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

Example 1:

• Input: node_1 = Node("Squirtle"), node_2 = Node("Pidgey"), node_3 = Node("Rattata"), node_4 = Node("Gengar") node_1.next = node_2, node_2.next = node_3, node_3.next = node_4
• Expected Output: node_1.value = "Squirtle", node_1.next.value = "Pidgey" node_2.value = "Pidgey", node_2.next.value = "Rattata" node_3.value = "Rattata", node_3.next.value = "Gengar" node_4.value = "Gengar", node_4.next = None
• Observed Output: node_1.value = "Squirtle", node_1.next.value = "Pidgey" node_2.value = "Pidgey", node_2.next.value = "Rattata" node_3.value = "Rattata", node_3.next.value = "Gengar" node_4.value = "Gengar", node_4.next = None

Example 2:

• Input: node_1 = Node("Squirtle")
• Expected Output: node_1.value = "Squirtle" node_1.next = None
• Observed Output: node_1.value = "Squirtle" node_1.next = None

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume n is the number of nodes created.

• Time Complexity: O(1) for creating and linking each node, as operations are constant-time.
• Space Complexity: O(1) additional space beyond the storage required for the nodes themselves.