Chase String - codepath/compsci_guides GitHub Wiki

Unit 5 Session 1 (Click for link to problem statements)

TIP102 Unit 5 Session 1 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Linked Lists, String Manipulation

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 should the function return if the list is empty?
  • The function should return an empty string.

HAPPY CASE
Input: dog -> cat -> mouse -> cheese
Output: "Spike chases Tom chases Jerry chases Gouda"
Explanation: The values of the nodes are concatenated with "chases" as the separator.

EDGE CASE
Input: (empty list)
Output: "
Explanation: The list is empty, so the function returns an empty string.

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:

• Traversal: We need to traverse the list to collect the values of the nodes.
• String Manipulation: We need to join the collected values with a separator.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Traverse the linked list to collect the values of the nodes in a list, then join the values with "chases" as the separator.

1) Initialize an empty list to store the node values.
2) Traverse the linked list from the head to the end.
3) Append each node's value to the list.
4) Join the values in the list with "chases" as the separator.
5) Return the resulting string.

⚠️ Common Mistakes

• Forgetting to handle the case where the list is empty.
• Not correctly traversing the entire list.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value, next=None):
        self.value = value
        self.next = next
        
def chase_list(head):
    current = head
    values = []
    while current:
        values.append(current.value)
        current = current.next
    return " chases ".join(values)

# Example usage
dog = Node("Spike")
cat = Node("Tom")
mouse = Node("Jerry")
cheese = Node("Gouda")

dog.next = cat
cat.next = mouse
mouse.next = cheese

print(chase_list(dog))  # Output: "Spike chases Tom chases Jerry chases Gouda"

5: R-eview

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

• Verify the correctness of the output string by comparing it to the expected result for various linked lists.

Example:

dog = Node("Spike")
cat = Node("Tom")
mouse = Node("Jerry")
cheese = Node("Gouda")

dog.next = cat
cat.next = mouse
mouse.next = cheese

print(chase_list(dog))  # Expected Output: "Spike chases Tom chases Jerry chases Gouda"

6: E-valuate

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

Time Complexity:

• O(N), where N is the number of nodes in the linked list. We traverse the entire list once to collect the values.

Space Complexity:

• O(N), as we store the node values in a list before joining them into a single string. This solution is efficient for converting a linked list into a concatenated string with minimal overhead.