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Unit 6 Session 1 (Click for link to problem statements)

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

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 15 mins
• 🛠️ Topics: Linked Lists, Debugging

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Q: What if the value to be removed appears more than once?
  • A: The function should only remove the first occurrence of the value.

HAPPY CASE
Input: Head = 1 -> 2 -> 3 -> 4, Value to remove = 2
Output: 1 -> 3 -> 4
Explanation: The node with value 2 is removed as expected.

EDGE CASE
Input: Head = 1 -> 1 -> 1 -> 4, Value to remove = 1
Output: 1 -> 1 -> 4
Explanation: Only the first occurrence of 1 is removed.

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.

This problem is a variant of linked list manipulation specifically focused on node deletion:

• Commonly involves checking and updating node references.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Identify where to put print statements to debug the function effectively.

1) Add a print statement at the beginning to print the initial list.
2) Add print statements before and after any conditional or value-changing operation to trace changes.
3) Include a print statement inside the loop to monitor the current and next node values.
4) Fix the bug by updating the removal logic when a match is detected.

⚠️ Common Mistakes

• Not handling the case where the value to be removed is at the head or the tail of the list.
• Incorrect pointer updates that could lead to losing track of the rest of the list.

4: I-mplement

Implement the code to solve the algorithm.

# Possible print statements
def remove_by_value(head, val):
    if not head:
        return None
    if head.value == val:
        return head.next  

    current = head
    while current.next:
	# print(f"Current: {current.value}")
	# print(f"Current Next: {current.next.value}")
        if current.next.value == val:
            current.next = current.next.next  # FIX: change from incorrect "current = current.next.next"
            # print(f"New Current Next: {current.next.value}")
            return head  
        current = current.next
    return head

5: R-eview

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

• Validate the function with several inputs to ensure it handles different cases correctly.

6: E-valuate

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

• Time Complexity: O(n) because we may need to traverse the entire list.
• Space Complexity: O(1) as no extra space is used aside from temporary pointers.