Merging Trail Segments

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TIP102 Unit 6 Session 2 Standard (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 25-35 mins
• 🛠️ Topics: Linked Lists, Summing, Traversal

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 does the problem ask for?
  • The problem asks to merge segments of a linked list between temporary markers (0s) by summing their values into a single marker. The final list should not contain any temporary markers.
• What should be returned?
  • The function should return the head of the merged linked list.

HAPPY CASE
Input: trail1 = Node(0, Node(3, Node(1, Node(0, Node(4, Node(5, Node(4, Node(2, Node(0)))))))))
Output: 4 -> 11
Explanation: The modified list contains the sum of the segments between `0`s: 
- 3 + 1 = 4
- 4 + 5 + 2 = 11

EDGE CASE
Input: trail2 = Node(0, Node(1, Node(0, Node(3, Node(0, Node(2, Node(2, Node(0))))))))
Output: 1 -> 3 -> 4
Explanation: The modified list contains the sum of the segments between `0`s:
- 1 = 1
- 3 = 3
- 2 + 2 = 4

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 involving Summing and Segmenting, we want to consider the following approaches:

• Traversal: Traverse the linked list while summing the values of nodes between 0s.
• Pointer Manipulation: Use pointers to build the new list as you traverse and sum the segments.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will traverse the linked list, summing the values between 0s and creating a new node in the result list for each segment's sum. The temporary markers (0s) will be skipped in the final list.

1) Initialize a new temporary head node for the result list.
2) Use a pointer `tail` to build the new list, starting from the temporary head node.
3) Traverse the linked list starting after the first `0` marker.
4) For each segment, sum the values until you encounter another `0`.
5) Create a new node in the result list with the sum, and move the `tail` pointer forward.
6) Continue until the end of the list is reached.
7) Return the node next to the temporary head (the new head of the list).

⚠️ Common Mistakes

• Forgetting to handle cases where there are consecutive 0s, or where a segment sums to 0.
• Incorrectly managing pointers, leading to loss of nodes or incorrect list structure.

4: I-mplement

Implement the code to solve the algorithm.

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

# Function to merge trail segments
def merge_trail(trailhead):
    if not trailhead:
        return None
    
    new_head = Node(0)  # Temp node to simplify list creation
    tail = new_head  # Pointer to the last node in the new list
    current = trailhead.next  # Start after the first 0
    segment_sum = 0
    
    while current:
        if current.value == 0:
            if segment_sum > 0:
                tail.next = Node(segment_sum)
                tail = tail.next
            segment_sum = 0
        else:
            segment_sum += current.value
        current = current.next
    
    return new_head.next

5: R-eview

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

• Example: Use the provided trail1 and trail2 linked lists to verify that the function correctly merges segments by summing the values between 0s.

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(N) because each node is visited exactly once.
• Space Complexity: O(1) because the algorithm uses a constant amount of extra space for pointers, excluding the space required for the new list.