Shared Music Taste

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

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 25-35 mins
• 🛠️ Topics: Linked Lists, Intersection, Pointer Manipulation

1: U-nderstand

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

• Q: What does the problem ask for?
  • A: The problem asks to find the intersection node of two linked lists, where the two lists merge.
• Q: What should be returned?
  • A: The function should return the node where the two lists intersect. If there is no intersection, return None.

HAPPY CASE
Input: 
    - playlist_a = Node('Song A', Node('Song B'))
    - playlist_b = Node('Song X', Node('Song Y', Node('Song Z')))
    - shared_segment = Node('Song M', Node('Song N', Node('Song O')))
    - playlist_a.next.next = shared_segment
    - playlist_b.next.next.next = shared_segment
Output: 
    - Song M
Explanation: 
    - Both playlists intersect at the node with value 'Song M'.

EDGE CASE
Input: 
    - playlist_a = Node('Song A')
    - playlist_b = Node('Song X')
Output: 
    - None
Explanation: 
    - There is no intersection between the two playlists.

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 Intersection, we want to consider the following approaches:

• Length Calculation: Calculate the lengths of both linked lists.
• Aligning: Align the start of the longer list with the shorter list.
• Simultaneous Traversal: Traverse both lists in tandem until an intersection is found.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will calculate the lengths of both linked lists, align the start of the longer list with the shorter list, and then traverse both lists in tandem to find the intersection point.

1) Initialize a helper function `get_length` to calculate the length of a linked list.
2) Calculate the lengths of both `playlist_a` and `playlist_b`.
3) Align the start of the longer list with the shorter list by moving the pointer of the longer list ahead by the difference in lengths.
4) Traverse both lists simultaneously until the intersection is found.
5) If an intersection is found, return the intersecting node. If no intersection is found, return `None`.

⚠️ Common Mistakes

• Forgetting to handle cases where the lists do not intersect.
• Incorrectly managing pointers, leading to loss of nodes or incorrect traversal.

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 playlist_overlap(playlist_a, playlist_b):
    # Step 1: Calculate the lengths of both linked lists
    def get_length(head):
        length = 0
        while head:
            length += 1
            head = head.next
        return length

    len_a = get_length(playlist_a)
    len_b = get_length(playlist_b)

    # Step 2: Align the start of the longer list
    while len_a > len_b:
        playlist_a = playlist_a.next
        len_a -= 1
    while len_b > len_a:
        playlist_b = playlist_b.next
        len_b -= 1

    # Step 3: Traverse both lists in tandem
    while playlist_a and playlist_b:
        if playlist_a == playlist_b:
            return playlist_a
        playlist_a = playlist_a.next
        playlist_b = playlist_b.next

    return None

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 playlist_a and playlist_b linked lists with a shared segment to verify that the function correctly identifies the intersection node.

6: E-valuate

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

Assume N and M represent the lengths of playlist_a and playlist_b respectively.

• Time Complexity: O(N + M) because we traverse both linked lists to calculate their lengths and then traverse them again to find the intersection.
• Space Complexity: O(1) because the algorithm uses a constant amount of extra space for pointers.