Maximum Protein Pair Stability - codepath/compsci_guides GitHub Wiki

TIP102 Unit 6 Session 1 Advanced (Click for link to problem statements)

Problem Highlights

  • 💡 Difficulty: Medium
  • ⏰ Time to complete: 20-30 mins
  • 🛠️ Topics: Linked Lists, Two Pointers, Reversal

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 determine the maximum "twin stability sum" in a linked list, where the "twin" of a node at index i is the node at index (n-1-i).
  • Q: What approach can be used?
    • A: Use the slow and fast pointer technique to find the middle of the list, reverse the second half, and then compare node values to find the maximum twin sum.
HAPPY CASE
Input: head1 = Node(5, Node(4, Node(2, Node(1))))
Output: 6
Explanation: The twin sums are calculated as:
- Node 0 + Node 3: 5 + 1 = 6
- Node 1 + Node 2: 4 + 2 = 6
The maximum twin sum is 6.

HAPPY CASE
Input: head2 = Node(4, Node(2, Node(2, Node(3))))
Output: 7
Explanation: The twin sums are calculated as:
- Node 0 + Node 3: 4 + 3 = 7
- Node 1 + Node 2: 2 + 2 = 4
The maximum twin sum is 7.

EDGE CASE
Input: head = None
Output: 0
Explanation: An empty list has no nodes, so the twin sum is 0.

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 Twin Sum Calculation, we want to consider the following approaches:

  • Two Pointers (Slow and Fast Pointer Technique): Use two pointers to find the middle of the list.
  • Reversal: Reverse the second half of the list to easily calculate the twin sums.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will use the slow and fast pointer technique to find the middle of the list. Then, we will reverse the second half of the list. After that, we will calculate the twin sums by iterating over the first half and the reversed second half, keeping track of the maximum twin sum.

1) Use slow and fast pointers to find the middle of the linked list.
2) Reverse the second half of the linked list.
3) Initialize a variable `max_stability` to keep track of the maximum twin sum.
4) Traverse the first half and the reversed second half of the list:
    a) Calculate the twin sum for each pair of nodes.
    b) Update `max_stability` with the maximum twin sum found.
5) Return the maximum twin sum.

⚠️ Common Mistakes

  • Forgetting to handle edge cases, such as an empty list.
  • Incorrectly reversing the second half of the list, leading to errors in twin sum calculations.

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 reverse a linked list
def reverse_list(head):
    prev = None
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

# Function to find the maximum twin stability sum
def max_protein_pair_stability(head):
    if not head:
        return 0

    # Step 1: Find the middle of the list using slow and fast pointers
    slow, fast = head, head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next

    # Step 2: Reverse the second half of the list
    second_half = reverse_list(slow)

    # Step 3: Calculate the twin sums and find the maximum
    max_stability = 0
    first_half = head
    while second_half:
        twin_sum = first_half.value + second_half.value
        max_stability = max(max_stability, twin_sum)
        first_half = first_half.next
        second_half = second_half.next

    return max_stability

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 head1 and head2 linked lists to verify that the function correctly calculates the maximum twin stability sum.

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 at most twice.
  • Space Complexity: O(1) because the algorithm uses a constant amount of extra space for pointers.