Segmenting Protein Chains for Analysis - codepath/compsci_guides GitHub Wiki

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

Problem Highlights

  • 💡 Difficulty: Medium
  • ⏰ Time to complete: 30-40 mins
  • 🛠️ Topics: Linked Lists, Segmentation

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 split a linked list into k consecutive segments where the segments are as equal in length as possible, with earlier segments being greater than or equal in size compared to later segments.
  • Q: How should the segments be structured?
    • A: If the linked list cannot be evenly divided, some segments should be smaller, and any remaining segments should be empty.
HAPPY CASE
Input: protein1 = Node('Ala', Node('Gly', Node('Leu', Node('Val', Node('Pro', Node('Ser', Node('Thr', Node('Cys')))))))), k = 3
Output:
Ala -> Gly -> Leu
Val -> Pro -> Ser
Thr -> Cys
Explanation: The linked list has been split into three segments with sizes 3, 3, and 2.

EDGE CASE
Input: protein2 = Node('Ala', Node('Gly', Node('Leu', Node('Val')))), k = 5
Output:
Ala
Gly
Leu
Val
Empty List
Explanation: Since `k` is greater than the length of the list, some segments will be empty.

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 Segmentation, we want to consider the following approaches:

  • Traversal and Segmentation: Traverse the list, splitting it into segments based on calculated segment sizes.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will traverse the linked list and split it into k segments. The sizes of the segments are calculated such that the first few segments are larger if the list cannot be evenly divided.

1) Calculate the total length of the linked list.
2) Determine the base size of each segment and the number of larger segments.
3) Traverse the list, creating each segment by linking the nodes accordingly.
4) Append each segment to the output list.
5) If there are fewer nodes than `k`, the remaining segments will be empty.
6) Return the list of `k` segments.

⚠️ Common Mistakes

  • Incorrectly calculating the segment sizes, leading to uneven or misaligned segments.
  • Failing to handle cases where k is greater than the length of the list, resulting in empty segments.

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 split the protein chain into k segments
def split_protein_chain(protein, k):
    # Step 1: Calculate the total length of the list
    total_length = 0
    current = protein
    while current:
        total_length += 1
        current = current.next

    # Step 2: Determine the size of each segment
    base_size = total_length // k
    larger_segments = total_length % k  # Segments that will be one node larger

    parts = []
    current = protein

    for i in range(k):
        head = current
        segment_size = base_size + (1 if i < larger_segments else 0)

        for j in range(segment_size - 1):
            if current:
                current = current.next
        
        if current:
            next_segment = current.next
            current.next = None
            current = next_segment

        parts.append(head)
    
    return parts

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 protein1 and protein2 linked lists to verify that the function correctly segments the list into the desired number of parts.

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 and k represents the number of segments.

  • Time Complexity: O(N) because each node is visited exactly once.
  • Space Complexity: O(k) because we store k segments in the output list.