Split Linked List in Parts - codepath/compsci_guides GitHub Wiki

Unit 12 Session 2 Standard (Click for link to problem statements)

Problem Highlights

  • 💡 Difficulty: Medium
  • ⏰ Time to complete: 25-30 mins
  • 🛠️ Topics: Linked Lists, Splitting, Division of Parts

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?

  • How do we split the linked list into k parts?

    • Each part should be as equal as possible, with the earlier parts being larger if necessary. If k is greater than the number of nodes, the extra parts will be None.

  • What should happen if the input list is shorter than k?

    • The result should contain some None values to make up for the missing parts.
HAPPY CASE
Input: list_1 = [1, 2, 3], k = 5
Output: [[1], [2], [3], None, None]
Explanation: The input is divided into 5 parts, with the extra parts being empty.

Input: list_2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
Output: [1, 2, 3, 4], [5, 6, 7], [8, 9, 10](/codepath/compsci_guides/wiki/1,-2,-3,-4],-[5,-6,-7],-[8,-9,-10)
Explanation: The list is split into 3 parts with size difference at most 1.

EDGE CASE
Input: list = [], k = 3
Output: [None, None, None]
Explanation: The input list is empty, so all parts are None.

Input: list = [1, 2, 3], k = 1
Output: [1, 2, 3](/codepath/compsci_guides/wiki/1,-2,-3)
Explanation: The whole list is returned as one part.

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 Splitting Problems, consider the following approaches:

  • Traversal and Counting: Traverse the list to determine its length.
  • Division Logic: Use division and remainder to decide how many nodes each part gets.
  • Iterative Splitting: Use pointers to split the list into multiple parts.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea:
First, find the total length of the linked list. Use integer division and modulus to determine how many nodes each part will get. If there are leftover nodes, distribute them across the initial parts.

1) Calculate the total length of the linked list.
2) Determine the size of each part: 
   - `part_size = total_length // k`
   - `extra_nodes = total_length % k`
3) Initialize an empty result list to store the parts.
4) Iterate `k` times to create each part:
   a) For each part, assign `part_size + 1` nodes if extra nodes are available.
   b) Terminate each part by setting the `next` pointer to None.
5) If there are fewer nodes than `k`, append `None` for the extra parts.
6) Return the result list.

⚠️ Common Mistakes

  • Forgetting to handle cases where k is greater than the number of nodes.
  • Not correctly terminating the parts with None pointers.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value=0, next=None):
        self.value = value
        self.next = next

def split_list(head, k):
    # Step 1: Find the total number of nodes in the linked list
    total_length = 0
    current = head
    while current:
        total_length += 1
        current = current.next
    
    # Step 2: Determine the size of each part
    part_size = total_length // k
    extra_nodes = total_length % k
    
    result = []
    current = head
    
    # Step 3: Split the list into parts
    for i in range(k):
        part_head = current
        # Determine the size of this part
        current_part_size = part_size + (1 if i < extra_nodes else 0)
        
        # Split the part
        for j in range(current_part_size - 1):
            if current:
                current = current.next
        
        # Terminate the current part and move to the next
        if current:
            next_part = current.next
            current.next = None  # Terminate this part
            current = next_part
        
        # Add the head of this part to the result
        result.append(part_head)
    
    # Step 4: If there are fewer nodes than parts, append None for the empty parts
    while len(result) < k:
        result.append(None)
    
    return result

5: R-eview

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

  • Input: list_1 = [1, 2, 3], k = 5

    • Total length: 3
    • Parts: [[1], [2], [3], None, None]
    • Output: [[1], [2], [3], None, None]

  • Input: list_2 = [1, 2, ..., 10], k = 3

    • Total length: 10
    • Parts: [1, 2, 3, 4], [5, 6, 7], 8, 9, 10
    • Output: [1, 2, 3, 4], [5, 6, 7], 8, 9, 10

6: E-valuate

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

Assume N is the length of the linked list.

  • Time Complexity: O(N) because we traverse the list to find its length and then split it.
  • Space Complexity: O(k) for storing the result list with k parts.