Introduction

In this blog post, we will explore how to solve the LeetCode problem "725" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given the head of a singly linked list and an integer k, split the linked list into k consecutive linked list parts. The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null. The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later. Return an array of the k parts. Example 1: Input: head = [1,2,3], k = 5 Output: [[1],[2],[3],[],[]] Explanation: The first element output[0] has output[0].val = 1, output[0].next = null. The last element output[4] is null, but its string representation as a ListNode is []. Example 2: Input: head = [1,2,3,4,5,6,7,8,9,10], k = 3 Output: [[1,2,3,4],[5,6,7],[8,9,10]] Explanation: The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts. Constraints: The number of nodes in the list is in the range [0, 1000]. 0 <= Node.val <= 1000 1 <= k <= 50

Explanation

Here's a solution to split a singly linked list into k parts with the specified constraints:

• Calculate Part Sizes: Determine the base size of each part and the number of parts that will have an extra node.
• Iterate and Split: Traverse the linked list, creating new sublists of the calculated sizes and adding them to the result array.

• Handle Remaining Null Parts: If k is greater than the list length, fill the remaining slots in the result array with nulls.

• Time Complexity: O(N + K), where N is the number of nodes in the linked list and K is the number of parts.

• Space Complexity: O(K), mainly for storing the array of K linked list parts.

Code

    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def splitListToParts(head: ListNode, k: int) -> list[ListNode]:
    """
    Splits a singly linked list into k consecutive linked list parts.

    Args:
        head: The head of the singly linked list.
        k: The number of parts to split the list into.

    Returns:
        A list of k linked list parts.
    """

    length = 0
    curr = head
    while curr:
        length += 1
        curr = curr.next

    base_size = length // k
    extra = length % k

    result = []
    curr = head
    for i in range(k):
        part_size = base_size + (1 if i < extra else 0)
        if part_size == 0:
            result.append(None)
            continue

        part_head = curr
        part_tail = None
        for j in range(part_size - 1):
            curr = curr.next
        part_tail = curr
        curr = curr.next

        part_tail.next = None  # Split the list
        result.append(part_head)

    return result