<p>Given the <code>head</code> of a singly linked list and an integer <code>k</code>, split the linked list into <code>k</code> consecutive linked list parts.</p>

<p>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.</p>

<p>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.</p>

<p>Return <em>an array of the </em><code>k</code><em> parts</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/split1-lc.jpg" style="width: 400px; height: 134px;" />
<pre>
<strong>Input:</strong> head = [1,2,3], k = 5
<strong>Output:</strong> [[1],[2],[3],[],[]]
<strong>Explanation:</strong>
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 [].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/split2-lc.jpg" style="width: 600px; height: 60px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,4,5,6,7,8,9,10], k = 3
<strong>Output:</strong> [[1,2,3,4],[5,6,7],[8,9,10]]
<strong>Explanation:</strong>
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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 1000]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
	<li><code>1 &lt;= k &lt;= 50</code></li>
</ul>

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def splitListToParts(self, head: Optional[ListNode], k: int) -> List[Optional[ListNode]]:
        n = 0

        # calculating the length of the ll
        curr = head
        while curr:
            n += 1
            curr = curr.next

        # calculating distribution of nodes in groupt and,
        # adding nodes in gropus simultaneously

        avg_distribution = n // k
        diff = n - (avg_distribution * k)
        prev = head
        res = [None] * k

        
        for i in range(k):
            
            group_size = avg_distribution
            if diff:
                diff -= 1
                group_size += 1

            if group_size == 0:
                break

            curr = prev
            res[i] = curr

            for i in range(max(group_size - 1, 0)):
                curr = curr.next

            prev = curr.next
            curr.next = None

        return res
16700