725. Split Linked List in Parts

Leetcode Challenge

?? Problem Overview:

Imagine you have a long chain of nodes (linked list) and you're tasked with breaking it into equal parts. The challenge? Some parts might be empty, and the length of each part should be as equal as possible. Sounds fun, right? ??

?? Problem:

Given the head of a singly linked list and an integer k, split the list into k consecutive parts. The goal is to make the sizes of these parts as equal as possible.

???? Example 1:

Input:

head = [1, 2, 3], k = 5        

OUTPUT:

[[1], [2], [3], [], []]        

???? 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]]        

?? Step-by-Step Solution Explanation:

1?? Find the Length of the Linked List

We first traverse the linked list to calculate its total length L.

?? Key insight: Knowing the total length helps us divide it evenly among the parts.

ListNode curr = head;
int L = 0;
while ( curr != null)
L++;
curr = curr.next;
}        

2?? Determine Base Size and Remainder

We calculate how many nodes each part should have using eachBucketNodes = L / k and handle any extra nodes using remainderNodes = L % k.

?? Key insight: The first remainderNodes parts will have one extra node.

int eachBucketNodes = L / k;
int remainderNodes = L % k;        

3?? Split the List into K Parts

We loop through the linked list, breaking it into k parts. The first few parts get the extra node if any, while the rest get equal-sized chunks.

ListNode[] result = new ListNode[k];
curr = head;
for (int i = 0; i < k && curr != null; i++) {
    result[i] = curr;
    for (int count = 1; count <= eachBucketNodes + (remainderNodes > 0 ? 1 : 0); count++) {
        prev = curr;
        curr = curr.next;
    }
    if (prev != null) prev.next = null;
    remainderNodes--;
}        

4?? Return the Resulting Parts:

Finally, we return the array of 'ListNode' parts as the solution.

?? Summary:

This approach uses efficient traversal and division to ensure the linked list is split evenly into parts with minimal overhead. It handles edge cases like when k is larger than the length of the list, ensuring some parts remain null.