Easy & less Complex... Linked list Solution..!
Intuition\nEasy & less Complex... Linked list Solution..!\n\n# Approach\nExplanation: \n1. Find the length: Traverse the list to calculate its length.\n\n2. Det
# Intuition\nEasy & less Complex... Linked list Solution..!\n\n# Approach\nExplanation: \n1. Find the length: Traverse the list to calculate its length.\n\n2. Determine part sizes: Compute the minimum size of each part (partSize = length / k) and distribute the remainder across the first few parts.\n\n3. Split the list: For each part, traverse currentPartSize nodes, cut off the rest of the list by setting the next pointer of the last node of the current part to null, and store the head of each part in the result array.\n4. Handle null parts: If the list has fewer nodes than k, some parts will be null.\n\n# Complexity\n- Time complexity: O(N) + O(N) = O(N)\n\n- Space complexity: List of linked List O(N)\n\n# Code\n```java []\nclass Solution {\n    public ListNode[] splitListToParts(ListNode head, int k) {\n        int length = 0;\n        ListNode current = head;\n        while(current != null){\n            length++;\n            current = current.next;\n        }\n        ListNode [] result = new ListNode[k];\n        current = head;\n        for(int i = 0;i<k;i++){\n            ListNode partHead = current;\n            ListNode prev = null;\n            int currentPartSize = (length / k) + (i < (length % k) ? 1 : 0);\n            for(int j = 0;j<currentPartSize;j++){\n                prev = current;\n                current = current.next;\n            }\n            if(prev != null){\n                prev.next = null;\n            }\n            result[i] = partHead;\n        }\n        return result;\n    }\n}\n```
['Java']