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Data Structures & Algorithms May 27, 2020 2 min read

Leetcode 23: Merge K Sorted Lists (Java)

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

  1. Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
  2. Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
  3. Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
  4. Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

Solution:

For this problem, the brute force solution would be passing all the lists and perform Merge Sort, but for the optimal solution we will use minHeap using PriorityQueue data structure.

Note: Please try to solve it yourself before looking into the solution below

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
        
        for(ListNode head: lists){
            while(head!=null){
                minHeap.offer(head.val);
                head = head.next;
            }
        }
        
        ListNode returnNode = new ListNode(-1);
        ListNode head = returnNode;
        
        while(!minHeap.isEmpty()){
            head.next = new ListNode(minHeap.remove());
            head = head.next;
        }
        
        return returnNode.next;
    }
}

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