In the realm of algorithms and data structures, the task of merging k sorted linked lists is a familiar challenge. In this blog, we’ll embark on a journey through a naive approach and then pivot to a more optimal solution. Our starting point is a Java solution that might seem intuitive but lacks the efficiency needed for larger datasets.

The Naive Approach: An Overview

Let’s dissect the given Java solution:

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ArrayList<ListNode> pointers = new ArrayList<ListNode>();

        // Assign head of all linked list to 
        for(int i=0;i<lists.length;i++){
            pointers.add(lists[i]);
        }

        ArrayList<ListNode> answer = new ArrayList<ListNode>();
       
        while(true) {
            int min = Integer.MAX_VALUE;
            int selectedIndex = -1;
            for(int i=0;i<lists.length;i++) {
                ListNode listNode = pointers.get(i);
                if (listNode == null) {
                    continue;
                }

                if (min > listNode.val) {
                    min = listNode.val;
                    selectedIndex = i;
                }
            }

            if (min == Integer.MAX_VALUE) {
                break;
            } else {
                answer.add(pointers.get(selectedIndex));
                pointers.set(selectedIndex, pointers.get(selectedIndex).next);
            }
        }
        ListNode answerHead = null;
        if (answer.size() > 0) {
            answerHead = answer.get(0);
        }

        for (int i=1;i<answer.size();i++) {
            answer.get(i-1).next = answer.get(i);
        }
        return answerHead;
    }
}

The Naive Steps:

1. Initialization:
   • Create an ArrayList to store pointers to the heads of the k linked lists.
   • Initialize an ArrayList to store the final merged list.
2. Iterative Selection:
   • Loop until all linked lists are traversed.
   • In each iteration, find the minimum value among the current heads of linked lists.
   • Add the selected node to the merged list.
3. Move Pointers:
   • Move the pointer of the selected list to its next node.
4. Assemble Result:
   • Assemble the merged list by connecting nodes.

Problem Link: https://leetcode.com/problems/merge-k-sorted-lists/

Identifying the Pitfalls

While this solution appears straightforward, it has notable inefficiencies:

• Repeated Linear Search:
  • The algorithm performs a linear search for the minimum value in each iteration, resulting in a time complexity of O(k * n^2), where k is the number of lists and n is the average length of the lists.
• Quadratic Time Complexity:
  • The overall time complexity becomes quadratic, making it suboptimal, especially for larger datasets.

A Glimpse into the Optimal Solution

The optimal solution for merging k sorted lists involves using a divide-and-conquer strategy, specifically employing the Merge Sort algorithm. Which I’m learning and will add that soon.

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if (lists == null || lists.length == 0) {
            return null;
        }
        return mergeSortedLists(lists, 0, lists.length - 1);
    }

    private ListNode mergeSortedLists(ListNode[] lists, int start, int end) {
        if (start >= end) {
            return lists[start];
        }
        int mid = start + (end - start) / 2;
        ListNode left = mergeSortedLists(lists, start, mid);
        ListNode right = mergeSortedLists(lists, mid + 1, end);
        return merge(left, right);
    }

    private ListNode merge(ListNode list1, ListNode list2) {
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;

        while (list1 != null && list2 != null) {
            if (list1.val < list2.val) {
                current.next = list1;
                list1 = list1.next;
            } else {
                current.next = list2;
                list2 = list2.next;
            }
            current = current.next;
        }

        current.next = (list1 != null) ? list1 : list2;

        return dummy.next;
    }
}

Key Intuition Points:

• Efficiency of Merge Sort:
  • Merge Sort ensures a stable and efficient merging process with a time complexity of O(n log k), where n is the total number of nodes across all lists.
• Divide-and-Conquer Strategy:
  • The algorithm divides the problem into smaller sub-problems, making it easier to solve and merge the results.
• Maintaining Sorted Order:
  • The merging process ensures that the overall sorted order is maintained, leveraging the inherent order within the individual linked lists.
• Avoiding Quadratic Complexity:
  • The initial naive approach might involve quadratic time complexity due to repeated linear searches, but this optimized solution minimizes time complexity and enhances efficiency.

In summary, the algorithm optimally utilizes the strengths of the Merge Sort algorithm and the divide-and-conquer strategy to efficiently merge k sorted linked lists, providing a robust solution for this common problem.