LeetCode 23 Solution: How to Efficiently Merge K Sorted Lists

The LeetCode 23 Solution is a common challenge that involves merging multiple sorted linked lists into a single sorted list. While this task may seem straightforward, finding an efficient solution is crucial, especially when the number of lists and their sizes vary significantly.

In this blog post, we’ll guide you through a clear and effective solution for LeetCode 23 using Python, with a focus on both performance and readability. We’ll also address important edge cases and explain the time and space complexities involved in the approach.

Problem Breakdown: Understanding LeetCode 23

The problem asks us to merge k linked lists, where each list is sorted in ascending order. The goal is to create one sorted linked list by merging all of them. For example, if the input lists are:

lists = [[1,4,5], [1,3,4], [2,6]]

The result should be:

1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6

You may also face edge cases, such as when no lists are provided, or when some lists are empty. For example:

• Input: lists = [] should return [].
• Input: lists = [[]] should also return [].

The challenge lies in merging all these lists efficiently, especially when k (the number of lists) can be as large as 10,000, and the combined size of all lists may reach 10,000 elements.

Approach to Solve LeetCode 23: Using a Min-Heap

To tackle this problem efficiently, we use a min-heap (also called a priority queue). A min-heap allows us to always extract the smallest element among a group of elements, which is exactly what we need to merge multiple sorted lists.

Here’s a step-by-step breakdown of the approach:

1. We start by pushing the first element of each linked list into the heap. The heap will automatically keep the smallest element at the top.
2. We then continuously extract the smallest element from the heap and add it to the merged list. After extracting an element, we push the next element from the same list (if it exists) into the heap.
3. This process continues until the heap is empty, at which point we have successfully merged all the lists into one sorted list.

By using the heap, we ensure that we always process the smallest available element from any of the lists, thus maintaining the overall order in the final merged list.

Python Code Implementation for Merge K Sorted Lists

Here’s how you can implement this solution in Python:

import heapq

class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def mergeKLists(lists):
    if not lists or len(lists) == 0:
        return None

    heap = []

    # Initialize heap with the first node of each list
    for i in range(len(lists)):
        if lists[i]:
            heapq.heappush(heap, (lists[i].val, i, lists[i]))

    dummy = ListNode()
    current = dummy

    while heap:
        val, i, node = heapq.heappop(heap)
        current.next = node
        current = current.next

        if node.next:
            heapq.heappush(heap, (node.next.val, i, node.next))

    return dummy.next

How the Code Works

• First, we check if the input lists is empty. If it is, we return None immediately because there is nothing to merge.
• We then initialize a min-heap and push the first element of each linked list into it. Each element in the heap is stored as a tuple of (value, index, node). This tuple ensures we can track the values and the original list the node came from.
• We create a dummy node that helps build the final merged list. The current pointer moves through the merged list as we add new nodes.
• Inside the loop, we pop the smallest element from the heap and add it to the merged list. If the popped node has a next node, we push that next node into the heap to ensure we keep adding elements from the same list.

Edge Cases and Special Scenarios

Handling edge cases is crucial in coding problems like this. For example:

• Empty Input: If there are no lists or all lists are empty, the function should return an empty list (None in this case).
• Single List: If there is only one non-empty list, the function should simply return that list without further processing.
• Varying List Lengths: If some lists are significantly shorter than others, the algorithm gracefully handles this by continuing to merge from other lists once a shorter list is fully processed.

Time and Space Complexity Analysis

• Time Complexity: The time complexity of our approach is O(n log k), where n is the total number of elements across all k lists. This is because we are performing n heap operations (inserting and removing nodes), and each heap operation takes O(log k) time.
• Space Complexity: The space complexity is O(k) for storing the heap, as at most we have k nodes in the heap at any time. Additionally, we need O(n) space to store the final merged list.

Conclusion: Efficiently Solving LeetCode 23 with a Min-Heap

The LeetCode 23 Solution to merge k sorted lists can be efficiently implemented using a min-heap. This ensures that the time complexity remains manageable even when the number of lists (k) and the total number of elements (n) are large. By following this approach, you can merge the lists in O(n log k) time, which is optimal for this problem.

If you encounter the merge k sorted lists problem in an interview or coding challenge, remember that using a min-heap is the key to solving it efficiently. This method ensures that we maintain the sorted order of the merged list without resorting to slower approaches like brute force concatenation and sorting.

By understanding and applying this approach, you’ll be well-prepared to handle similar problems in the future.

Happy Coding!

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Looking for more LeetCode solutions? Stay tuned for upcoming guides on tackling other LeetCode challenges in Python. Whether you’re preparing for technical interviews or aiming to improve your coding skills, exploring a variety of problems will enhance your understanding and proficiency.

You can find this problem on LeedCode, click here.