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Leetcode #23: Merge k Sorted Lists

In this guide, we solve Leetcode #23 Merge k Sorted Lists in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it.

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort

Intuition

We need to repeatedly access the smallest or largest element as the input changes.

A heap provides fast insertions and removals while keeping order.

Approach

Push candidates into the heap as you scan, and pop when you need the best element.

Keep the heap size bounded if the problem requires a top-k structure.

Steps:

  • Push candidates into a heap.
  • Pop the best candidate when needed.
  • Maintain heap size or invariants.

Example

Input: lists = [[1,4,5],[1,3,4],[2,6]] Output: [1,1,2,3,4,4,5,6] Explanation: The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted linked list: 1->1->2->3->4->4->5->6

Python Solution

# Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: def mergeKLists(self, lists: List[Optional[ListNode]]) -> Optional[ListNode]: setattr(ListNode, "__lt__", lambda a, b: a.val < b.val) pq = [head for head in lists if head] heapify(pq) dummy = cur = ListNode() while pq: node = heappop(pq) if node.next: heappush(pq, node.next) cur.next = node cur = cur.next return dummy.next

Complexity

The time complexity is O(n×log⁡k)O(n \times \log k)O(n×logk), and the space complexity is O(k)O(k)O(k). The space complexity is O(k)O(k)O(k).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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