Efficiently Combine Multiple Sorted Linked Lists
Merging k sorted linked lists into a single sorted list is a common coding challenge that tests your understanding of linked lists and efficient algorithms. Let’s break down different approaches, including brute force and optimal solutions, to solve this problem.
Problem Overview
Given k sorted linked lists, merge them into one sorted linked list efficiently while maintaining order.
Brute Force Approach (O(N log N))
A straightforward approach is to collect all values, sort them, and reconstruct the list.
import java.util.*;
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
List<Integer> values = new ArrayList<>();
// Collect all values
for (ListNode list : lists) {
while (list != null) {
values.add(list.val);
list = list.next;
}
}
// Sort the values
Collections.sort(values);
// Reconstruct linked list
ListNode dummy = new ListNode();
ListNode current = dummy;
for (int val : values) {
current.next = new ListNode(val);
current = current.next;
}
return dummy.next;
}
}
🔴 Time Complexity: O(N log N) (sorting dominates).
🟢 Space Complexity: O(N) (stores all values).
Better Approach: Pairwise Merging (O(kN))
Merge two lists at a time, reducing problem size iteratively.
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode merged = null;
for (ListNode list : lists) {
merged = mergeTwoLists(merged, list);
}
return merged;
}
private ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode();
ListNode tail = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
tail.next = l1;
l1 = l1.next;
} else {
tail.next = l2;
l2 = l2.next;
}
tail = tail.next;
}
tail.next = (l1 != null) ? l1 : l2;
return dummy.next;
}
}
🔴 Time Complexity: O(kN) (k merges).
🟢 Space Complexity: O(1) (modifies lists in place).
Best Approach: Min-Heap (Priority Queue) – O(N log k)
Using a priority queue ensures optimal merging.
import java.util.PriorityQueue;
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
PriorityQueue<ListNode> minHeap = new PriorityQueue<>((a, b) -> a.val - b.val);
// Add all non-null lists to the heap
for (ListNode list : lists) {
if (list != null) {
minHeap.add(list);
}
}
ListNode dummy = new ListNode();
ListNode tail = dummy;
while (!minHeap.isEmpty()) {
ListNode smallest = minHeap.poll();
tail.next = smallest;
tail = tail.next;
if (smallest.next != null) {
minHeap.add(smallest.next);
}
}
return dummy.next;
}
}
🟢 Time Complexity: O(N log k) (heap operations).
🟢 Space Complexity: O(k) (heap storage).
Best Practices for Coding Interviews
- Clarify Inputs: Are lists sorted? Can they be empty?
- Handle Edge Cases:
- Empty input array
- Lists with varying lengths
- Single list input
- Understand Tradeoffs: Brute force is easier but inefficient, while min-heap is optimal.
FAQs
Q: Why use a priority queue?
A: It efficiently maintains the smallest node among k lists at all times.
Q: What if lists are unsorted?
A: Sorting beforehand increases complexity—this problem assumes sorted lists.
Q: How does pairwise merging compare to heap merging?
A: Pairwise merging is O(kN), while heap merging is O(N log k), making it much faster for large k.
Final Thoughts
While the brute-force approach is simple, it’s inefficient for large inputs. Using a min-heap provides the best performance and scalability, making it the go-to method in coding interviews.
(Keywords: Merge k sorted lists Java, LeetCode 23 solution, linked list merging, brute force merging, priority queue merge, time complexity analysis, coding interview tips)
Master efficient merging techniques—your next coding interview depends on it! 🚀
