Mastering the Merge: K Sorted Lists Interview Strategies
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Understanding the Problem: Merging K Sorted Lists
The challenge presented is to merge an array of k linked lists, each sorted in ascending order, into a single sorted linked list.
Example Scenarios
Example 1:
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked lists provided are:
1 -> 4 -> 5
1 -> 3 -> 4
2 -> 6
Merging them results in:
1 -> 1 -> 2 -> 3 -> 4 -> 4 -> 5 -> 6
Example 2:
Input: lists = []
Output: []
Example 3:
Input: lists = [[]]
Output: []
Constraints
- k == lists.length
- 0 <= k <= 10^4
- 0 <= lists[i].length <= 500
- -10^4 <= lists[i][j] <= 10^4
- Each lists[i] is sorted in ascending order.
- The sum of lists[i].length will not exceed 10^4.
Methodologies for Solution
Approach 1: Brute Force
To tackle this, one can traverse all the linked lists, gather the values into an array, sort this array, and then create a new sorted linked list from the sorted values.
class Solution(object):
def mergeKLists(self, lists):
""" :type lists: List[ListNode] :rtype: ListNode
"""
self.nodes = []
head = point = ListNode(0)
for l in lists:
while l:
self.nodes.append(l.val)
l = l.next
for x in sorted(self.nodes):
point.next = ListNode(x)
point = point.next
return head.next
Approach 2: Pairwise Comparison
This method involves comparing the heads of all k linked lists to identify the node with the smallest value, which is then added to the final sorted linked list.
Time Complexity: O(n log n)
Space Complexity: O(n)
Optimized Approach 3: Using a Priority Queue
By utilizing a priority queue, this approach enhances the comparison efficiency while keeping the overall logic similar to the previous method.
from Queue import PriorityQueue
class Solution(object):
def mergeKLists(self, lists):
""" :type lists: List[ListNode] :rtype: ListNode
"""
head = point = ListNode(0)
q = PriorityQueue()
for l in lists:
if l:
q.put((l.val, l))while not q.empty():
val, node = q.get()
point.next = ListNode(val)
point = point.next
node = node.next
if node:
q.put((node.val, node))return head.next
C++ and Java Implementations
C++ and Java also provide elegant solutions using similar logic with priority queues, ensuring optimal performance.
C++ Example:
class cmp {
public:
bool operator()(ListNode * l1, ListNode * l2) {
return l1->val > l2->val;}
};
Java Example:
public class Solution {
public ListNode mergeKLists(List<ListNode> lists) {
if (lists == null || lists.size() == 0) return null;
PriorityQueue<ListNode> queue = new PriorityQueue<ListNode>(lists.size(), new Comparator<ListNode>() {
@Override
public int compare(ListNode o1, ListNode o2) {
return Integer.compare(o1.val, o2.val);}
});
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
for (ListNode node : lists) {
if (node != null) {
queue.add(node);}
}
while (!queue.isEmpty()) {
tail.next = queue.poll();
tail = tail.next;
if (tail.next != null) {
queue.add(tail.next);}
}
return dummy.next;
}
}
Complexity Analysis
- Time Complexity: O(n log k)
- Space Complexity: O(n)
Stay tuned for more engaging interview questions!
— As a senior software engineer at MANNG, I look forward to sharing more informative content.
Chapter 2: Video Tutorials
Dive deeper into the merging of k sorted lists with these video tutorials:
This video covers a challenging FAANG interview question on merging k sorted linked lists, providing insights into solving Leetcode problem 23.
In this tutorial, learn how to implement the merging of k sorted lists in Python, demonstrating practical coding techniques for the Leetcode problem.