What is the time complexity of Merge k Sorted Lists?

The optimal solution has a time complexity of O(Nlogk).

What data structures are used to solve Merge k Sorted Lists?

This problem uses Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Merge k Sorted Lists?

Insight 1: The key is to efficiently find the smallest element among the heads of the k lists at each step. Insight 2: Using a min-heap (priority queue) is an efficient way to track the smallest elements across all lists.

Which companies ask LeetCode 23. Merge k Sorted Lists in interviews?

This problem is frequently asked at Adobe, Airbnb, Amazon, Anduril, Apple and 36 more companies.

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Merge k Sorted Lists - LeetCode 23 Solution

Q: How to solve Merge k Sorted Lists?

The provided solution uses a min-heap (priority queue) to efficiently merge the k sorted linked lists. The algorithm initializes the min-heap with the head nodes of each non-empty linked list. Then, it repeatedly extracts the smallest element from the heap, adds it to the merged list, and inserts the next node from the corresponding list into the heap. A dummy head node is used to simplify the construction of the merged list.

Merge k Sorted Lists - Complete Solution Guide

Merge k Sorted Lists is LeetCode problem 23, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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. 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 are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted linked list: 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 <

Detailed Explanation

The problem asks us to merge k sorted linked lists into one sorted linked list. We are given an array of linked lists, where each linked list is sorted in ascending order. The task is to combine all the elements from these k lists into a single, sorted linked list, and return the head of this merged list. The input can be an empty array of lists, or lists containing empty lists, both of which should return an empty list as the result.

Solution Approach

The provided solution uses a min-heap (priority queue) to efficiently merge the k sorted linked lists. The algorithm initializes the min-heap with the head nodes of each non-empty linked list. Then, it repeatedly extracts the smallest element from the heap, adds it to the merged list, and inserts the next node from the corresponding list into the heap. A dummy head node is used to simplify the construction of the merged list.

Step-by-Step Algorithm

Step 1: Initialize a min-heap (priority queue) to store the head nodes of the k linked lists. Also, use a counter to differentiate between nodes with the same value in the heap and ensure the heap sorts correctly.
Step 2: Iterate through the input array of linked lists. For each non-empty list, add its head node to the min-heap. The min-heap will automatically maintain the smallest element at the top.
Step 3: Create a dummy head node for the merged list. This simplifies the insertion process.
Step 4: While the min-heap is not empty, extract the smallest element (node) from the heap.
Step 5: Append the extracted node to the merged list.
Step 6: If the extracted node has a next node in its original list, add the next node to the min-heap.
Step 7: Repeat steps 4-6 until the min-heap is empty.
Step 8: Return the next node of the dummy head, which is the head of the merged linked list.

Key Insights

Insight 1: The key is to efficiently find the smallest element among the heads of the k lists at each step.
Insight 2: Using a min-heap (priority queue) is an efficient way to track the smallest elements across all lists.
Insight 3: A dummy head node simplifies the process of building the merged linked list.

Complexity Analysis

Time Complexity: O(Nlogk)

Space Complexity: O(k)

Topics

This problem involves: Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort.

Companies

Asked at: Adobe, Airbnb, Amazon, Anduril, Apple, Bloomberg, ByteDance, Cisco, Citadel, Dell, Deloitte, Docusign, DoorDash, Hubspot, IXL, Indeed, LinkedIn, Meta, Microsoft, MongoDB, NetApp, Nutanix, Nvidia, Nykaa, Oracle, Palantir Technologies, Rivian, Salesforce, Samsung, Snowflake, SoFi, TikTok, Two Sigma, Uber, Verkada, Walmart Labs, Warnermedia, X, Yahoo, Yandex, eBay.

Merge k Sorted Lists | LeetCode 23

Hard

Linked List Divide and Conquer Heap (Priority Queue)Merge Sort

Time: O(Nlogk)Space: O(k)

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
import java.util.PriorityQueue;

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNodeWrapper> minHeap = new PriorityQueue<>((a, b) -> a.val - b.val);
        int counter = 0;
        
        for (ListNode head : lists) {
            if (head != null) {
                minHeap.offer(new ListNodeWrapper(head.val, counter, head));
                counter++;
            }
        }
        
        ListNode dummyHead = new ListNode();
        ListNode current = dummyHead;
        
        while (!minHeap.isEmpty()) {
            ListNodeWrapper wrapper = minHeap.poll();
            int val = wrapper.val;
            ListNode node = wrapper.node;
            
            current.next = node;
            current = current.next;
            
            if (node.next != null) {
                minHeap.offer(new ListNodeWrapper(node.next.val, counter, node.next));
                counter++;
            }
        }
        
        return dummyHead.next;
    }

    private static class ListNodeWrapper {
        int val;
        int counter;
        ListNode node;

        public ListNodeWrapper(int val, int counter, ListNode node) {
            this.val = val;
            this.counter = counter;
            this.node = node;
        }
    }
}

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Problem Info

Description

Problem 23: Merge k Sorted Lists - Detailed explanation and solution approaches.

Topics

Linked List Divide and Conquer Heap (Priority Queue)Merge Sort

Companies

AdobeAirbnbAmazonAndurilAppleBloombergByteDanceCisco

Stats

Difficulty:Hard

Languages:4

Companies:41

Problem #:23

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize a min-heap (priority queue) to store the head nodes of the k linked lists. Also, use a counter to differentiate between nodes with the same value in the heap and ensure the heap sorts correctly.
2
Step 2: Iterate through the input array of linked lists. For each non-empty list, add its head node to the min-heap. The min-heap will automatically maintain the smallest element at the top.
3
Step 3: Create a dummy head node for the merged list. This simplifies the insertion process.
4
Step 4: While the min-heap is not empty, extract the smallest element (node) from the heap.
5
Step 5: Append the extracted node to the merged list.
6
Step 6: If the extracted node has a next node in its original list, add the next node to the min-heap.
7
Step 7: Repeat steps 4-6 until the min-heap is empty.
8
Step 8: Return the next node of the dummy head, which is the head of the merged linked list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(Nlogk)

💾

Space Complexity

O(k)

📝

Why?

The time complexity is determined by the heap operations. Each node goes into the heap once (logk) and out once (logk), and there are N nodes total. The space complexity is dominated by the size of the heap, which can be at most k (one node from each list).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Merge Two Lists at a Time

Repeatedly merge pairs of lists until only one list remains. This is similar to merge sort.

Time: O(kN), where k is the number of lists and N is the average length of lists. Each merge of two lists takes O(N) time, and we repeat this k-1 times in the worst case.Space: O(1) or O(N) depending on implementation. An in-place merge would use O(1), but merging into a new list might use O(N).

Trade-offs: This approach is simpler to implement but less efficient than the min-heap approach when k is large. The time complexity degrades linearly with k, whereas the min-heap approach's time complexity degrades logarithmically with k.

Divide and Conquer (Merge Sort)

Recursively divide the list of lists into two halves, merge each half, and then merge the two resulting sorted lists.

Time: O(N log k), where N is the total number of nodes and k is the number of lists. Similar to the heap-based solution.Space: O(log k) due to the recursive call stack, or O(1) for iterative implementation, plus O(N) to create the merged list during each merge operation.

Trade-offs: Has similar time complexity to the min-heap, potentially greater constant factors due to recursion overhead. An iterative divide and conquer with a merge sort approach may prove valuable.

✨

Key Insights

1Insight 1: The key is to efficiently find the smallest element among the heads of the k lists at each step.
2Insight 2: Using a min-heap (priority queue) is an efficient way to track the smallest elements across all lists.
3Insight 3: A dummy head node simplifies the process of building the merged linked list.

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Empty input array of lists (lists is empty). The solution correctly handles this by returning an empty list (NULL in C, None in Python, null in Java).
Edge case 2: An input array containing empty lists (lists contains only empty lists). The solution correctly handles this by adding only the non-empty heads to the heap. If all heads are empty, the heap remains empty and an empty list is returned.
Edge case 3: One or more lists are exhausted before others. The algorithm correctly handles this by checking if a node has a 'next' node before adding it to the heap. If there is no 'next' node, nothing gets added to the heap from that list.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the input array is empty or contains empty lists. This can lead to null pointer exceptions or incorrect results.

✗Common mistake 2: Not properly maintaining the min-heap property. This can lead to incorrect ordering in the merged list.

✗Common mistake 3: Forgetting to advance the pointer in the original linked list after extracting a node. This will result in duplicate nodes in the merged list.

✗Common mistake 4: Using inefficient heap implementation, leading to higher time complexity. Using built-in heap/priority queue data structures in the respective languages avoids this.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Database systems merging sorted result sets from multiple indexes.

🏦Real-world scenario 2: Combining sorted log files from different servers into a single sorted log file for analysis.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 24, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

How does the fast/slow pointer technique work?

Fast pointer moves 2 steps while slow moves 1 step. When fast reaches the end, slow is at middle. If there's a cycle, they will meet. This technique is used for: finding middle node, detecting cycles, finding cycle start point, and checking palindromes.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Merge k Sorted Lists"?

For this hard-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether divide and conquer techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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