What is the time complexity of Sort List?

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

What data structures are used to solve Sort List?

This problem uses Linked List, Two Pointers, Divide and Conquer, Sorting, Merge Sort. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sort List?

Insight 1: Merge Sort is an efficient sorting algorithm suitable for linked lists because it doesn't require random access, which is inefficient in linked lists. Insight 2: Using the Fast/Slow pointer technique is crucial for efficiently finding the middle of the linked list without knowing its length beforehand.

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Sort List - LeetCode 148 Solution

Q: What is the space complexity of Sort List?

The space complexity is O(log n).

Q: Which companies ask LeetCode 148. Sort List in interviews?

This problem is frequently asked at ByteDance, Lyft, Oracle, Palantir Technologies, TikTok and 1 more companies.

Sort List - Complete Solution Guide

Sort List is LeetCode problem 148, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given the head of a linked list, return the list after sorting it in ascending order . Example 1: Input: head = [4,2,1,3] Output: [1,2,3,4] Example 2: Input: head = [-1,5,3,4,0] Output: [-1,0,3,4,5] Example 3: Input: head = [] Output: [] Constraints: The number of nodes in the list is in the range [0, 5 * 10 4 ] . -10 5 <= Node.val <= 10 5 Follow up: Can you sort the linked list in O(n logn) time and O(1) memory (i.e. constant space)?

Detailed Explanation

The problem asks us to sort a singly linked list in ascending order. The input is the `head` of the linked list, and the output should be the `head` of the sorted linked list. The problem specifies that the number of nodes in the list can range from 0 to 5 * 104, and the value of each node can be between -105 and 105. The follow-up question challenges us to achieve this in O(n log n) time and O(1) space, suggesting the use of Merge Sort.

Solution Approach

The provided solution uses the Merge Sort algorithm to sort the linked list. Merge Sort is a divide-and-conquer algorithm that recursively divides the list into smaller sublists until each sublist contains only one element (which is considered sorted). Then, it repeatedly merges the sublists to produce new sorted sublists until there is only one sorted list remaining. The algorithm consists of three main parts: splitting the list, recursively sorting the halves, and merging the sorted halves.

Step-by-Step Algorithm

Step 1: Base Case: Check if the list is empty or has only one element. If so, it's already sorted, so return the head.
Step 2: Split the List: Use the fast and slow pointer technique to find the middle of the list. The slow pointer moves one step at a time, and the fast pointer moves two steps at a time. When the fast pointer reaches the end, the slow pointer will be at the middle of the list. Disconnect the first half from the second half by setting slow.next to None.
Step 3: Recursive Sorting: Recursively call `sortList` on both the first half (from head to slow) and the second half (from mid to the end).
Step 4: Merge: Merge the two sorted halves using the `merge` function. This function creates a dummy node and iterates through both lists, appending the smaller value to the tail of the dummy node's list. Once one list is exhausted, append the remaining elements of the other list. Finally, return dummy.next, which is the head of the merged sorted list.

Key Insights

Insight 1: Merge Sort is an efficient sorting algorithm suitable for linked lists because it doesn't require random access, which is inefficient in linked lists.
Insight 2: Using the Fast/Slow pointer technique is crucial for efficiently finding the middle of the linked list without knowing its length beforehand.
Insight 3: Recursion is a natural way to implement the divide-and-conquer strategy of Merge Sort.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(log n)

Topics

This problem involves: Linked List, Two Pointers, Divide and Conquer, Sorting, Merge Sort.

Companies

Asked at: ByteDance, Lyft, Oracle, Palantir Technologies, TikTok, Yahoo.

Sort List | LeetCode 148

Medium

Linked List Two Pointers Divide and Conquer Sorting Merge Sort

Time: O(n log n)Space: O(log n)

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; }
 * }
 */
class Solution {
    public ListNode sortList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }

        // Split the list into two halves
        ListNode slow = head;
        ListNode fast = head.next;
        while (fast != null && fast.next != null) {
            slow = slow.next;
            fast = fast.next.next;
        }

        ListNode mid = slow.next;
        slow.next = null;

        // Recursively sort both halves
        ListNode left = sortList(head);
        ListNode right = sortList(mid);

        // Merge the sorted halves
        return merge(left, right);
    }

    public ListNode merge(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;
        }

        if (l1 != null) {
            tail.next = l1;
        } else if (l2 != null) {
            tail.next = l2;
        }

        return dummy.next;
    }
}

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Problem Info

Description

Problem 148: Sort List - Detailed explanation and solution approaches.

Topics

Linked List Two Pointers Divide and Conquer Sorting Merge Sort

Companies

ByteDanceLyftOraclePalantir TechnologiesTikTokYahoo

Stats

Difficulty:Medium

Languages:4

Companies:6

Problem #:148

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

4 steps

1
Step 1: Base Case: Check if the list is empty or has only one element. If so, it's already sorted, so return the head.
2
Step 2: Split the List: Use the fast and slow pointer technique to find the middle of the list. The slow pointer moves one step at a time, and the fast pointer moves two steps at a time. When the fast pointer reaches the end, the slow pointer will be at the middle of the list. Disconnect the first half from the second half by setting slow.next to None.
3
Step 3: Recursive Sorting: Recursively call `sortList` on both the first half (from head to slow) and the second half (from mid to the end).
4
Step 4: Merge: Merge the two sorted halves using the `merge` function. This function creates a dummy node and iterates through both lists, appending the smaller value to the tail of the dummy node's list. Once one list is exhausted, append the remaining elements of the other list. Finally, return dummy.next, which is the head of the merged sorted 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(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(log n)

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is O(n log n) because the list is divided into halves log n times, and each merge operation takes O(n) time. The space complexity is O(log n) due to the recursive calls. Each recursive call adds a frame to the call stack. In the worst case, the depth of the recursion is log n.

📖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

Convert to Array, Sort, Convert Back

Convert the linked list to an array, sort the array using a built-in sorting algorithm (e.g., Python's `sorted`), and then create a new linked list from the sorted array.

Time: O(n log n)Space: O(n)

Trade-offs: This approach is simpler to implement but requires O(n) extra space to store the array. It is generally less efficient than in-place Merge Sort due to the overhead of array creation and conversion.

Insertion Sort

Iterate through the linked list and insert each node into its correct position in a sorted sublist.

Time: O(n^2)Space: O(1)

Trade-offs: Insertion sort has O(n^2) time complexity, making it inefficient for larger lists. However, it has O(1) space complexity and can be more efficient than Merge Sort for very small lists.

✨

Key Insights

1Insight 1: Merge Sort is an efficient sorting algorithm suitable for linked lists because it doesn't require random access, which is inefficient in linked lists.
2Insight 2: Using the Fast/Slow pointer technique is crucial for efficiently finding the middle of the linked list without knowing its length beforehand.
3Insight 3: Recursion is a natural way to implement the divide-and-conquer strategy of Merge Sort.

📚

Key Terms Explained

For Students

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Linked List

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty List (head is None). The base case `if not head or not head.next:` handles this case by returning None.
Edge case 2: List with only one element. The base case `if not head or not head.next:` handles this case by returning the head.
Edge case 3: List with two elements. The algorithm correctly splits, sorts, and merges.
Edge case 4: List with all elements having the same value. The merge function still correctly merges the two sorted halves.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting the base case for recursion. Without the base case, the recursion will continue infinitely, leading to a stack overflow.

✗Common mistake 2: Incorrectly splitting the list, leading to an infinite loop or incorrect sorting. The Fast/Slow pointer technique needs to be implemented correctly to ensure the list is split in the middle.

✗Common mistake 3: Not handling the remaining elements in the merge step. After one list is exhausted, you must append the remaining elements of the other list to the tail.

✗Common mistake 4: Modifying the `next` pointer of the original list nodes when merging instead of creating a new list with correct ordering. This will make the algorithm destructive.

✅

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: Sorting playlist tracks by title or artist in a music player application.

🏦Real-world scenario 2: Sorting database records represented as a linked list for efficient retrieval.

🎮Real-world scenario 3: Implementing priority queues where elements are stored in a linked list and need to be kept sorted based on priority.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 27, 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 13

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

@TechLeadJul 6

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Sort List"?

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

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