What is the time complexity of Middle of the Linked List?

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

What data structures are used to solve Middle of the Linked List?

This problem uses Linked List, Two Pointers. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Middle of the Linked List?

Insight 1: The 'fast and slow pointer' technique is efficient for solving linked list problems involving finding the middle or detecting cycles. Insight 2: The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end of the list, the slow pointer will be at the middle.

Which companies ask LeetCode 876. Middle of the Linked List in interviews?

This problem is frequently asked at Bosch, Intuit, Qualcomm, Walmart Labs, Zoho.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Middle of the Linked List - LeetCode 876 Solution

Q: How to solve Middle of the Linked List?

The provided solutions all use the "fast and slow pointer" technique (also known as the "tortoise and hare" algorithm). We initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list. The `slow` pointer moves one node at a time, while the `fast` pointer moves two nodes at a time. This means the `fast` pointer traverses the list twice as fast as the `slow` pointer. The loop continues as long as the `fast` pointer and the node after it are not null. When the `fast` pointer reaches the end of the list (or the node before the end if the list has an even number of nodes), the `slow` pointer will be pointing to the middle node (or the second middle node in case of even length).

Middle of the Linked List - Complete Solution Guide

Middle of the Linked List is LeetCode problem 876, a Easy 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 singly linked list, return the middle node of the linked list . If there are two middle nodes, return the second middle node. Example 1: Input: head = [1,2,3,4,5] Output: [3,4,5] Explanation: The middle node of the list is node 3. Example 2: Input: head = [1,2,3,4,5,6] Output: [4,5,6] Explanation: Since the list has two middle nodes with values 3 and 4, we return the second one. Constraints: The number of nodes in the list is in the range [1, 100] . 1 <= Node.val <= 100

Detailed Explanation

The problem asks us to find the middle node of a given singly linked list. If the list has an odd number of nodes, there will be only one middle node. If the list has an even number of nodes, there will be two middle nodes, and we need to return the *second* one. For example, in the list [1, 2, 3, 4, 5], the middle node is 3. In the list [1, 2, 3, 4, 5, 6], the middle nodes are 3 and 4, and we return 4.

Solution Approach

The provided solutions all use the "fast and slow pointer" technique (also known as the "tortoise and hare" algorithm). We initialize two pointers, `slow` and `fast`, both pointing to the head of the linked list. The `slow` pointer moves one node at a time, while the `fast` pointer moves two nodes at a time. This means the `fast` pointer traverses the list twice as fast as the `slow` pointer. The loop continues as long as the `fast` pointer and the node after it are not null. When the `fast` pointer reaches the end of the list (or the node before the end if the list has an even number of nodes), the `slow` pointer will be pointing to the middle node (or the second middle node in case of even length).

Step-by-Step Algorithm

Step 1: Initialize two pointers, `slow` and `fast`, to the head of the linked list.
Step 2: Iterate through the linked list using a `while` loop. The loop continues as long as `fast` is not null and `fast.next` is not null.
Step 3: Inside the loop, move the `slow` pointer one node forward (`slow = slow.next`).
Step 4: Also inside the loop, move the `fast` pointer two nodes forward (`fast = fast.next.next`).
Step 5: After the loop terminates, the `slow` pointer will be pointing to the middle node (or the second middle node if the list has an even length).
Step 6: Return the `slow` pointer.

Key Insights

Insight 1: The 'fast and slow pointer' technique is efficient for solving linked list problems involving finding the middle or detecting cycles.
Insight 2: The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end of the list, the slow pointer will be at the middle.
Insight 3: Properly handling the edge cases where the list is empty or contains only one node is crucial for a robust solution.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List, Two Pointers.

Companies

Asked at: Bosch, Intuit, Qualcomm, Walmart Labs, Zoho.

Middle of the Linked List | LeetCode 876

Easy

Linked List Two Pointers

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

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 middleNode(ListNode head) {
        ListNode slow = head;
        ListNode fast = head;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 876: Middle of the Linked List - Detailed explanation and solution approaches.

Topics

Linked List Two Pointers

Companies

BoschIntuitQualcommWalmart LabsZoho

Stats

Difficulty:Easy

Languages:4

Companies:5

Problem #:876

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

6 steps

1
Step 1: Initialize two pointers, `slow` and `fast`, to the head of the linked list.
2
Step 2: Iterate through the linked list using a `while` loop. The loop continues as long as `fast` is not null and `fast.next` is not null.
3
Step 3: Inside the loop, move the `slow` pointer one node forward (`slow = slow.next`).
4
Step 4: Also inside the loop, move the `fast` pointer two nodes forward (`fast = fast.next.next`).
5
Step 5: After the loop terminates, the `slow` pointer will be pointing to the middle node (or the second middle node if the list has an even length).
6
Step 6: Return the `slow` pointer.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because in the worst-case scenario (even length list), the fast pointer will traverse the entire list, which requires n/2 iterations of the while loop for the slow pointer. The space complexity is O(1) because the algorithm uses a constant amount of extra space, independent of the input size (the length of the linked 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

Calculate Length First

First, iterate through the linked list to calculate its length. Then, iterate again to reach the middle node (length / 2 or (length+1)/2 based on even/odd length).

Time: O(n), where n is the length of the linked list.Space: O(1), constant space.

Trade-offs: This approach requires two passes through the linked list. The 'fast and slow pointer' method only requires one pass, making it slightly more efficient. This approach is conceptually simpler to understand for beginners.

Store Nodes in an Array/List

Iterate through the linked list and store each node in an array or list. Then, access the element at the middle index (length / 2) to retrieve the middle node.

Time: O(n), where n is the length of the linked list.Space: O(n), requires extra space to store the nodes.

Trade-offs: This approach has a space complexity of O(n) because it requires storing all the nodes in an array or list. It's less efficient than the 'fast and slow pointer' method, but may be easier to implement initially. Not ideal for very large linked lists due to memory constraints.

✨

Key Insights

1Insight 1: The 'fast and slow pointer' technique is efficient for solving linked list problems involving finding the middle or detecting cycles.
2Insight 2: The fast pointer moves twice as fast as the slow pointer. When the fast pointer reaches the end of the list, the slow pointer will be at the middle.
3Insight 3: Properly handling the edge cases where the list is empty or contains only one node is crucial for a robust solution.

📚

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.

Edge Cases

Edge case 1: Empty list (head is null). The code correctly handles this because the while loop condition `fast != null && fast.next != null` will immediately evaluate to false, and the `slow` pointer (which is `head`, which is null) will be returned. Thus, a null pointer exception is avoided.
Edge case 2: List with only one node. The while loop condition `fast != null && fast.next != null` will also immediately evaluate to false since `fast.next` will be null. The `slow` pointer (which is `head`) will be returned, which is the correct behavior.
Edge case 3: List with two nodes. The slow pointer moves to the second node in the list, then the fast pointer will be at null, thus the while loop breaks and the second node is returned.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the `while` loop condition. For example, using `fast.next != null && fast != null` will cause a `NullPointerException` if the list has an even number of nodes, because `fast.next.next` will be accessed when `fast.next` is null.

✗Common mistake 2: Forgetting to initialize both `slow` and `fast` pointers to the head of the list. Initializing them to null or some other incorrect value will lead to incorrect results.

✗Common mistake 3: Not considering the edge cases of empty or single-node lists. A robust solution should handle these cases gracefully.

✅

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: Media players use a similar concept to determine the midpoint of a song or video file, allowing for features like skipping to the middle or marking a point for resuming playback.

🏦Real-world scenario 2: In data streaming applications, this approach can be adapted to efficiently identify the median element of a stream of data, providing valuable insights into the distribution of the data.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: November 4, 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

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 13

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 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.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 "Middle of the Linked List"?

For this easy-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.

🤖 Need help with this problem?