What is the time complexity of Linked List Cycle?

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

What data structures are used to solve Linked List Cycle?

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

What are the key insights for solving Linked List Cycle?

Insight 1: Using two pointers, one moving twice as fast as the other, is crucial for detecting cycles. If a cycle exists, the fast pointer will eventually lap the slow pointer. Insight 2: The Floyd's Tortoise and Hare algorithm is the most efficient approach for this problem, leveraging the two-pointer technique to achieve O(1) space complexity.

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Linked List Cycle - LeetCode 141 Solution

Q: Which companies ask LeetCode 141. Linked List Cycle in interviews?

This problem is frequently asked at Autodesk, Cisco, DE Shaw, EPAM Systems, Goldman Sachs and 10 more companies.

Linked List Cycle - Complete Solution Guide

Linked List Cycle is LeetCode problem 141, 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 head , the head of a linked list, determine if the linked list has a cycle in it. There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter . Return true if there is a cycle in the linked list . Otherwise, return false . Example 1: Input: head = [3,2,0,-4], pos = 1 Output:

Detailed Explanation

The problem asks you to determine if a given singly linked list contains a cycle (a loop). The input is the head of the linked list. The output is a boolean value: `true` if a cycle exists, and `false` otherwise. The problem doesn't provide the index of the node where the cycle begins (if one exists); you need to detect the cycle's presence without this information. The linked list nodes contain integer values.

Solution Approach

The provided code implements Floyd's Tortoise and Hare algorithm. It uses two pointers, `slow` and `fast`, initialized 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. The algorithm iterates until either the `fast` pointer reaches the end of the list (indicating no cycle) or the `slow` and `fast` pointers meet (indicating a cycle).

Step-by-Step Algorithm

Step 1: Initialize two pointers, `slow` and `fast`, to the head of the linked list.
Step 2: Check for the base case: If the head is null or the next node is null, there is no cycle, return `false`.
Step 3: Move `slow` one step forward and `fast` two steps forward in each iteration.
Step 4: If `slow` and `fast` pointers meet at any point, a cycle is detected, return `true`.
Step 5: If the `fast` pointer reaches the end of the list (becomes null or its next node is null), there is no cycle, return `false`.

Key Insights

Insight 1: Using two pointers, one moving twice as fast as the other, is crucial for detecting cycles. If a cycle exists, the fast pointer will eventually lap the slow pointer.
Insight 2: The Floyd's Tortoise and Hare algorithm is the most efficient approach for this problem, leveraging the two-pointer technique to achieve O(1) space complexity.
Insight 3: Handling edge cases such as an empty list or a list with only one node is essential to avoid null pointer exceptions or incorrect results.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, Linked List, Two Pointers.

Companies

Asked at: Autodesk, Cisco, DE Shaw, EPAM Systems, Goldman Sachs, Intel, Oracle, Qualcomm, SAP, Samsung, Uber, Walmart Labs, Wipro, Yahoo, ZScaler.

Linked List Cycle | LeetCode 141

Easy

Hash Table Linked List Two Pointers

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

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return false;
            }
            slow = slow.next;
            fast = fast.next.next;
        }
        return true;
    }
}

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Problem Info

Description

Problem 141: Linked List Cycle - Detailed explanation and solution approaches.

Topics

Hash Table Linked List Two Pointers

Companies

AutodeskCiscoDE ShawEPAM SystemsGoldman SachsIntelOracleQualcomm

Stats

Difficulty:Easy

Languages:4

Companies:15

Problem #:141

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

5 steps

1
Step 1: Initialize two pointers, `slow` and `fast`, to the head of the linked list.
2
Step 2: Check for the base case: If the head is null or the next node is null, there is no cycle, return `false`.
3
Step 3: Move `slow` one step forward and `fast` two steps forward in each iteration.
4
Step 4: If `slow` and `fast` pointers meet at any point, a cycle is detected, return `true`.
5
Step 5: If the `fast` pointer reaches the end of the list (becomes null or its next node is null), there is no cycle, return `false`.

💡

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 algorithm iterates through the linked list at most twice. In the worst case (no cycle), the fast pointer reaches the end of the list, traversing approximately 2n nodes. The space complexity is O(1) because we only use a constant amount of extra space for the `slow` and `fast` pointers, regardless of the list's size.

📖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

Hash Table Approach

Store each node's address in a hash table. If you encounter a node whose address is already in the table, a cycle is detected.

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

Trade-offs: This approach is less efficient in terms of space complexity compared to Floyd's algorithm. It is simpler to understand but uses extra space proportional to the number of nodes in the list.

✨

Key Insights

1Insight 1: Using two pointers, one moving twice as fast as the other, is crucial for detecting cycles. If a cycle exists, the fast pointer will eventually lap the slow pointer.
2Insight 2: The Floyd's Tortoise and Hare algorithm is the most efficient approach for this problem, leveraging the two-pointer technique to achieve O(1) space complexity.
3Insight 3: Handling edge cases such as an empty list or a list with only one node is essential to avoid null pointer exceptions or incorrect results.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

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 == null). The solution correctly handles this by returning `false` immediately.
Edge case 2: List with only one node (head != null, head.next == null). The solution correctly handles this by returning `false` because a cycle requires at least two nodes.
Edge case 3: Cycle starts at the head. The algorithm will detect the cycle in the first few iterations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge cases of an empty list or a list with only one node. This can lead to null pointer exceptions.

✗Common mistake 2: Incorrectly implementing the two-pointer movement, leading to an incorrect detection of cycles or infinite loops.

✅

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: Detecting deadlocks in operating systems. A deadlock can be represented as a cycle in a graph of processes and resources.

🏦Real-world scenario 2: Finding loops in computer networks. Network routing protocols can utilize cycle detection to prevent routing loops and ensure efficient data transmission.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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 20, 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

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 "Linked List Cycle"?

For this easy-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether linked list techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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