Detect if a Linked List is Circular in DSA Typescript - Build from Scratch

Detect if a Linked List is Circular

📖 Scenario: Imagine you have a chain of connected train cars. Sometimes, the last car connects back to an earlier car, making a loop. We want to check if the train cars form such a loop.

🎯 Goal: You will create a simple linked list and write code to check if it forms a circular loop or not.

📋 What You'll Learn

Create a linked list with nodes containing numbers

Add a variable to help detect circularity

Write a function to check if the linked list is circular

Print the result as true or false

💡 Why This Matters

🌍 Real World

Detecting loops in linked lists is important in computer programs that manage chains of data, like train cars, playlists, or network packets.

💼 Career

Understanding how to detect circular references helps in debugging memory leaks and designing efficient data structures in software development.

1

Create a simple linked list

Create a class called Node with a value property of type number and a next property that can be another Node or null. Then create three nodes with values 1, 2, and 3. Link them so that the first node points to the second, and the second points to the third. Store the first node in a variable called head.

DSA Typescript

// Your code here to create Node class and linked list nodes

Hint

Define the Node class first. Then create three nodes and link them by setting the next property.

2

Add a helper variable for detection

Create a variable called slow and set it to head. Also create a variable called fast and set it to head. These will help us move through the list at different speeds.

DSA Typescript

class Node {
    value: number;
    next: Node | null;
    constructor(value: number) {
        this.value = value;
        this.next = null;
    }
}

const node1 = new Node(1);
const node2 = new Node(2);
const node3 = new Node(3);

node1.next = node2;
node2.next = node3;

const head = node1;

// Create slow and fast pointers

Hint

Set both slow and fast to start at head.

3

Write function to detect circular linked list

Write a function called isCircular that takes head as input. Inside, use a while loop that continues as long as fast and fast.next are not null. Move slow one step forward and fast two steps forward. If at any point slow equals fast, return true. If the loop ends, return false.

DSA Typescript

class Node {
    value: number;
    next: Node | null;
    constructor(value: number) {
        this.value = value;
        this.next = null;
    }
}

const node1 = new Node(1);
const node2 = new Node(2);
const node3 = new Node(3);

node1.next = node2;
node2.next = node3;

const head = node1;

let slow = head;
let fast = head;

// Write isCircular function below

Hint

Use two pointers moving at different speeds. If they meet, the list is circular.

4

Print the result of circular check

Use console.log to print the result of calling isCircular(head).

DSA Typescript

class Node {
    value: number;
    next: Node | null;
    constructor(value: number) {
        this.value = value;
        this.next = null;
    }
}

const node1 = new Node(1);
const node2 = new Node(2);
const node3 = new Node(3);

node1.next = node2;
node2.next = node3;

const head = node1;

let slow = head;
let fast = head;

function isCircular(head: Node | null): boolean {
    let slow = head;
    let fast = head;
    while (fast !== null && fast.next !== null) {
        slow = slow!.next!;
        fast = fast.next.next;
        if (slow === fast) {
            return true;
        }
    }
    return false;
}

// Print the result of isCircular

Hint

Call isCircular(head) and print the result using console.log.

Practice

1. What does it mean if a linked list is circular?

easy

A. The last node points back to one of the previous nodes, forming a loop

B. The linked list has no nodes

C. The linked list nodes are sorted in ascending order

D. The linked list has only one node

Solution

1. Step 1: Understand linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Define circular linked list

   If the last node points back to any previous node, it creates a loop, making the list circular.

3. Final Answer:

   The last node points back to one of the previous nodes, forming a loop -> Option A

4. Quick Check:

   Circular linked list = last node points back [OK]

Hint: Circular means last node links back to earlier node [OK]

Common Mistakes:

• Thinking circular means sorted
• Confusing empty list with circular
• Assuming single node list is always circular

2. Which TypeScript code snippet correctly checks if a linked list is circular using two pointers?

easy

A. for(let node = head; node != null; node = node.next) { if(node.next === head) return true; } return false;

B. let slow = head; while(slow) { slow = slow.next; if(slow === head) return true; } return false;

C. let slow = head; let fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; if(slow === fast) return true; } return false;

D. let current = head; while(current.next) { current = current.next; } return current === null;

Solution

1. Step 1: Understand Floyd's cycle detection

   Use two pointers: slow moves 1 step, fast moves 2 steps. If they meet, list is circular.

2. Step 2: Analyze code snippets

   let slow = head; let fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; if(slow === fast) return true; } return false; correctly implements this logic with proper conditions and pointer moves.

3. Final Answer:

   let slow = head; let fast = head; while(fast && fast.next) { slow = slow.next; fast = fast.next.next; if(slow === fast) return true; } return false; -> Option C

4. Quick Check:

   Two pointers meet means circular [OK]

Hint: Use slow and fast pointers to detect loop [OK]

Common Mistakes:

• Moving fast pointer only one step
• Not checking fast and fast.next for null
• Returning false too early

3. What will be the output of this TypeScript code snippet?

class Node { constructor(public val: number, public next: Node | null = null) {} }
const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = head.next;
function isCircular(head: Node | null): boolean {
  let slow = head;
  let fast = head;
  while (fast && fast.next) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow === fast) return true;
  }
  return false;
}
console.log(isCircular(head));

medium

A. true

B. false

C. undefined

D. Runtime error

Solution

1. Step 1: Understand the linked list structure

   Nodes 1 -> 2 -> 3, then 3 points back to node 2, creating a loop.

2. Step 2: Trace the two-pointer loop detection

   Slow and fast pointers will meet inside the loop, so function returns true.

3. Final Answer:

   true -> Option A

4. Quick Check:

   Loop exists, so output is true [OK]

Hint: Loop in list makes function return true [OK]

Common Mistakes:

• Assuming no loop because list looks linear
• Confusing node references
• Ignoring pointer equality check

4. Identify the bug in this TypeScript function meant to detect a circular linked list:

function isCircular(head: Node | null): boolean {
  let slow = head;
  let fast = head;
  while (fast !== null) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow === fast) return true;
  }
  return false;
}

medium

A. The function should use a for loop instead of while

B. The slow pointer should move two steps instead of one

C. The function should return false inside the loop

D. The while loop should check fast and fast.next for null before moving pointers

Solution

1. Step 1: Analyze pointer movement safety

   fast.next.next is accessed without checking if fast.next is null, risking runtime error.

2. Step 2: Correct loop condition

   Loop should check both fast and fast.next are not null before moving pointers to avoid errors.

3. Final Answer:

   The while loop should check fast and fast.next for null before moving pointers -> Option D

4. Quick Check:

   Check fast and fast.next to avoid errors [OK]

Hint: Always check fast and fast.next before advancing fast [OK]

Common Mistakes:

• Not checking fast.next before fast.next.next
• Moving slow pointer incorrectly
• Returning false too early

5. Given a singly linked list where the last node might point to any node in the list or null, how would you modify the detection function to also return the node where the cycle begins if it is circular?
Which approach correctly finds the start of the loop?

hard

A. Count the number of nodes and return the middle node as the start of the loop

B. After detecting the cycle, reset one pointer to head and move both pointers one step at a time until they meet; the meeting point is the start of the loop

C. Use a hash set to store visited nodes and return the first repeated node found

D. Reverse the linked list and check if the head changes to detect the loop start

Solution

1. Step 1: Detect cycle using two pointers

   Use Floyd's algorithm to find if slow and fast pointers meet inside the loop.

2. Step 2: Find loop start node

   Reset one pointer to head, move both pointers one step at a time; their meeting point is the loop start.

3. Final Answer:

   After detecting the cycle, reset one pointer to head and move both pointers one step at a time until they meet; the meeting point is the start of the loop -> Option B

4. Quick Check:

   Reset pointer and move both to find loop start [OK]

Hint: Reset pointer after cycle detection to find loop start [OK]

Common Mistakes:

• Assuming middle node is loop start
• Using node count instead of pointers
• Trying to reverse list to find loop