Detect if a Linked List is Circular in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start traversing the linked list from the head node.

DSA Javascript

function isCircular(head) {
  let current = [1];
  while (current !== null) {
    current = current.next;
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Ahead

Bnull

Chead.next

Dcurrent

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning null instead of head.

Assigning head.next which skips the first node.

We start traversing from the head node, so current should be assigned head.

2fill in blank

medium

Complete the condition to detect if the linked list is circular by checking if the next node is the head.

DSA Javascript

function isCircular(head) {
  let current = head;
  while (current !== null) {
    if (current.next === [1]) {
      return true;
    }
    current = current.next;
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Anull

Bhead

Ccurrent

Dcurrent.next

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if next is null instead of head.

Comparing with current or current.next incorrectly.

If the next node points back to the head, the list is circular.

3fill in blank

hard

Fix the error in the loop condition to prevent infinite loops when the list is circular.

DSA Javascript

function isCircular(head) {
  let current = head;
  while (current !== [1]) {
    if (current.next === head) {
      return true;
    }
    current = current.next;
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Acurrent

Bhead

Cnull

Dundefined

Attempts:

3 left

💡 Hint

Common Mistakes

Using head in the condition causes infinite loop for circular lists.

Using current or undefined is incorrect for loop termination.

The loop should continue until current becomes null, which means the list ends and is not circular.

4fill in blank

hard

Fill both blanks to correctly detect a circular linked list using a fast and slow pointer approach.

DSA Javascript

function isCircular(head) {
  let slow = head;
  let fast = head;
  while (fast !== null && fast.next !== null) {
    slow = slow.[1];
    fast = fast.[2].next;
    if (slow === fast) {
      return true;
    }
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Anext

Bprev

Cnext.next

Dprev.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'prev' which does not exist in singly linked lists.

Using 'next.next' as a property name which is invalid.

Slow pointer moves one step (next) and fast pointer moves two steps (next.next), but since fast is updated as fast = fast.next.next, the code uses fast = fast.next.next. Here, the blanks are for the property names, so both should be next.

5fill in blank

hard

Fill all three blanks to create a function that detects circular linked list using a Set to track visited nodes.

DSA Javascript

function isCircular(head) {
  const visited = new Set();
  let current = head;
  while (current !== null) {
    if (visited.[1](current)) {
      return true;
    }
    visited.[2](current);
    current = current.[3];
  }
  return false;
}

Drag options to blanks, or click blank then click option'

Ahas

Badd

Cnext

Ddelete

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'delete' instead of 'add' to add nodes.

Using wrong property instead of 'next' to move forward.

We check if the current node is already in the set using has. If yes, the list is circular. Otherwise, add the node to the set with add and move to the next node using next.

Practice

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

easy

A. The linked list has only one node

B. The linked list has no nodes

C. The linked list nodes are sorted in ascending order

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

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 D

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 list
• Confusing empty list with circular list
• Assuming single node list is always circular

2. Which of the following is the correct way to check if a linked list is circular in JavaScript?

easy

A. Count nodes and compare with array length

B. Use two pointers moving at different speeds and check if they meet

C. Check if head is null only

D. Sort the linked list nodes

Solution

1. Step 1: Understand Floyd's cycle detection

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

2. Step 2: Eliminate wrong options

   Checking head null only doesn't detect loops. Counting nodes or sorting is unrelated.

3. Final Answer:

   Use two pointers moving at different speeds and check if they meet -> Option B

4. Quick Check:

   Two-pointer cycle detection = correct method [OK]

Hint: Two pointers meeting means circular list [OK]

Common Mistakes:

• Only checking if head is null
• Trying to sort linked list to detect cycle
• Counting nodes without loop detection

3. What will be the output of the following code snippet?

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

const head = new Node(1);
const second = new Node(2);
const third = new Node(3);
head.next = second;
second.next = third;
third.next = second; // creates a loop

function isCircular(head) {
  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. false

B. undefined

C. true

D. TypeError

Solution

1. Step 1: Analyze linked list structure

   Nodes 1 -> 2 -> 3, and third node points back to second, creating a loop.

2. Step 2: Trace two-pointer loop detection

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

3. Final Answer:

   true -> Option C

4. Quick Check:

   Loop detected = true [OK]

Hint: Loop causes pointers to meet, returning true [OK]

Common Mistakes:

• Assuming no loop because list looks linear
• Confusing pointer equality check
• Expecting false without tracing pointers

4. Find the error in this code that tries to detect a circular linked list:

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

medium

A. The while condition should check if fast and fast.next exist

B. The comparison should use === instead of ==

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

D. The function should return false inside the loop

Solution

1. Step 1: Check while loop condition safety

   fast.next.next is accessed without confirming fast is not null, can cause error if fast is null.

2. Step 2: Correct condition to avoid runtime error

   Condition should be while (fast && fast.next) to safely access fast.next.next.

3. Final Answer:

   The while condition should check if fast and fast.next exist -> Option A

4. Quick Check:

   Check fast and fast.next before accessing next nodes [OK]

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

Common Mistakes:

• Not checking if fast is null before accessing fast.next.next
• Using == instead of === (not critical here)
• Moving slow pointer two steps incorrectly

5. Given a linked list where the last node points back to the head, how can you modify the detection function to also return the node where the loop starts?

function detectLoopStart(head) {
  let slow = head;
  let fast = head;
  while (fast && fast.next) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow === fast) {
      // Loop detected
      slow = head;
      while (slow !== fast) {
        slow = slow.next;
        fast = fast.next;
      }
      return slow;
    }
  }
  return null;
}

hard

A. The function returns the node where the loop starts by resetting slow to head and moving both pointers one step

B. The function returns true if a loop exists but does not find the start node

C. The function incorrectly moves fast pointer one step after detection

D. The function will always return null because slow and fast never meet

Solution

1. Step 1: Detect loop using two pointers

   Slow and fast pointers meet inside the loop confirming a cycle.

2. Step 2: Find loop start node

   Reset slow to head, move slow and fast one step at a time until they meet again; this node is loop start.

3. Final Answer:

   The function returns the node where the loop starts by resetting slow to head and moving both pointers one step -> Option A

4. Quick Check:

   Reset slow and move both pointers to find loop start [OK]

Hint: Reset slow to head after detection to find loop start [OK]

Common Mistakes:

• Returning true instead of loop start node
• Moving fast pointer incorrectly after detection
• Assuming slow and fast never meet