Detect if a Linked List is Circular in DSA Javascript

Mental Model

A circular linked list loops back to an earlier node instead of ending with null. We want to find if such a loop exists.

Analogy: Imagine walking down a path that either ends or loops back to a previous spot. Detecting a circular list is like noticing if you start walking in circles.

1 -> 2 -> 3 -> 4 -> null

vs

1 -> 2 -> 3 -> 4 ->
          ↑      ↓
          ←------

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> 4 -> points back to node 2 (circular)

Goal: Detect if the list has a cycle (loop) or ends normally

Step 1: Initialize two pointers: slow at 1, fast at 1

[slow->1] -> 2 -> 3 -> 4 -> points back to 2
[fast->1] -> 2 -> 3 -> 4 -> points back to 2

Why: Start both pointers at head to begin cycle detection

Step 2: Move slow by 1 step to 2, fast by 2 steps to 3

1 -> [slow->2] -> 3 -> 4 -> points back to 2
1 -> 2 -> [fast->3] -> 4 -> points back to 2

Why: Slow moves one node, fast moves two nodes to detect cycle faster

Step 3: Move slow by 1 step to 3, fast by 2 steps to 2 (loop back)

1 -> 2 -> [slow->3] -> 4 -> points back to 2
1 -> 2 -> 3 -> 4 -> [fast->2] (loop back)

Why: Fast pointer loops back due to cycle, slow moves forward

Step 4: Move slow by 1 step to 4, fast by 2 steps to 4

1 -> 2 -> 3 -> [slow->4] -> points back to 2
1 -> 2 -> 3 -> [fast->4] -> points back to 2

Why: Both pointers meet at node 4, confirming cycle

Result:

Cycle detected because slow and fast pointers met at node 4

Annotated Code

DSA Javascript

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

function isCircular(head) {
  if (!head) return false; // empty list has no cycle

  let slow = head;
  let fast = head;

  while (fast !== null && fast.next !== null) {
    slow = slow.next; // move slow by 1
    fast = fast.next.next; // move fast by 2

    if (slow === fast) {
      return true; // cycle detected
    }
  }

  return false; // reached end, no cycle
}

// Driver code
const head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);
// Creating a cycle: last node points back to node with value 2
head.next.next.next.next = head.next;

console.log(isCircular(head));

if (!head) return false; // empty list has no cycle

handle empty list edge case

while (fast !== null && fast.next !== null) {

continue loop while fast pointer can move ahead

slow = slow.next; // move slow by 1

advance slow pointer by one node

fast = fast.next.next; // move fast by 2

advance fast pointer by two nodes

if (slow === fast) { return true; }

if pointers meet, cycle exists

return false; // reached end, no cycle

no cycle found if fast pointer reaches null

Output

true

Complexity Analysis

Time: O(n) because each node is visited at most twice by slow and fast pointers

Space: O(1) because only two pointers are used regardless of list size

vs Alternative: Compared to using a hash set to track visited nodes (O(n) space), this uses constant space making it more efficient

Edge Cases

empty list

returns false immediately because no nodes exist

DSA Javascript

if (!head) return false; // empty list has no cycle

single node pointing to itself

detects cycle because slow and fast meet at the single node

DSA Javascript

if (slow === fast) { return true; }

list with no cycle

returns false after fast pointer reaches null

DSA Javascript

while (fast !== null && fast.next !== null)

When to Use This Pattern

When you see a problem asking if a linked list loops back to an earlier node, use the two-pointer (slow and fast) technique to detect cycles efficiently.

Common Mistakes

Mistake: Not checking if fast or fast.next is null before moving fast pointer
Fix: Add condition in while loop: while (fast !== null && fast.next !== null)

Mistake: Comparing slow and fast pointers before moving them
Fix: Move slow and fast pointers first, then check if they meet

Summary

Detects if a linked list has a cycle by using two pointers moving at different speeds.

Use when you need to know if a linked list loops back to an earlier node instead of ending.

If two pointers moving at different speeds meet, a cycle exists.

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