Detect if a Linked List is Circular in DSA Javascript - Execution Trace

Concept Flow - Detect if a Linked List is Circular

Start at head

↓

Set slow = head, fast = head

↓

Loop: fast and fast.next exist?

↓

Move slow by 1

↓

Move fast by 2

↓

Check if slow == fast

↓

Circular detected

↓

Continue loop

↓

If loop ends

→No circular, return false

Start with two pointers moving at different speeds; if they meet, the list is circular; if fast pointer reaches end, it's not.

Execution Sample

DSA Javascript

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;
}

Checks if a linked list has a cycle by moving two pointers at different speeds and seeing if they meet.

Execution Table

Step	Operation	slow Pointer Node	fast Pointer Node	Pointer Changes	Visual State
1	Initialize slow and fast at head	Node 1	Node 1	slow = head, fast = head	1 -> 2 -> 3 -> 4 -> 5 -> null
2	Move slow by 1, fast by 2	Node 2	Node 3	slow = slow.next, fast = fast.next.next	1 -> 2 -> 3 -> 4 -> 5 -> null
3	Move slow by 1, fast by 2	Node 3	Node 5	slow = slow.next, fast = fast.next.next	1 -> 2 -> 3 -> 4 -> 5 -> null
4	Check loop condition (fast && fast.next)	Node 3	Node 5	Condition false (fast.next === null), loop ends	1 -> 2 -> 3 -> 4 -> 5 -> null
5	Loop ends, no cycle found	Node 3	Node 5	Return false	1 -> 2 -> 3 -> 4 -> 5 -> null

💡 fast.next === null (fast at Node 5), so no cycle detected

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
slow	Node 1	Node 2	Node 3	Node 3	Node 3
fast	Node 1	Node 3	Node 5	Node 5	Node 5

Key Moments - 3 Insights

Why do we move fast pointer by two steps and slow pointer by one?

Moving fast pointer twice as fast ensures that if there is a cycle, fast will eventually catch slow inside the loop, as shown in steps 2 and 3 of the execution_table.

What happens if the list is not circular?

The fast pointer will reach the last node (fast.next === null) first, causing the loop to stop and return false, as seen in step 4 and 5 where fast is at Node 5.

Why do we check 'fast && fast.next' in the loop condition?

To avoid errors when accessing fast.next.next, we must ensure fast and fast.next exist; this prevents trying to access properties of null, as shown before step 4.

Visual Quiz - 3 Questions

Look at the execution_table, what node is the slow pointer at after step 3?

A Node 2

B Node 3

C Node 4

D Node 5

💡 Hint

Check the 'slow Pointer Node' column at step 3 in the execution_table.

At which step does the loop end because fast.next is null?

A Step 4

B Step 3

C Step 2

D Step 5

💡 Hint

Look at the 'Pointer Changes' column and see when the condition fails.

If the linked list had a cycle, what would happen in the execution_table?

A slow and fast pointers would never meet

B fast pointer would reach null

C slow and fast pointers would meet at some node

D slow pointer would reach null first

💡 Hint

Refer to the concept_flow where meeting of slow and fast pointers indicates a cycle.

Concept Snapshot

Detect Circular Linked List:
- Use two pointers: slow (1 step), fast (2 steps)
- Move pointers until fast or fast.next is null (no cycle)
- If slow == fast at any point, cycle exists
- Return true if cycle found, else false
- Efficient O(n) time, O(1) space

Full Transcript

To detect if a linked list is circular, we use two pointers starting at the head. The slow pointer moves one node at a time, while the fast pointer moves two nodes at a time. If the list has a cycle, the fast pointer will eventually catch up to the slow pointer inside the loop. If the list is not circular, the fast pointer will reach the end (null) first. We check for fast and fast.next to avoid errors when moving the fast pointer two steps. This method efficiently detects cycles in linear time and constant space.

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