Detect if a Linked List is Circular in DSA Typescript - Time & Space Complexity

Time Complexity: Detect if a Linked List is Circular

O(n)

Understanding Time Complexity

We want to know how long it takes to check if a linked list loops back on itself.

How does the time needed grow as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code uses two pointers moving at different speeds to detect a loop in the linked list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves two pointers through the list.
• How many times: At most, it runs once per node until pointers meet or list ends.

How Execution Grows With Input

As the list grows, the number of steps grows roughly in a straight line with the number of nodes.

Input Size (n)	Approx. Operations
10	Up to 10 steps
100	Up to 100 steps
1000	Up to 1000 steps

Pattern observation: The time grows linearly as the list gets longer.

Final Time Complexity

Time Complexity: O(n)

This means the time to detect a loop grows in direct proportion to the number of nodes in the list.

Common Mistake

[X] Wrong: "The fast pointer moves twice as fast, so the time is half the list length."

[OK] Correct: Even though the fast pointer moves faster, the slow pointer still moves step by step, so the total steps depend on the list size, not just half.

Interview Connect

Understanding this method shows you can analyze pointer movement and loops, a key skill for many linked list problems.

Self-Check

"What if we used only one pointer moving one step at a time? How would the time complexity change?"

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