Detect if a Linked List is Circular in DSA Java - 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.

public boolean isCircular(Node head) {
    if (head == null) return false;
    Node slow = head;
    Node 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 checks if a linked list has a cycle by moving two pointers at different speeds.

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 gets longer, the number of steps grows roughly with the number of nodes.

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

Pattern observation: The steps increase linearly as the list size grows.

Final Time Complexity

Time Complexity: O(n)

This means the time to detect a cycle grows in direct proportion to the list length.

Common Mistake

[X] Wrong: "The fast pointer moves twice as fast, so the time is O(n/2), which is constant time."

[OK] Correct: Even though the fast pointer moves faster, the time still grows with the number of nodes, so it is linear, not constant.

Interview Connect

Understanding this helps you explain how to efficiently find cycles, a common problem in coding interviews.

Self-Check

"What if we used only one pointer and checked visited nodes with extra memory? How would the time complexity change?"

Practice

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

easy

A. The linked list nodes are sorted in ascending order

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

C. The linked list has no nodes

D. The linked list contains 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 B

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 of the following Java code snippets correctly checks if a linked list is circular using two pointers?

easy

A. while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; }

B. while (slow != null) { slow = slow.next; if (slow == fast) return true; }

C. for (int i = 0; i < size; i++) { if (node == null) return false; node = node.next; }

D. if (head == null) return true; else return false;

Solution

1. Step 1: Identify Floyd's cycle detection logic

   Two pointers move at different speeds: slow moves one step, fast moves two steps.

2. Step 2: Check for meeting point

   If slow and fast pointers meet, the list is circular; otherwise, if fast reaches null, it's not circular.

3. Final Answer:

   while (fast != null && fast.next != null) { slow = slow.next; fast = fast.next.next; if (slow == fast) return true; } -> Option A

4. Quick Check:

   Two pointers meet = circular [OK]

Hint: Use two pointers moving at different speeds [OK]

Common Mistakes:

• Not checking fast.next for null
• Using only one pointer to detect loop
• Returning true when list is empty

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

class Node { int data; Node next; Node(int d) { data = d; next = null; } }
public class Test {
  public static boolean isCircular(Node head) {
    if (head == null) return false;
    Node slow = head, fast = head.next;
    while (fast != null && fast.next != null) {
      if (slow == fast) return true;
      slow = slow.next;
      fast = fast.next.next;
    }
    return false;
  }
  public static void main(String[] args) {
    Node head = new Node(1);
    head.next = new Node(2);
    head.next.next = new Node(3);
    head.next.next.next = head; // circular link
    System.out.println(isCircular(head));
  }
}

medium

A. Runtime exception

B. false

C. Compilation error

D. true

Solution

1. Step 1: Understand the linked list structure

   The list has nodes 1 -> 2 -> 3, and then 3 points back to 1, making it circular.

2. Step 2: Trace the isCircular method

   Slow starts at 1, fast at 2. They move until they meet because of the loop, so method returns true.

3. Final Answer:

   true -> Option D

4. Quick Check:

   Loop present = true [OK]

Hint: If fast and slow meet, list is circular [OK]

Common Mistakes:

• Starting fast pointer incorrectly
• Not creating circular link properly
• Confusing output with false

4. Identify the bug in this code snippet that tries to detect if a linked list is circular:

public boolean isCircular(Node head) {
  Node slow = head, fast = head;
  while (fast != null && fast.next != null) {
    slow = slow.next;
    fast = fast.next.next;
    if (slow.next == fast.next) return true;
  }
  return false;
}

medium

A. slow pointer should start at head.next

B. fast pointer should move only one step at a time

C. The condition should compare slow and fast, not slow.next and fast.next

D. The loop condition should be while (slow != null)

Solution

1. Step 1: Analyze pointer comparison

   The code compares slow.next and fast.next, but to detect loop, slow and fast themselves must be compared.

2. Step 2: Correct the comparison

   Change if condition to if (slow == fast) to detect when pointers meet inside the loop.

3. Final Answer:

   The condition should compare slow and fast, not slow.next and fast.next -> Option C

4. Quick Check:

   Compare slow and fast pointers directly [OK]

Hint: Compare slow and fast nodes directly to detect loop [OK]

Common Mistakes:

• Comparing next nodes instead of current nodes
• Moving fast pointer incorrectly
• Wrong loop condition

5. Given a linked list where the last node points to the head node, how can you modify the detection method to also return the length of the loop?

hard

A. After detecting the loop with slow == fast, keep slow fixed and move fast until it meets slow again, counting steps

B. Count nodes from head to tail without checking for loop

C. Return the total number of nodes visited before fast reaches null

D. Use a hash set to store nodes and count unique nodes only

Solution

1. Step 1: Detect loop using two pointers

   Use Floyd's cycle detection to find when slow equals fast inside the loop.

2. Step 2: Count loop length

   Keep slow pointer fixed, move fast pointer one step at a time, incrementing count until fast meets slow again.

3. Final Answer:

   After detecting the loop with slow == fast, keep slow fixed and move fast until it meets slow again, counting steps -> Option A

4. Quick Check:

   Count steps in loop after detection [OK]

Hint: Count nodes in loop by moving one pointer until it meets again [OK]

Common Mistakes:

• Counting nodes before loop detection
• Using total nodes instead of loop length
• Not moving pointer correctly to count loop