Detect if a Linked List is Circular in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this code detecting circular linked list?

Given the following Java code that checks if a linked list is circular, what will be printed?

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; this.next = null; }
}

public class Test {
    public static boolean isCircular(Node head) {
        if (head == null) return false;
        Node temp = head.next;
        while (temp != null && temp != head) {
            temp = temp.next;
        }
        return temp == head;
    }

    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));
    }
}

A false

B true

C NullPointerException

D Compilation error

💡 Hint

Think about how the loop checks if the next node points back to the head.

The linked list is made circular by pointing the last node's next to head. The method traverses nodes until it finds head again, so it returns true.

❓ Predict Output

intermediate

What does this code print when the list is not circular?

Analyze the output of the following Java code that checks if a linked list is circular when the list is linear (not circular).

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; this.next = null; }
}

public class Test {
    public static boolean isCircular(Node head) {
        if (head == null) return false;
        Node temp = head.next;
        while (temp != null && temp != head) {
            temp = temp.next;
        }
        return temp == head;
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        // no circular link
        System.out.println(isCircular(head));
    }
}

A false

B true

C NullPointerException

D Compilation error

💡 Hint

The list ends with a null next pointer, so the loop stops without finding head again.

Since the list is linear and does not loop back to head, the method returns false.

🔧 Debug

advanced

What error occurs in this circular detection code?

Consider this Java method to detect if a linked list is circular. What error will it cause when run on a non-circular list?

DSA Java

public static boolean isCircular(Node head) {
    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;
}

A No error, returns correct result

B Infinite loop

C Compilation error

D NullPointerException

💡 Hint

This is Floyd's cycle detection algorithm. Check how fast and slow pointers move.

The code safely checks fast and fast.next for null before moving pointers, so no NullPointerException occurs. It returns false if no cycle.

❓ Predict Output

advanced

What is the output of this code with a single node circular list?

What will this Java code print when the linked list has only one node that points to itself?

DSA Java

class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; this.next = null; }
}

public class Test {
    public static boolean isCircular(Node head) {
        if (head == null) return false;
        Node temp = head.next;
        while (temp != null && temp != head) {
            temp = temp.next;
        }
        return temp == head;
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = head; // circular link to itself
        System.out.println(isCircular(head));
    }
}

A Compilation error

B false

C NullPointerException

D true

💡 Hint

A single node pointing to itself forms a circular list.

The method detects the next node is head itself, so it returns true.

🧠 Conceptual

expert

Why is Floyd's cycle detection algorithm preferred for circular linked list detection?

Which of the following is the main advantage of Floyd's cycle detection algorithm over the simple traversal method for detecting circular linked lists?

A It requires extra memory to store visited nodes in a hash set.

B It modifies the linked list nodes to mark visited nodes.

C It uses two pointers moving at different speeds to detect cycles efficiently without extra memory.

D It only works for singly linked lists with integer data.

💡 Hint

Think about memory usage and pointer movement in Floyd's algorithm.

Floyd's algorithm uses two pointers moving at different speeds to detect cycles without extra memory or modifying the list.

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