Detect if a Linked List is Circular in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the linked list is empty before starting the loop.

DSA Java

if (head == [1]) {
    return false;
}

Drag options to blanks, or click blank then click option'

Anull

Bhead

Cthis

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if head equals head (always true).

Using tail instead of head.

We check if head is null to know if the list is empty. If it is, it cannot be circular.

2fill in blank

medium

Complete the code to move the slow pointer one step forward.

DSA Java

slow = slow.[1];

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Ahead

Bnext

Cprev

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which points backward.

Using head or tail which are not node pointers.

To move forward in a linked list, we use the next pointer.

3fill in blank

hard

Fix the error in the while loop condition to avoid null pointer exceptions.

DSA Java

while (fast != null && fast.[1] != null) {
    // loop body
}

Drag options to blanks, or click blank then click option'

Aprev

Bhead

Cnext

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Checking fast.prev which does not exist in singly linked list.

Checking fast.head or fast.tail which are invalid.

We check fast.next to ensure fast can move two steps safely.

4fill in blank

hard

Fill both blanks to move fast pointer two steps ahead safely.

DSA Java

fast = fast.[1];
fast = fast.[2];

Drag options to blanks, or click blank then click option'

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev which is not in singly linked list.

Using head or tail which are not node pointers.

Fast pointer moves two steps by following next twice.

5fill in blank

hard

Fill all four blanks to complete the circular linked list detection using Floyd's cycle algorithm.

DSA Java

Node slow = head;
Node fast = head;
while (fast != null && fast.[1] != null) {
    slow = slow.[2];
    fast = fast.[3].[4];
    if (slow == fast) {
        return true;
    }
}
return false;

Drag options to blanks, or click blank then click option'

Anext

Bprev

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev pointers which do not exist in singly linked list.

Not checking fast.next before moving fast.

Floyd's cycle detection moves slow by one step and fast by two steps using next pointers.

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