Detect Cycle in Linked List Floyd's Algorithm in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind Floyd's Cycle Detection Algorithm in a linked list?

It uses two pointers moving at different speeds (slow and fast). If there is a cycle, the fast pointer will eventually meet the slow pointer inside the cycle.

beginner

In Floyd's Algorithm, what speeds do the two pointers move at?

The slow pointer moves one step at a time, and the fast pointer moves two steps at a time.

intermediate

Why does the fast pointer eventually meet the slow pointer if a cycle exists?

Because the fast pointer moves faster, it will loop around and catch up to the slow pointer inside the cycle.

beginner

What does it mean if the fast pointer reaches the end (null) in Floyd's Algorithm?

It means there is no cycle in the linked list because the fast pointer reached the end without meeting the slow pointer.

intermediate

How can you find the starting node of the cycle after detecting a cycle with Floyd's Algorithm?

After the pointers meet, reset one pointer to the head and move both pointers one step at a time. The node where they meet again is the start of the cycle.

In Floyd's Cycle Detection Algorithm, what happens if the fast pointer reaches null?

A There is no cycle in the linked list

B There is a cycle in the linked list

C The slow pointer moves faster

D The algorithm restarts

If the fast pointer reaches null, it means the list ends and no cycle exists.

What is the speed of the slow pointer in Floyd's Algorithm?

A Two steps at a time

B One step at a time

C Three steps at a time

D Variable speed

The slow pointer moves one node at a time.

Why does the fast pointer move faster than the slow pointer?

A To skip nodes

B To reach the end faster

C To catch the slow pointer if a cycle exists

D To avoid cycles

The fast pointer moves faster to detect if it meets the slow pointer inside a cycle.

After detecting a cycle, how do you find the cycle's starting node?

A Reset one pointer to head and move both one step until they meet

B Move both pointers two steps until they meet

C Move slow pointer to null

D Restart the algorithm

Resetting one pointer to head and moving both one step at a time finds the cycle start.

What data structure is Floyd's Cycle Detection Algorithm used on?

A Queue

B Array

C Stack

D Linked List

Floyd's Algorithm is specifically designed to detect cycles in linked lists.

Explain how Floyd's Cycle Detection Algorithm works step-by-step.

Think about two runners on a track moving at different speeds.

Describe how to find the starting node of a cycle after detecting it with Floyd's Algorithm.

Imagine walking from the start and from the meeting point at the same speed.

Practice

1. What is the main idea behind Floyd's Cycle Detection Algorithm in a linked list?

easy

A. Use two pointers moving at different speeds to detect a cycle

B. Count the number of nodes and check if it exceeds a limit

C. Reverse the linked list and check if it matches the original

D. Use a hash set to store visited nodes

Solution

1. Step 1: Understand Floyd's Algorithm concept

   It uses two pointers: a slow pointer moving one step at a time and a fast pointer moving two steps at a time.

2. Step 2: Detect cycle by pointer meeting

   If there is a cycle, the fast pointer will eventually meet the slow pointer inside the loop.

3. Final Answer:

   Use two pointers moving at different speeds to detect a cycle -> Option A

4. Quick Check:

   Two pointers at different speeds detect cycle [OK]

Hint: Two pointers moving at different speeds detect cycle [OK]

Common Mistakes:

• Thinking counting nodes detects cycle
• Assuming reversing list helps detect cycle
• Using extra memory instead of pointers

2. Which of the following is the correct way to initialize the slow and fast pointers in Floyd's cycle detection in Java?

easy

A. ListNode slow = head; ListNode fast = head.next;

B. ListNode slow = head.next; ListNode fast = head.next.next;

C. ListNode slow = head; ListNode fast = head;

D. ListNode slow = null; ListNode fast = head;

Solution

1. Step 1: Initialize both pointers at head

   Both slow and fast pointers start at the head node to begin traversal.

2. Step 2: Move pointers at different speeds

   Slow moves one step, fast moves two steps from the same start point.

3. Final Answer:

   ListNode slow = head; ListNode fast = head; -> Option C

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head for Floyd's algorithm [OK]

Common Mistakes:

• Starting fast pointer ahead causing null errors
• Initializing slow pointer as null
• Starting pointers at different nodes incorrectly

3. What will be the output of the following Java code snippet detecting cycle using Floyd's algorithm?

class ListNode {
  int val;
  ListNode next;
  ListNode(int val) { this.val = val; this.next = null; }
}

public class Test {
  public static boolean hasCycle(ListNode head) {
    ListNode slow = head, fast = head;
    while (fast != null && fast.next != null) {
      slow = slow.next;
      fast = fast.next.next;
      if (slow == fast) return true;
    }
    return false;
  }

  public static void main(String[] args) {
    ListNode a = new ListNode(1);
    ListNode b = new ListNode(2);
    ListNode c = new ListNode(3);
    a.next = b; b.next = c; c.next = b;
    System.out.println(hasCycle(a));
  }
}

medium

A. false

B. true

C. Compilation error

D. Runtime exception

Solution

1. Step 1: Analyze the linked list structure

   Node c points back to node b, creating a cycle in the list.

2. Step 2: Floyd's algorithm detects cycle

   Slow and fast pointers will meet inside the cycle, so method returns true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Cycle exists, so output is true [OK]

Hint: Cycle means slow and fast pointers meet [OK]

Common Mistakes:

• Assuming no cycle because list looks linear
• Confusing output with false
• Ignoring pointer movement conditions

4. Identify the bug in this Floyd's cycle detection code snippet:

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

medium

A. The return false statement is missing

B. slow pointer should move two steps instead of one

C. The while loop condition is incorrect

D. fast pointer should start at head, not head.next

Solution

1. Step 1: Check pointer initialization

   Fast pointer starts at head.next, which can cause missing cycle detection if list has only one node.

2. Step 2: Correct initialization for reliable detection

   Both slow and fast pointers should start at head to ensure proper cycle detection.

3. Final Answer:

   fast pointer should start at head, not head.next -> Option D

4. Quick Check:

   Start both pointers at head [OK]

Hint: Initialize both pointers at head to avoid missing cycles [OK]

Common Mistakes:

• Starting fast pointer ahead causing missed cycles
• Changing slow pointer speed incorrectly
• Incorrect loop condition usage

5. Given a linked list where a cycle exists, how can Floyd's algorithm be extended to find the exact node where the cycle begins? Choose the correct approach.

hard

A. After detecting cycle, reset slow to head and move both pointers one step until they meet

B. Count the number of nodes in the cycle and move slow pointer that many steps

C. Reverse the linked list and check for repeated nodes

D. Use a hash set to store nodes and find the first repeated node

Solution

1. Step 1: Detect cycle using Floyd's algorithm

   When slow and fast pointers meet, a cycle is confirmed.

2. Step 2: Reset slow pointer to head and move both pointers one step

   Move slow and fast pointers one step at a time; the node where they meet is the cycle start.

3. Final Answer:

   After detecting cycle, reset slow to head and move both pointers one step until they meet -> Option A

4. Quick Check:

   Reset slow and move both pointers to find cycle start [OK]

Hint: Reset slow to head, move both pointers one step to find cycle start [OK]

Common Mistakes:

• Trying to find cycle start without resetting slow pointer
• Counting nodes instead of pointer movement
• Using extra memory unnecessarily