Detect Cycle in Linked List Floyd's Algorithm in DSA Java - Time & Space Complexity

Time Complexity: Detect Cycle in Linked List Floyd's Algorithm

O(n)

Understanding Time Complexity

We want to understand how long it takes to check if a linked list has a cycle using Floyd's Algorithm.

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 hasCycle(ListNode head) {
    ListNode slow = head;
    ListNode fast = head;
    while (fast != null && fast.next != null) {
        slow = slow.next;
        fast = fast.next.next;
        if (slow == fast) {
            return true;
        }
    }
    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 moves the slow pointer one step and the fast pointer two steps through the list.
• How many times: At most, the loop runs until the fast pointer reaches the end or the two pointers meet inside a cycle.

How Execution Grows With Input

As the list length n grows, the number of steps the pointers take grows roughly in a straight line.

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

Pattern observation: The steps increase roughly one-to-one with the list size.

Final Time Complexity

Time Complexity: O(n)

This means the time to detect a cycle grows linearly with the number of nodes in the list.

Common Mistake

[X] Wrong: "Because there are two pointers, the time is doubled or squared."

[OK] Correct: Both pointers move through the list together, so the total steps still grow linearly, not squared.

Interview Connect

Understanding this linear time check is a key skill for working with linked lists and detecting cycles efficiently.

Self-Check

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

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