Detect Cycle in Linked List Floyd's Algorithm in DSA Go - Execution Trace

Concept Flow - Detect Cycle in Linked List Floyd's Algorithm

Start: slow=head, fast=head

↓

Move slow by 1 step

↓

Move fast by 2 steps

↓

Check if fast or fast.next is null?

Yes→No cycle, return false

No↓

Check if slow == fast?

Yes→Cycle detected, return true

↩Back to Move slow by 1 step

The algorithm uses two pointers moving at different speeds to detect if a cycle exists by checking if they meet.

Execution Sample

DSA Go

slow := head
fast := head
for fast != nil && fast.next != nil {
  slow = slow.next
  fast = fast.next.next
  if slow == fast {
    return true
  }
}
return false

Moves two pointers through the list at different speeds to find if they meet, indicating a cycle.

Execution Table

Step	Operation	slow Node	fast Node	Pointer Changes	Visual State
1	Initialize slow=head, fast=head	Node 1	Node 1	slow → Node 1, fast → Node 1	1 → 2 → 3 → 2 (cycle: 3 back to 2) slow@1 fast@1
2	Check loop condition (fast != nil && fast.next != nil)	Node 1	Node 1	Condition: true	No change
3	Move: slow = slow.next, fast = fast.next.next	Node 2	Node 3	slow → Node 2, fast → Node 3	1 → 2 → 3 → 2 (cycle) slow@2 fast@3
4	Check if slow == fast?	Node 2	Node 3	No (2 != 3)	No cycle yet
5	Check loop condition again	Node 2	Node 3	Condition: true (fast.next=2 != nil)	No change
6	Move: slow = slow.next →3, fast = fast.next.next →3	Node 3	Node 3	slow → Node 3, fast → Node 3	1 → 2 → 3 → 2 (cycle) slow@3 fast@3
7	Check if slow == fast?	Node 3	Node 3	Yes! Cycle detected, return true	Pointers meet: cycle confirmed

💡 slow == fast at Node 3, cycle detected in the linked list

Variable Tracker

Variable	Start	After Step 3	After Step 6	Final
slow	Node 1	Node 2	Node 3	Node 3
fast	Node 1	Node 3	Node 3	Node 3

Key Moments - 3 Insights

Why do we move fast pointer by 2 steps and slow by 1 step?

Moving fast pointer twice as fast ensures that if a cycle exists, fast will eventually catch up to slow inside the cycle, as shown in steps 3 and 6.

Why do we check if fast or fast.next is nil before moving pointers?

If fast or fast.next is nil, it means we've reached the end of the list with no cycle, so the algorithm stops early. Here, cycle prevented reaching nil.

What does it mean when slow equals fast during traversal?

When slow equals fast, it means both pointers met inside the cycle, indicating a cycle exists, as in step 7.

Visual Quiz - 3 Questions

Look at the execution_table at step 3, where is the fast pointer?

A Node 2

B Node 3

C Node 1

D nil

💡 Hint

Check the 'fast Node' column in execution_table row for step 3

At which step does the algorithm detect the cycle and return true?

A Step 4

B Step 6

C Step 7

D Step 3

💡 Hint

Look at the 'Operation' column and 'exit_note' in execution_table

If the linked list had no cycle, what would happen to the fast pointer?

A It would reach nil or fast.next nil

B It would meet slow inside the list

C It would stop at head

D Slow would overtake fast

💡 Hint

Refer to the concept_flow diagram's 'no cycle' path

Concept Snapshot

Detect Cycle in Linked List using Floyd's Algorithm:
- Use two pointers: slow (1 step), fast (2 steps)
- Move pointers until fast or fast.next is nil (no cycle)
- If slow == fast at any point, cycle detected
- Return true if cycle, false otherwise

Full Transcript

This visualization shows Floyd's cycle detection algorithm in a linked list with a cycle. Two pointers start at the head: slow moves one node at a time, fast moves two nodes. They meet at Node 3 inside the cycle. The execution table traces pointer positions step-by-step, demonstrating cycle detection. Key moments explain the speed difference and meeting condition.

Practice

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

easy

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

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

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

D. Use a hash map to store visited nodes

Solution

1. Step 1: Understand Floyd's Algorithm concept

   It uses two pointers, one moving one step at a time (slow), and another moving two steps at a time (fast).

2. Step 2: Detect cycle by pointer meeting

   If the fast pointer meets the slow pointer, it means there is a cycle in the linked list.

3. Final Answer:

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

4. Quick Check:

   Two pointers at different speeds = cycle detection [OK]

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

Common Mistakes:

• Thinking counting nodes detects cycle
• Assuming reversing list helps detect cycle
• Using extra memory like hash map is not Floyd's method

2. Which of the following Go code snippets correctly initializes the two pointers for Floyd's cycle detection in a linked list named head?

easy

A. slow, fast := head, head

B. slow, fast := head.next, head.next.next

C. slow, fast := head, head.next

D. slow, fast := nil, head

Solution

1. Step 1: Initialize both pointers at head

   Floyd's algorithm starts both slow and fast pointers at the head node.

2. Step 2: Move pointers inside loop, not at initialization

   Moving fast or slow pointer ahead at initialization can skip nodes or cause errors.

3. Final Answer:

   slow, fast := head, head -> Option A

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head before loop [OK]

Common Mistakes:

• Starting fast pointer ahead causing nil pointer errors
• Initializing slow pointer ahead skipping first node
• Setting slow pointer to nil initially

3. What will be the output of the following Go code snippet detecting a cycle in a linked list?

type Node struct {
  val  int
  next *Node
}

func hasCycle(head *Node) bool {
  slow, fast := head, head
  for fast != nil && fast.next != nil {
    slow = slow.next
    fast = fast.next.next
    if slow == fast {
      return true
    }
  }
  return false
}

// Linked list setup
n1 := &Node{val: 1}
n2 := &Node{val: 2}
n3 := &Node{val: 3}
n4 := &Node{val: 4}
n1.next = n2
n2.next = n3
n3.next = n4
n4.next = n2 // cycle here

fmt.Println(hasCycle(n1))

medium

A. true

B. false

C. runtime error

D. infinite loop

Solution

1. Step 1: Analyze linked list structure

   The list nodes are connected 1->2->3->4, and 4 points back to 2, creating a cycle.

2. Step 2: Trace Floyd's algorithm behavior

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

3. Final Answer:

   true -> Option A

4. Quick Check:

   Cycle exists, so output is true [OK]

Hint: Cycle means pointers meet, so return true [OK]

Common Mistakes:

• Assuming no cycle because list looks linear
• Expecting false without tracing pointers
• Confusing cycle detection with list length

4. Identify the bug in this Go code snippet for detecting a cycle in a linked list:

func hasCycle(head *Node) bool {
  slow, fast := head, head
  for fast != nil {
    slow = slow.next
    fast = fast.next.next
    if slow == fast {
      return true
    }
  }
  return false
}

medium

A. The slow pointer should move two steps instead of one

B. The loop condition should check fast.next != nil to avoid nil pointer dereference

C. The fast pointer should move one step instead of two

D. The function should return false inside the loop when pointers don't meet

Solution

1. Step 1: Check loop condition safety

   The loop only checks fast != nil, but fast.next could be nil causing fast.next.next to panic.

2. Step 2: Correct loop condition

   Loop should check both fast != nil and fast.next != nil to safely move fast pointer two steps.

3. Final Answer:

   The loop condition should check fast.next != nil to avoid nil pointer dereference -> Option B

4. Quick Check:

   Check fast and fast.next != nil before moving fast [OK]

Hint: Always check fast.next != nil before fast.next.next [OK]

Common Mistakes:

• Ignoring fast.next nil check causing runtime panic
• Moving slow pointer two steps incorrectly
• Returning false too early inside loop

5. Given a linked list where the cycle starts at node with value 3, how can Floyd's algorithm be extended to find the exact node where the cycle begins?

hard

A. Count nodes until pointers meet and return the count as cycle start position

B. Reverse the linked list and check where reversal breaks to find cycle start

C. Use a hash map to store nodes and find the first repeated node

D. After detecting the cycle, reset one pointer to head and move both one step until they meet at cycle start

Solution

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

   First, detect if a cycle exists by moving slow and fast pointers until they meet.

2. Step 2: Find cycle start node

   Reset one pointer to head, then move both pointers one step at a time; the node where they meet is the cycle start.

3. Final Answer:

   After detecting the cycle, reset one pointer to head and move both one step until they meet at cycle start -> Option D

4. Quick Check:

   Reset pointer + move one step = cycle start [OK]

Hint: Reset pointer to head after cycle detection to find start [OK]

Common Mistakes:

• Trying to count nodes instead of resetting pointer
• Using extra memory instead of pointer method
• Reversing list which breaks structure