Detect Cycle in Linked List Floyd's Algorithm in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the slow pointer at the head of the linked list.

DSA Go

slow := [1]

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

Anil

Bhead

Chead.next

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Starting slow at nil causes no traversal.

Starting slow at head.next skips the first node.

The slow pointer should start at the head of the linked list to begin traversal.

2fill in blank

medium

Complete the code to move the fast pointer two steps ahead safely.

DSA Go

if fast != nil && fast.next != nil {
    fast = [1]
}

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

Afast.next.next

Bfast.next

Cfast

Dfast.next.next.next

Attempts:

3 left

💡 Hint

Common Mistakes

Moving fast only one step slows detection.

Moving fast three steps causes runtime errors.

The fast pointer moves two nodes ahead each time to detect cycles efficiently.

3fill in blank

hard

Fix the error in the condition that checks if slow and fast pointers meet.

DSA Go

if [1] == [2] {
    return true
}

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

Afast

Bslow

Cslow.next

Dfast.next

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing slow to fast.next misses the meeting point.

Comparing slow.next to fast causes incorrect detection.

The condition should compare slow and fast pointers to detect a cycle.

4fill in blank

hard

Fill both blanks to correctly move slow and fast pointers inside the loop.

DSA Go

slow = [1]
fast = [2]

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

Aslow.next

Bfast.next.next

Cfast.next

Dslow

Attempts:

3 left

💡 Hint

Common Mistakes

Moving slow two steps causes skipping nodes.

Moving fast one step slows detection.

Slow pointer moves one step, fast pointer moves two steps each iteration.

5fill in blank

hard

Fill all three blanks to complete the cycle detection function's loop condition and return statement.

DSA Go

for [1] != nil && [2] != nil && [2].next != nil {
    // move pointers
}
return [3]

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

Aslow

Bfast

Cfalse

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Using true as return when no cycle exists.

Not checking fast.next causes runtime errors.

The loop continues while slow and fast pointers are valid. Return false if no cycle is found.

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