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DSA Goprogramming~5 mins

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

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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 loop.
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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.
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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.
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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.
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intermediate
How can you find the starting node of the cycle after detecting a cycle using Floyd's Algorithm?
Reset one pointer to the head of the list and move both pointers one step at a time. The node where they meet is the start of the cycle.
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In Floyd's Cycle Detection Algorithm, what happens if the fast pointer reaches null?
AA cycle is detected
BThe slow pointer moves faster
CThere is no cycle in the linked list
DThe algorithm restarts
What is the speed of the slow pointer in Floyd's Algorithm?
AOne step at a time
BTwo steps at a time
CThree steps at a time
DVariable speed
Why does Floyd's Algorithm use two pointers with different speeds?
ATo detect if the fast pointer catches the slow pointer inside a cycle
BTo sort the linked list
CTo reverse the linked list
DTo count the number of nodes
After detecting a cycle, how do you find the cycle's starting node?
ACount the nodes in the cycle
BMove both pointers two steps at a time
CReverse the linked list
DMove one pointer to head and advance both one step until they meet
What data structure is Floyd's Cycle Detection Algorithm used on?
AArray
BLinked List
CStack
DQueue
Explain how Floyd's Cycle Detection Algorithm works step-by-step.
Think about how the pointers move and meet.
You got /5 concepts.
    Describe how to find the starting node of a cycle after detecting it with Floyd's Algorithm.
    Focus on pointer movement after cycle detection.
    You got /3 concepts.

      Practice

      (1/5)
      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
      Recommended

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