Detect Cycle in Linked List Floyd's Algorithm in DSA Python

Mental Model

Use two pointers moving at different speeds to find if a loop exists in a linked list.

Analogy: Imagine two runners on a circular track: one runs fast and the other slow. If the track loops, the fast runner will eventually catch the slow runner.

head -> 1 -> 2 -> 3 -> 4 -> 5 -> null
slow ↑
fast ↑

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> 4 -> 5, with a cycle from 5 back to 3

Goal: Detect if the linked list has a cycle using Floyd's algorithm

Step 1: Initialize slow and fast pointers at head (node 1)

1 -> 2 -> 3 -> 4 -> 5 ->
↑slow, ↑fast (both at 1)

Why: Start both pointers at the beginning to begin traversal

Step 2: Move slow by 1 step to node 2, fast by 2 steps to node 3

1 -> 2 -> 3 -> 4 -> 5 ->
   ↑slow
         ↑fast

Why: Slow moves one step, fast moves two steps to detect cycle faster

Step 3: Move slow by 1 step to node 3, fast by 2 steps to node 5

1 -> 2 -> 3 -> 4 -> 5 ->
      ↑slow
               ↑fast

Why: Pointers continue moving; fast pointer moves twice as fast

Step 4: Move slow by 1 step to node 4, fast moves 2 steps but cycles back to node 4

1 -> 2 -> 3 -> 4 -> 5 ->
           ↑slow
           ↑fast

Why: Fast pointer loops back due to cycle; slow pointer approaches fast

Step 5: Slow and fast pointers meet at node 4, cycle detected

1 -> 2 -> 3 -> 4 -> 5 ->
           ↑slow, ↑fast (meeting point)

Why: Meeting point confirms presence of cycle

Result:

1 -> 2 -> 3 -> 4 -> 5 -> (cycle back to 3)
Cycle detected: True

Annotated Code

DSA Python

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

def detect_cycle(head):
    slow = head
    fast = head
    while fast and fast.next:
        slow = slow.next  # move slow by 1
        fast = fast.next.next  # move fast by 2
        if slow == fast:
            return True  # cycle detected
    return False  # no cycle

# Create linked list 1->2->3->4->5 with cycle from 5 to 3
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)
head.next.next.next.next = Node(5)
head.next.next.next.next.next = head.next.next  # cycle here

print("Cycle detected:", detect_cycle(head))

slow = slow.next # move slow by 1

advance slow pointer by one node

fast = fast.next.next # move fast by 2

advance fast pointer by two nodes

if slow == fast:

check if pointers meet indicating a cycle

return True # cycle detected

return true when cycle is found

Output

Cycle detected: True

Complexity Analysis

Time: O(n) because both pointers traverse the list at most once until they meet or reach the end

Space: O(1) because only two pointers are used regardless of list size

vs Alternative: Compared to using a hash set to track visited nodes (O(n) space), Floyd's algorithm uses constant space making it more efficient

Edge Cases

empty list (head is None)

function returns False immediately as there is no cycle

DSA Python

while fast and fast.next:

single node without cycle

function returns False as fast.next is None

DSA Python

while fast and fast.next:

two nodes with cycle (second node points back to first)

function detects cycle when slow and fast meet

DSA Python

if slow == fast:

When to Use This Pattern

When you need to detect a cycle in a linked list efficiently, Floyd's algorithm is the go-to pattern because it uses two pointers moving at different speeds to find a loop without extra space.

Common Mistakes

Mistake: Moving fast pointer by one step instead of two
Fix: Advance fast pointer by two steps to ensure faster traversal and correct cycle detection

Mistake: Not checking if fast or fast.next is None before moving pointers
Fix: Add condition 'while fast and fast.next' to avoid runtime errors

Summary

Detects if a linked list has a cycle by using two pointers moving at different speeds.

Use it when you want to find loops in linked lists efficiently without extra memory.

The key insight is that a fast pointer will eventually catch a slow pointer if a cycle exists.

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 and check for repeats

Solution

1. Step 1: Understand Floyd's algorithm concept

   It uses two pointers, one moving fast (two steps) and one slow (one step).

2. Step 2: Detect cycle by pointer meeting

   If the fast pointer meets the slow pointer, a cycle exists in the linked list.

3. Final Answer:

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

4. Quick Check:

   Two pointers at different speeds = cycle detection [OK]

Hint: Two pointers, slow and fast, meet if cycle exists [OK]

Common Mistakes:

• Thinking counting nodes detects cycle
• Assuming reversing list helps detect cycle
• Confusing hash set method with Floyd's algorithm

2. Which of the following is the correct way to move the fast pointer in Floyd's cycle detection?

easy

A. fast = fast.next

B. fast = fast.next.next

C. fast = fast.next.next.next

D. fast = fast.next.prev

Solution

1. Step 1: Understand fast pointer movement

   The fast pointer moves two steps at a time to catch up with the slow pointer.

2. Step 2: Identify correct syntax

   Moving two nodes ahead is done by fast = fast.next.next.

3. Final Answer:

   fast = fast.next.next -> Option B

4. Quick Check:

   Fast pointer moves two steps = fast.next.next [OK]

Hint: Fast pointer moves two nodes ahead using next.next [OK]

Common Mistakes:

• Moving fast pointer only one step
• Trying to move fast pointer three steps
• Using invalid pointer like prev in singly linked list

3. What will be the output of the following code snippet?

class Node:
    def __init__(self, val):
        self.val = val
        self.next = None

head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = head.next  # cycle here

def has_cycle(head):
    slow = head
    fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow == fast:
            return True
    return False

print(has_cycle(head))

medium

A. True

B. None

C. False

D. Error

Solution

1. Step 1: Analyze linked list structure

   Nodes 1->2->3 and then 3 points back to 2, creating a cycle.

2. Step 2: Trace Floyd's algorithm

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

3. Final Answer:

   True -> Option A

4. Quick Check:

   Cycle exists, so output = True [OK]

Hint: Cycle means slow and fast pointers meet, so return True [OK]

Common Mistakes:

• Ignoring cycle creation line
• Assuming no cycle returns True
• Confusing None return with False

4. Find the error in the following Floyd's cycle detection code:

def detect_cycle(head):
    slow = head
    fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next
        if slow == fast:
            return True
    return False

medium

A. No return statement for cycle detection

B. slow pointer moves two steps instead of one

C. Missing initialization of slow pointer

D. fast pointer moves only one step instead of two

Solution

1. Step 1: Check pointer movements

   Fast pointer should move two steps (fast = fast.next.next), but here it moves one step.

2. Step 2: Understand impact on cycle detection

   Moving one step slows fast pointer, so it may never meet slow pointer correctly.

3. Final Answer:

   fast pointer moves only one step instead of two -> Option D

4. Quick Check:

   Fast pointer must move two steps [OK]

Hint: Fast pointer must jump two nodes, not one [OK]

Common Mistakes:

• Moving fast pointer one step
• Confusing slow pointer movement
• Forgetting to check fast and fast.next

5. Given a linked list where the last node points back to the head (forming a cycle), how can Floyd's algorithm be adapted to find the node where the cycle begins?

hard

A. Use a hash set to store nodes and return the first repeated node

B. Count nodes until slow and fast meet, then return count as cycle start

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

D. Reverse the linked list and check where reversal breaks

Solution

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

   Use slow and fast pointers to find meeting point inside cycle.

2. Step 2: Find cycle start node

   Reset slow to head, then move slow and fast one step at a time until they meet; meeting point is cycle start.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Trying to count nodes to find cycle start
• Using reversal which doesn't help here
• Confusing hash set method with Floyd's approach