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

slow = [1]

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

Ahead

BNone

Ctail

Dslow

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow to None or tail instead of head.

The slow pointer starts 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 in the linked list.

DSA Python

fast = fast[1]next[2]next

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

A.

B->

C,

D::

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect operators like '->' or commas.

In Python, to access the next node twice, use dot notation twice: fast.next.next.

3fill in blank

hard

Fix the error in the while loop condition to check if fast and fast.next exist.

DSA Python

while fast is not None and fast[1]next is not None:

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

A->

B,

C.

D::

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->' which is not valid in Python.

Use dot notation to access the 'next' attribute in Python: fast.next.

4fill in blank

hard

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

DSA Python

slow = slow[1]next
fast = fast[2]next.next

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

A.

B->

C,

D::

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->' or commas instead of dot notation.

Use dot notation to move slow one step and fast two steps ahead.

5fill in blank

hard

Fill both blanks to complete the cycle detection check inside the loop.

DSA Python

if slow == fast:
    return [1]

return [2]

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

ATrue

BFalse

CNone

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning None or 0 instead of True/False.

If slow meets fast, a cycle exists, so return True. If loop ends, return False.

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