Detect if a Linked List is Circular 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'

Aslow

BNone

Ctail

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing slow to None will cause the loop to never start.

Using tail instead of head will skip the list start.

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 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 '->' is syntax from other languages like C++.

Using commas or double colons are invalid here.

In Python, to access the next node, use dot notation twice: fast.next.next moves two steps ahead.

3fill in blank

hard

Fix the error in the condition to check if the fast pointer or its next node is None.

DSA Python

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

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

Ahead

BNone

Cfast

Dslow

Attempts:

3 left

💡 Hint

Common Mistakes

Checking against head or slow is incorrect here.

Using variable names instead of None causes errors.

We check if fast is not None to ensure the pointer has not reached the end of the list.

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[3]next

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

A.

B->

C::

D,

Attempts:

3 left

💡 Hint

Common Mistakes

Using '->' or other symbols causes syntax errors.

Using commas or double colons is invalid.

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

5fill in blank

hard

Fill all three blanks to return True if slow meets fast, else False after the loop.

DSA Python

if slow == [1]:
    return [2]
return [3]

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

Afast

BTrue

CFalse

DNone

Attempts:

3 left

💡 Hint

Common Mistakes

Returning None instead of False when no cycle is found.

Comparing slow with wrong variable.

If slow equals fast, the list is circular, so return True; otherwise, return False.

Practice

1. What does it mean if a linked list is circular?

easy

A. The linked list nodes are sorted in ascending order

B. The linked list has no nodes

C. The linked list has only one node

D. The last node points back to one of the previous nodes, forming a loop

Solution

1. Step 1: Understand linked list structure

   A linked list is a chain of nodes where each node points to the next node.

2. Step 2: Define circular linked list

   If the last node points back to any previous node, it creates a loop, making the list circular.

3. Final Answer:

   The last node points back to one of the previous nodes, forming a loop -> Option D

4. Quick Check:

   Circular linked list means loop exists [OK]

Hint: Circular means last node links back inside list [OK]

Common Mistakes:

• Thinking circular means sorted
• Confusing empty list with circular
• Assuming single node list is always circular

2. Which of the following Python code snippets correctly checks if a linked list is circular?

easy

A. def is_circular(head):
if not head:
return False
slow = head
fast = head.next
while fast and fast.next:
if slow == fast:
return True
slow = slow.next
fast = fast.next.next
return False

B. def is_circular(head):
current = head
while current:
current = current.next
return True

C. def is_circular(head):
return head.next == None

D. def is_circular(head):
return head == None

Solution

1. Step 1: Understand Floyd's cycle detection

   Using two pointers (slow and fast) moving at different speeds detects loops if they meet.

2. Step 2: Analyze code correctness

   def is_circular(head): if not head: return False slow = head fast = head.next while fast and fast.next: if slow == fast: return True slow = slow.next fast = fast.next.next return False uses slow and fast pointers correctly and returns True if they meet, else False.

3. Final Answer:

   Code using slow and fast pointers to detect loop -> Option A

4. Quick Check:

   Floyd's cycle detection = def is_circular(head): if not head: return False slow = head fast = head.next while fast and fast.next: if slow == fast: return True slow = slow.next fast = fast.next.next return False [OK]

Hint: Use two pointers moving at different speeds [OK]

Common Mistakes:

• Not checking fast pointer's next before accessing
• Assuming reaching None means circular
• Returning True without loop detection

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)
second = Node(2)
third = Node(3)
head.next = second
second.next = third
third.next = head  # Creates a circular linked list

def is_circular(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(is_circular(head))

medium

A. True

B. False

C. None

D. Runtime Error

Solution

1. Step 1: Understand list structure

   Nodes 1->2->3, and third points back to head, making a loop.

2. Step 2: Trace Floyd's cycle detection

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

3. Final Answer:

   True -> Option A

4. Quick Check:

   Loop detected = True [OK]

Hint: Loop means slow and fast pointers meet [OK]

Common Mistakes:

• Ignoring the loop created by third.next = head
• Assuming function returns False by default
• Confusing None return with False

4. Find the error in the following code that tries to detect if a linked list is circular:

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

medium

A. The function should return True outside the loop unconditionally

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

C. The while loop condition should check if fast and fast.next exist

D. The fast pointer should start at head.next

Solution

1. Step 1: Analyze while loop condition

   Current condition checks fast.next and fast.next.next but does not check if fast itself is None.

2. Step 2: Correct condition to avoid runtime error

   Should check if fast and fast.next exist before accessing fast.next.next to prevent errors.

3. Final Answer:

   The while loop condition should check if fast and fast.next exist -> Option C

4. Quick Check:

   Check fast and fast.next to avoid errors [OK]

Hint: Always check pointer exists before accessing next [OK]

Common Mistakes:

• Not checking if fast is None before fast.next
• Moving slow pointer incorrectly
• Returning True too early

5. You have a linked list where the last node points to the third node, making it circular. How would you modify the detection function to also return the node where the loop starts?

hard

A. Return the node where slow and fast first meet during detection

B. Use Floyd's algorithm: after detecting loop, reset slow to head and move slow and fast one step until they meet; that node is loop start

C. Return the head node always as loop start

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

Solution

1. Step 1: Detect loop using Floyd's cycle detection

   Use slow and fast pointers to find if loop exists by meeting inside the loop.

2. Step 2: Find loop start node

   Reset slow to head, then move slow and fast one step at a time; the node where they meet is the loop start.

3. Final Answer:

   Use Floyd's algorithm with reset slow pointer to find loop start -> Option B

4. Quick Check:

   Reset slow and move both pointers to find loop start [OK]

Hint: After loop detection, reset slow to head and move both one step [OK]

Common Mistakes:

• Returning meeting point as loop start without reset
• Always returning head as loop start
• Not handling loop detection before finding start