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.