Find Middle Element of Linked List in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a new node with the given value.

DSA Python

class Node:
    def __init__(self, data):
        self.data = [1]
        self.next = None

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

ANone

Bvalue

Cnode

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Using a variable name not defined in the constructor.

Assigning None instead of the input data.

The constructor assigns the input parameter 'data' to the node's data attribute.

2fill in blank

medium

Complete the code to move the slow pointer one step forward in the linked list.

DSA Python

slow = slow[1]

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

A.prev

B.next

C.next.next

D.data

Attempts:

3 left

💡 Hint

Common Mistakes

Moving two steps instead of one.

Trying to access a non-existent 'prev' attribute.

To move one step forward, we use slow.next.

3fill in blank

hard

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

DSA Python

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

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

A.next

B.next.next

C.data

D.prev

Attempts:

3 left

💡 Hint

Common Mistakes

Checking fast.next.next instead of fast.next.

Checking fast.prev which does not exist.

We must check if fast.next is not None before moving fast two steps.

4fill in blank

hard

Fill both blanks to create a function that returns the middle node's data.

DSA Python

def find_middle(head):
    slow = head
    fast = head
    while fast is not None and fast[1] is not None:
        slow = slow[2]
        fast = fast.next.next
    return slow.data

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

A.next

B.prev

C.next.next

D.data

Attempts:

3 left

💡 Hint

Common Mistakes

Using .prev which does not exist.

Moving slow two steps instead of one.

We check fast.next and move slow one step with slow.next.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps nodes to their data if data is even.

DSA Python

result = [1]: [2] for node in nodes if node.data [3] 0

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

Anode

Bnode.data

C% 2 ==

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of '% 2 =='.

Swapping key and value in the dictionary.

We map node to node.data and check if node.data % 2 == 0 for even data.