Reverse a Singly Linked List Iterative in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the previous pointer to None.

DSA Python

def reverse_linked_list(head):
    prev = [1]
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

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ANone

Bhead

Ccurrent

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev to head instead of None

Using current instead of None

We start with prev as None because the new tail of the reversed list points to nothing.

2fill in blank

medium

Complete the code to save the next node before changing the current node's next pointer.

DSA Python

def reverse_linked_list(head):
    prev = None
    current = head
    while current:
        [1] = current.next
        current.next = prev
        prev = current
        current = next_node
    return prev

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Acurrent

Bhead

Cnext_node

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Not saving next node before changing links

Using prev or current instead of next_node

We save the next node in next_node before changing links to avoid losing the rest of the list.

3fill in blank

hard

Fix the error in updating the current pointer to the next node.

DSA Python

def reverse_linked_list(head):
    prev = None
    current = head
    while current:
        next_node = current.next
        current.next = prev
        prev = current
        current = [1]
    return prev

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Ahead

Bnext_node

Cprev

Dcurrent.next

Attempts:

3 left

💡 Hint

Common Mistakes

Setting current to prev or current.next

Using head instead of next_node

After reversing the link, we move current to the saved next_node to continue traversal.

4fill in blank

hard

Fill both blanks to complete the loop condition and return statement correctly.

DSA Python

def reverse_linked_list(head):
    prev = None
    current = head
    while [1]:
        next_node = current.next
        current.next = prev
        prev = current
        current = next_node
    return [2]

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Acurrent

Bprev

Chead

Dnext_node

Attempts:

3 left

💡 Hint

Common Mistakes

Using head or next_node in loop or return

Returning current instead of prev

The loop runs while current is not None, and we return prev which is the new head of the reversed list.

5fill in blank

hard

Fill all three blanks to create a dictionary comprehension that maps node values to their reversed positions.

DSA Python

def map_reversed_positions(head):
    reversed_list = reverse_linked_list(head)
    position_map = { [1]: index for index, node in enumerate(iter_nodes(reversed_list)) if node.value [2] 0 and node.value [3] 100 }
    return position_map

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Anode.value

B>

C<

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using index as key

Wrong comparison operators

Mixing up > and <

The dictionary maps each node's value to its index if the value is greater than 0 and less than 100.