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DSA Pythonprogramming~10 mins

Delete a Node from Circular Linked List in DSA Python - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to move to the next node in the circular linked list.

DSA Python
current = head
current = current[1]
Drag options to blanks, or click blank then click option'
A.prev
B.child
C.next
D.parent
Attempts:
3 left
💡 Hint
Common Mistakes
Using '.prev' instead of '.next' to move forward.
Using attributes that don't exist like '.child' or '.parent'.
2fill in blank
medium

Complete the code to check if the list has only one node.

DSA Python
if head[1] head.next:
Drag options to blanks, or click blank then click option'
A<
B!=
C>
D==
Attempts:
3 left
💡 Hint
Common Mistakes
Using '!=' which checks for inequality instead of equality.
Using comparison operators like '>' or '<' which are not valid for node references.
3fill in blank
hard

Fix the error in the code to correctly delete the head node from the circular linked list.

DSA Python
head.data = head.next.data
head.next = head[1]
Drag options to blanks, or click blank then click option'
A.prev
B.next.next
C.next
D.next.prev
Attempts:
3 left
💡 Hint
Common Mistakes
Linking head.next to head.next which causes no change.
Using '.prev' which is not relevant here.
Using '.next.prev' which is invalid in this context.
4fill in blank
hard

Fill both blanks to correctly find the node before the one to delete and update links.

DSA Python
prev = head
while prev.next[1] != current:
    prev = prev.next
prev.next = current[2]
Drag options to blanks, or click blank then click option'
A.next
B.prev
C.next.next
D.next.prev
Attempts:
3 left
💡 Hint
Common Mistakes
Using '.prev' which is not used in singly linked circular lists.
Linking prev.next to current.prev which is invalid.
Incorrect loop condition causing infinite loops.
5fill in blank
hard

Fill all three blanks to complete the function that deletes a node with given key from circular linked list.

DSA Python
def delete_node(head, key):
    if head is None:
        return None
    current = head
    prev = None
    while True:
        if current.data == [1]:
            if prev is not None:
                prev.next = current[2]
            else:
                head = current[3]
            return head
        prev = current
        current = current.next
        if current == head:
            break
    return head
Drag options to blanks, or click blank then click option'
Akey
B.next
Dhead.next
Attempts:
3 left
💡 Hint
Common Mistakes
Using wrong variable names for key comparison.
Not updating head when deleting the head node.
Incorrectly linking nodes causing broken list.