Insert at Beginning of Circular Linked List in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Go

newNode := &Node{data: [1]

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

Avalue

Bnext

Cnil

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'head' instead of the value for the new node's data.

Setting data to nil which is invalid for an int field.

We create a new node with the data field set to the given value.

2fill in blank

medium

Complete the code to handle the empty list case by pointing newNode's next to itself.

DSA Go

if head == nil {
    newNode.next = [1]
    head = newNode
}

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

Anext

Bhead

Cnil

DnewNode

Attempts:

3 left

💡 Hint

Common Mistakes

Setting next to nil breaks the circular property.

Setting next to head when head is nil causes nil pointer errors.

In a circular linked list with one node, the node points to itself.

3fill in blank

hard

Fix the error in the loop that finds the last node in the circular list.

DSA Go

last := head
for last.next != [1] {
    last = last.next
}

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

Ahead

Bnil

CnewNode

Dlast

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing last.next to nil causes infinite loop.

Comparing last.next to newNode is incorrect before insertion.

The last node points back to head in a circular linked list.

4fill in blank

hard

Fill both blanks to insert newNode at the beginning and update the last node's next pointer.

DSA Go

newNode.next = [1]
last.next = [2]

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

Ahead

BnewNode

Cnil

Dlast

Attempts:

3 left

💡 Hint

Common Mistakes

Setting last.next to head instead of newNode breaks the list.

Setting newNode.next to nil breaks circular property.

newNode points to old head; last node points to newNode to maintain circularity.

5fill in blank

hard

Fill both blanks to update head and complete the insertion at beginning.

DSA Go

head = [1]
return [2]

// Function signature: func insertAtBeginning(head *Node, value int) *Node {
// Node struct: type Node struct { data int; next *Node }
// }

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

AnewNode

Bhead

Cnil

Dlast

Attempts:

3 left

💡 Hint

Common Mistakes

Returning old head instead of newNode.

Returning nil causes loss of list reference.

Head is updated to newNode; function returns newNode; no extra return needed.

Practice

1. What happens when you insert a new node at the beginning of a circular linked list?

easy

A. The new node becomes the head and points to the old head, maintaining the circular link.

B. The new node is added at the end and does not affect the head.

C. The list breaks and becomes a linear linked list.

D. The new node replaces the last node in the list.

Solution

1. Step 1: Understand the circular linked list structure

   A circular linked list has its last node pointing back to the head, forming a circle.

2. Step 2: Insert new node at beginning

   The new node is linked to the current head, and the last node's next pointer is updated to this new node, making it the new head.

3. Final Answer:

   The new node becomes the head and points to the old head, maintaining the circular link. -> Option A

4. Quick Check:

   New head points to old head, circle intact = A [OK]

Hint: New head points to old head; last node points to new head [OK]

Common Mistakes:

• Not updating last node's next pointer
• Breaking the circular link
• Adding node at end instead of beginning

2. Which of the following Go code snippets correctly inserts a new node at the beginning of a circular linked list?

easy

A. newNode.next = nil
head = newNode

B. head.next = newNode
newNode.next = head

C. newNode.next = head
last.next = newNode
head = newNode

D. last.next = head
head = newNode
newNode.next = last

Solution

1. Step 1: Link new node to current head

   Assign newNode.next = head to point new node to the current head.

2. Step 2: Update last node to point to new node and update head

   Set last.next = newNode to maintain circular link, then update head = newNode.

3. Final Answer:

   newNode.next = head last.next = newNode head = newNode -> Option C

4. Quick Check:

   Correct pointer updates = B [OK]

Hint: Link new node to head, update last node, then head [OK]

Common Mistakes:

• Setting newNode.next to nil breaks circular link
• Not updating last.next to new node
• Incorrect order of pointer assignments

3. What will be the printed output of the circular linked list after inserting node with value 10 at the beginning, given the list initially contains 20 -> 30 -> 40 -> (back to 20)?

func printList(head *Node) {
    if head == nil {
        fmt.Println("List is empty")
        return
    }
    temp := head
    for {
        fmt.Printf("%d -> ", temp.data)
        temp = temp.next
        if temp == head {
            break
        }
    }
    fmt.Println("null")
}

// Insert 10 at beginning
newNode := &Node{data: 10}
// Assume last points to node with 40
newNode.next = head
last.next = newNode
head = newNode

printList(head)

medium

A. List is empty

B. 20 -> 30 -> 40 -> 10 -> null

C. 10 -> 20 -> 30 -> 40 -> 10 -> null

D. 10 -> 20 -> 30 -> 40 -> null

Solution

1. Step 1: Insert new node 10 at beginning

   New node 10 points to old head 20, last node 40 points to new node 10, head updated to 10.

2. Step 2: Print list starting from new head

   Traverse from 10: 10 -> 20 -> 30 -> 40, then stop at head again.

3. Final Answer:

   10 -> 20 -> 30 -> 40 -> null -> Option D

4. Quick Check:

   New head 10 followed by old nodes = A [OK]

Hint: New node prints first, then old nodes in order [OK]

Common Mistakes:

• Printing extra node causing repeated values
• Not updating last.next causing infinite loop
• Confusing order of nodes after insertion

4. Given the following Go code to insert at beginning of a circular linked list, what is the error?

func insertAtBeginning(head *Node, data int) *Node {
    newNode := &Node{data: data}
    if head == nil {
        newNode.next = newNode
        return newNode
    }
    temp := head
    for temp.next != nil {
        temp = temp.next
    }
    newNode.next = head
    temp.next = newNode
    head = newNode
    return head
}

medium

A. The function should not return head.

B. The loop condition should check for temp.next != head, not nil.

C. head should not be updated inside the function.

D. newNode.next should be nil when list is empty.

Solution

1. Step 1: Analyze loop condition for circular list

   In circular linked list, last node's next points to head, so loop should check temp.next != head, not nil.

2. Step 2: Identify impact of wrong condition

   Checking temp.next != nil causes infinite loop or crash because next is never nil in circular list.

3. Final Answer:

   The loop condition should check for temp.next != head, not nil. -> Option B

4. Quick Check:

   Loop must stop at head, not nil [OK]

Hint: In circular list, last.next points to head, not nil [OK]

Common Mistakes:

• Using nil instead of head in loop condition
• Not updating head after insertion
• Returning wrong pointer from function

5. You have a circular linked list with nodes 5 -> 15 -> 25 -> (back to 5). You want to insert a new node with value 1 at the beginning. After insertion, you want to print the list starting from the new head. Which sequence of pointer updates correctly achieves this without breaking the circular structure?

hard

A. Set newNode.next = head; find last node and set last.next = newNode; update head = newNode

B. Set newNode.next = nil; update head = newNode; last.next = head

C. Update head = newNode; set newNode.next = head; last.next = newNode

D. Set last.next = newNode; newNode.next = head; head = newNode.next

Solution

1. Step 1: Link new node to current head

   Assign newNode.next = head so new node points to old head.

2. Step 2: Update last node to point to new node and update head

   Find last node (with value 25) and set last.next = newNode to maintain circular link, then update head = newNode.

3. Final Answer:

   Set newNode.next = head; find last node and set last.next = newNode; update head = newNode -> Option A

4. Quick Check:

   New head points to old head; last points to new head [OK]

Hint: Link new node to head, last to new node, then update head [OK]

Common Mistakes:

• Setting newNode.next to nil breaks circle
• Updating head before linking nodes
• Not updating last.next to new node