Insert at End of Circular Linked List in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA C

struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
newNode->data = [1];
newNode->next = NULL;

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

Atemp

BNULL

Chead

Ddata

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning NULL or pointer variables instead of data to newNode->data.

The new node's data field should be set to the given data value.

2fill in blank

medium

Complete the code to handle the case when the list is empty.

DSA C

if (head == NULL) {
    newNode->next = newNode;
    head = [1];
    return head;
}

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

AnewNode

Bhead

CNULL

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning head to NULL or an unrelated pointer.

When the list is empty, head should point to the new node.

3fill in blank

hard

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

DSA C

struct Node* temp = head;
while (temp->next != [1]) {
    temp = temp->next;
}

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

ANULL

Bhead

CnewNode

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Checking for NULL instead of head in the loop condition.

In a circular linked list, the last node points back to head, not NULL.

4fill in blank

hard

Fill both blanks to insert the new node at the end and maintain circularity.

DSA C

temp->next = [1];
newNode->next = [2];

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

AnewNode

Bhead

Ctemp

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Breaking the circular link by pointing to NULL or wrong nodes.

The last node should point to the new node, and the new node should point back to head.

5fill in blank

hard

Fill all three blanks to complete the insert function returning the updated head.

DSA C

struct Node* insertEnd(struct Node* head, int data) {
    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = data;
    if (head == NULL) {
        newNode->next = newNode;
        head = [1];
        return head;
    }
    struct Node* temp = head;
    while (temp->next != [2]) {
        temp = temp->next;
    }
    temp->next = [3];
    newNode->next = head;
    return head;
}

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

AnewNode

BNULL

Chead

Dtemp

Attempts:

3 left

💡 Hint

Common Mistakes

Using NULL instead of head in the loop condition.

Not linking the last node to the new node.

Assign head to newNode when list is empty; loop until temp->next is head; link temp->next to newNode.

Practice

1. What happens when you insert a new node at the end of a circular linked list that currently has 3 nodes?

easy

A. The new node becomes the head node and points to the second node.

B. The new node becomes the last node and points back to the head node.

C. The new node is inserted in the middle of the list.

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

Solution

1. Step 1: Understand circular linked list structure

   In a circular linked list, the last node points back to the head node, forming a circle.

2. Step 2: Insert new node at end

   The new node is linked after the current last node, and its next pointer points to the head node, maintaining the circle.

3. Final Answer:

   The new node becomes the last node and points back to the head node. -> Option B

4. Quick Check:

   New node points to head, preserving circular link [OK]

Hint: New end node always points to head in circular list [OK]

Common Mistakes:

• Forgetting to link new node back to head
• Inserting node at wrong position
• Breaking the circular link

2. Which of the following is the correct syntax to create a new node in C for a circular linked list?

easy

A. struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));

B. newNode = malloc(sizeof(Node));

C. struct Node newNode = malloc(sizeof(struct Node));

D. Node* newNode = new Node();

Solution

1. Step 1: Allocate memory for new node

   In C, use malloc with sizeof(struct Node) and cast to (struct Node*).

2. Step 2: Correct pointer declaration

   Declare newNode as a pointer to struct Node to hold the allocated memory address.

3. Final Answer:

   struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); -> Option A

4. Quick Check:

   Proper malloc and pointer cast in C [OK]

Hint: Use malloc with sizeof(struct Node) and cast to struct Node* [OK]

Common Mistakes:

• Missing (struct Node*) cast
• Declaring newNode as struct Node instead of pointer
• Using C++ style new operator in C

3. Given the following code snippet for inserting at the end of a circular linked list, what will be the printed list after insertion?

struct Node {
  int data;
  struct Node* next;
};

void printList(struct Node* head) {
  struct Node* temp = head;
  if (head != NULL) {
    do {
      printf("%d -> ", temp->data);
      temp = temp->next;
    } while (temp != head);
  }
  printf("NULL\n");
}

int main() {
  struct Node* head = (struct Node*)malloc(sizeof(struct Node));
  struct Node* second = (struct Node*)malloc(sizeof(struct Node));
  struct Node* third = (struct Node*)malloc(sizeof(struct Node));

  head->data = 1; head->next = second;
  second->data = 2; second->next = third;
  third->data = 3; third->next = head;

  // Insert new node with data 4 at end
  struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
  newNode->data = 4;
  newNode->next = head;
  third->next = newNode;

  printList(head);
  return 0;
}

medium

A. 1 -> 2 -> 3 -> 4 -> 1 -> ... (infinite loop)

B. 1 -> 2 -> 3 -> NULL

C. 3 -> 4 -> 1 -> 2 -> NULL

D. 1 -> 2 -> 3 -> 4 -> NULL

Solution

1. Step 1: Understand list before insertion

   Initial list is 1 -> 2 -> 3 -> back to 1 (circular).

2. Step 2: Insert new node at end

   New node with data 4 points to head; third node points to new node, extending list to 1 -> 2 -> 3 -> 4 -> back to 1.

3. Step 3: Print list

   printList prints nodes until it reaches head again, so output is 1 -> 2 -> 3 -> 4 -> NULL.

4. Final Answer:

   1 -> 2 -> 3 -> 4 -> NULL -> Option D

5. Quick Check:

   New node added at end, circular link maintained [OK]

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

Common Mistakes:

• Printing infinite loop by not stopping at head
• Forgetting to update last node's next pointer
• Incorrect print format

4. Identify the error in the following code snippet that attempts to insert a new node at the end of a circular linked list:

void insertEnd(struct Node** head, int data) {
  struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
  newNode->data = data;
  newNode->next = *head;

  struct Node* temp = *head;
  while (temp->next != NULL) {
    temp = temp->next;
  }
  temp->next = newNode;
}

medium

A. The function should not use double pointer for head.

B. newNode->next should be NULL instead of *head.

C. The while loop condition should check for temp->next != *head instead of NULL.

D. Memory allocation for newNode is missing.

Solution

1. Step 1: Analyze loop condition in circular list

   In a circular linked list, last node's next points to head, not NULL.

2. Step 2: Correct loop condition

   Loop should continue until temp->next equals head, not NULL, to find last node.

3. Final Answer:

   The while loop condition should check for temp->next != *head instead of NULL. -> Option C

4. Quick Check:

   Loop must detect circular end by comparing to head [OK]

Hint: In circular list, last node points to head, not NULL [OK]

Common Mistakes:

• Using NULL to detect end in circular list
• Setting newNode->next to NULL
• Not updating last node's next pointer

5. You have a circular linked list with only one node where head points to itself. You want to insert a new node at the end. Which sequence of steps correctly updates the list?

hard

A. Create new node, set newNode->next = head->next, set head->next = newNode.

B. Create new node, set newNode->next = head, set head->next = newNode, then update head to newNode.

C. Create new node, set newNode->next = NULL, set head->next = newNode.

D. Create new node, set newNode->next = newNode, set head = newNode.

Solution

1. Step 1: Understand single node circular list

   The single node has head->next pointing to itself (head).

2. Step 2: Insert at end

   Set newNode->next = head->next (which is head), then head->next = newNode. This makes head -> newNode -> head, maintaining circularity without changing head.

3. Final Answer:

   Create new node, set newNode->next = head->next, set head->next = newNode. -> Option A

4. Quick Check:

   head -> newNode -> head [OK]

Hint: For single node circular: newNode->next = head->next; head->next = newNode [OK]

Common Mistakes:

• Setting newNode->next to NULL breaking circularity
• Unnecessarily updating head to newNode
• Incorrectly linking new node to itself