Delete from End of Doubly Linked List in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the list is empty before deletion.

DSA C

if (head == [1]) {
    printf("List is empty\n");
    return;
}

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

A0

BNULL

Chead

Dtail

Attempts:

3 left

💡 Hint

Common Mistakes

Checking if head == 0 instead of NULL

Checking tail instead of head

Using an uninitialized pointer

We check if head is NULL to determine if the list is empty.

2fill in blank

medium

Complete the code to move to the last node in the list.

DSA C

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

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

Anext

Bprev

Cdata

Dhead

Attempts:

3 left

💡 Hint

Common Mistakes

Using prev instead of next to traverse forward

Checking data instead of next pointer

Using head inside the loop incorrectly

We use the next pointer to move forward to the last node.

3fill in blank

hard

Fix the error in updating the previous node's next pointer after deleting the last node.

DSA C

if (temp->prev != NULL) {
    temp->prev->[1] = NULL;
} else {
    head = NULL;
}

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

Aprev

Bhead

Cdata

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Setting prev pointer to NULL instead of next

Setting head pointer inside the if block incorrectly

Not handling the case when temp->prev is NULL

We set the previous node's next pointer to NULL to mark it as the new end.

4fill in blank

hard

Fill both blanks to correctly free the last node and update the list.

DSA C

struct Node* toDelete = temp;
temp = temp->prev;
free([1]);
temp->[2] = NULL;

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

AtoDelete

Btemp

Cprev

Dnext

Attempts:

3 left

💡 Hint

Common Mistakes

Freeing temp instead of toDelete

Setting prev pointer to NULL instead of next

Not updating pointers after free

We free the node stored in toDelete and set temp's next pointer to NULL.

5fill in blank

hard

Fill all three blanks to handle deletion when only one node exists.

DSA C

if (head->[1] == NULL) {
    free(head);
    head = [2];
    tail = [3];
}

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

Anext

BNULL

Chead

Dprev

Attempts:

3 left

💡 Hint

Common Mistakes

Checking prev instead of next

Setting head or tail to head instead of NULL

Not freeing the node before resetting pointers

If the only node's next is NULL, free it and set head and tail to NULL.

Practice

1. What happens to the tail pointer when you delete the last node from a doubly linked list?

easy

A. It becomes NULL immediately without checking.

B. It remains unchanged.

C. It points to the head node.

D. It moves to the previous node of the deleted node.

Solution

1. Step 1: Understand the tail pointer role

   The tail pointer always points to the last node in the doubly linked list.

2. Step 2: Update tail after deletion

   When the last node is deleted, the tail pointer must move to the previous node to keep pointing to the new last node.

3. Final Answer:

   It moves to the previous node of the deleted node. -> Option D

4. Quick Check:

   Tail points to previous node after deletion [OK]

Hint: Tail always points to last node; update to previous on delete [OK]

Common Mistakes:

• Setting tail to NULL without checking if list is empty
• Not updating tail pointer after deletion
• Pointing tail to head incorrectly

2. Which of the following is the correct syntax to delete the last node in a doubly linked list in C?

easy

A. tail = tail->prev; free(tail->next); tail->next = NULL;

B. free(tail); tail = tail->prev; tail->next = NULL;

C. tail->prev = tail; free(tail); tail = NULL;

D. tail->next = NULL; free(tail); tail = tail->prev;

Solution

1. Step 1: Free the last node correctly

   We first move tail to previous node, then free the old last node which is tail->next.

2. Step 2: Update tail's next pointer

   After freeing, set tail->next to NULL to mark new end of list.

3. Final Answer:

   tail = tail->prev; free(tail->next); tail->next = NULL; -> Option A

4. Quick Check:

   Move tail first, free old node, update next pointer [OK]

Hint: Move tail first, then free old node, update next pointer [OK]

Common Mistakes:

• Freeing tail before moving pointer
• Not setting tail->next to NULL
• Incorrect pointer assignments causing memory errors

3. What will be the output of the following code snippet after deleting the last node?

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

Node* tail = NULL;
// List: 10 <-> 20 <-> 30
// tail points to node with data 30

// Delete last node
Node* temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

// Print list from head
Node* head = tail;
while(head->prev) head = head->prev;
while(head) {
  printf("%d ", head->data);
  head = head->next;
}

medium

A. 10 20

B. 10 20 30

C. 20 30

D. 10 30

Solution

1. Step 1: Understand list before deletion

   Initial list is 10 <-> 20 <-> 30, tail points to 30.

2. Step 2: Delete last node and update pointers

   Last node 30 is freed, tail moves to 20, tail->next set to NULL.

3. Step 3: Print list from head

   Head is found by moving to start (10), printing nodes 10 and 20.

4. Final Answer:

   10 20 -> Option A

5. Quick Check:

   Last node removed, list prints 10 20 [OK]

Hint: After deletion, tail moves back; print from head to see updated list [OK]

Common Mistakes:

• Printing old tail node after deletion
• Not updating tail->next to NULL
• Starting print from tail instead of head

4. Identify the error in the following code snippet that deletes the last node of a doubly linked list:

void deleteEnd(Node** tail) {
  if (*tail == NULL) return;
  Node* temp = *tail;
  *tail = (*tail)->prev;
  free(temp);
  (*tail)->next = NULL;
}

medium

A. Not freeing the node properly.

B. Dereferencing tail without checking if list becomes empty.

C. Not updating the head pointer.

D. Using double pointer for tail unnecessarily.

Solution

1. Step 1: Check for empty list after deletion

   If the list had only one node, after deletion tail becomes NULL, so (*tail)->next causes error.

2. Step 2: Fix by checking tail before dereference

   Must check if *tail is not NULL before setting (*tail)->next = NULL.

3. Final Answer:

   Dereferencing tail without checking if list becomes empty. -> Option B

4. Quick Check:

   Check tail NULL before accessing next pointer [OK]

Hint: Always check if tail is NULL before accessing its members [OK]

Common Mistakes:

• Dereferencing NULL pointer after deletion
• Ignoring empty list edge case
• Not freeing memory causing leaks

5. You have a doubly linked list with nodes: 5 <-> 15 <-> 25 <-> 35. After deleting the last node twice, what will be the printed list from head?

// Initial tail points to 35
// Delete last node twice
Node* temp;
temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

temp = tail;
tail = tail->prev;
free(temp);
tail->next = NULL;

// Print from head
Node* head = tail;
while(head->prev) head = head->prev;
while(head) {
  printf("%d ", head->data);
  head = head->next;
}

hard

A. 5 15 25

B. 15 25

C. 5 15

D. 25 35

Solution

1. Step 1: Understand initial list and tail

   List is 5 <-> 15 <-> 25 <-> 35, tail points to 35.

2. Step 2: Delete last node once

   Delete 35, tail moves to 25, tail->next = NULL.

3. Step 3: Delete last node second time

   Delete 25, tail moves to 15, tail->next = NULL.

4. Step 4: Print list from head

   Head found by moving to start (5), print nodes 5 and 15.

5. Final Answer:

   5 15 -> Option C

6. Quick Check:

   Two deletions remove last two nodes, list prints 5 15 [OK]

Hint: Each delete moves tail back; print from head to see current list [OK]

Common Mistakes:

• Not updating tail->next to NULL after each deletion
• Printing from tail instead of head
• Assuming deleted nodes still exist