Delete from End of Doubly Linked List in DSA C - Time & Space Complexity

Time Complexity: Delete from End of Doubly Linked List

O(n)

Understanding Time Complexity

We want to understand how long it takes to remove the last item from a doubly linked list.

How does the time needed change as the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Delete node from end of doubly linked list
void deleteEnd(Node** head) {
    if (*head == NULL) return;
    Node* temp = *head;
    if (temp->next == NULL) {
        free(temp);
        *head = NULL;
        return;
    }
    while (temp->next != NULL) {
        temp = temp->next;
    }
    temp->prev->next = NULL;
    free(temp);
}
    

This code removes the last node from a doubly linked list by walking to the end and unlinking it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop that moves through nodes to find the last one.
• How many times: It runs once for each node until the end, so as many times as the list length.

How Execution Grows With Input

As the list gets longer, the time to reach the last node grows in a straight line.

Input Size (n)	Approx. Operations
10	About 10 steps to reach the end
100	About 100 steps to reach the end
1000	About 1000 steps to reach the end

Pattern observation: The steps increase directly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to delete the last node grows linearly with the list size.

Common Mistake

[X] Wrong: "Deleting the last node is always quick and takes constant time."

[OK] Correct: Without a pointer to the last node, we must walk through the whole list, so time grows with list length.

Interview Connect

Knowing how linked list operations scale helps you explain your code choices clearly and confidently.

Self-Check

"What if we kept a pointer to the last node? How would the time complexity change?"

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