Delete from End of Doubly Linked List in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output after deleting last node from doubly linked list

What is the printed state of the doubly linked list after deleting the last node?

DSA C++

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

void deleteFromEnd(Node*& head) {
    if (!head) return;
    if (!head->next) {
        delete head;
        head = nullptr;
        return;
    }
    Node* temp = head;
    while (temp->next) {
        temp = temp->next;
    }
    temp->prev->next = nullptr;
    delete temp;
}

void printList(Node* head) {
    while (head) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
}

int main() {
    Node* head = new Node{1, nullptr, nullptr};
    head->next = new Node{2, head, nullptr};
    head->next->next = new Node{3, head->next, nullptr};
    deleteFromEnd(head);
    printList(head);
    return 0;
}

A 2 -> 3 -> null

B 1 -> 2 -> 3 -> null

C 1 -> null

D 1 -> 2 -> null

💡 Hint

Think about what happens to the last node and its previous node's next pointer.

Deleting from the end removes the last node (3). The previous node (2) now points to null, so the list ends at 2.

❓ Predict Output

intermediate

Output after deleting last node from single-node doubly linked list

What is the printed state of the doubly linked list after deleting the last node when the list has only one node?

DSA C++

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

void deleteFromEnd(Node*& head) {
    if (!head) return;
    if (!head->next) {
        delete head;
        head = nullptr;
        return;
    }
    Node* temp = head;
    while (temp->next) {
        temp = temp->next;
    }
    temp->prev->next = nullptr;
    delete temp;
}

void printList(Node* head) {
    while (head) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
}

int main() {
    Node* head = new Node{10, nullptr, nullptr};
    deleteFromEnd(head);
    printList(head);
    return 0;
}

A null

B 10 -> null

C 10

D 0 -> null

💡 Hint

Deleting the only node should leave the list empty.

When the list has only one node, deleting from the end deletes that node and sets head to nullptr, so the list is empty.

🔧 Debug

advanced

Identify the error in deleteFromEnd function

What error will occur when running this deleteFromEnd function on a doubly linked list with multiple nodes?

DSA C++

void deleteFromEnd(Node*& head) {
    if (!head) return;
    Node* temp = head;
    while (temp->next) {
        temp = temp->next;
    }
    temp->next = nullptr;
    delete temp;
}

A Syntax error due to missing semicolon

B No error, function works correctly

C Runtime error due to deleting node but not updating previous node's next pointer

D Memory leak because node is not deleted

💡 Hint

Check what happens to the previous node's next pointer after deletion.

The function deletes the last node but does not update the previous node's next pointer to nullptr, leaving a dangling pointer and causing runtime errors.

❓ Predict Output

advanced

Output after deleting last node twice from doubly linked list

What is the printed state of the doubly linked list after deleting the last node twice in a row?

DSA C++

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

void deleteFromEnd(Node*& head) {
    if (!head) return;
    if (!head->next) {
        delete head;
        head = nullptr;
        return;
    }
    Node* temp = head;
    while (temp->next) {
        temp = temp->next;
    }
    temp->prev->next = nullptr;
    delete temp;
}

void printList(Node* head) {
    while (head) {
        std::cout << head->data << " -> ";
        head = head->next;
    }
    std::cout << "null" << std::endl;
}

int main() {
    Node* head = new Node{5, nullptr, nullptr};
    head->next = new Node{10, head, nullptr};
    head->next->next = new Node{15, head->next, nullptr};
    deleteFromEnd(head);
    deleteFromEnd(head);
    printList(head);
    return 0;
}

A 5 -> null

B 5 -> 10 -> null

C 10 -> 15 -> null

D null

💡 Hint

Each deletion removes the last node, so after two deletions, one node remains.

First deletion removes 15, second removes 10, leaving 5 as the only node.

🧠 Conceptual

expert

Why is updating the previous node's next pointer necessary when deleting from end?

In a doubly linked list, why must we update the previous node's next pointer to nullptr when deleting the last node?

A To make the list circular by linking last node to head

B To avoid leaving a dangling pointer that points to deleted memory, preventing undefined behavior

C To increase the size of the list by adding a new node

D To reset the head pointer to nullptr

💡 Hint

Think about what happens if the previous node still points to the deleted node.

If the previous node's next pointer is not set to nullptr, it points to memory that was freed, causing crashes or unpredictable behavior.

Practice

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

easy

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

B. It becomes null immediately

C. It points to the head node

D. It remains unchanged

Solution

1. Step 1: Understand the tail pointer role

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

2. Step 2: Effect of deleting the last node

   When the last node is deleted, the tail must move to the previous node to remain the last node.

3. Final Answer:

   It moves to the previous node of the deleted node -> Option A

4. Quick Check:

   Tail updates 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->prev = tail; delete tail; tail = nullptr;

B. delete tail; tail = tail->prev; tail->next = nullptr;

C. tail = tail->prev; delete tail->next; tail->next = nullptr;

D. tail->next = nullptr; delete tail; tail = tail->prev;

Solution

1. Step 1: Identify correct order of operations

   First, move tail pointer to previous node, then delete the old last node, and finally set new tail's next to nullptr.

2. Step 2: Match with options

   tail = tail->prev; delete tail->next; tail->next = nullptr; correctly moves tail, deletes old node, and updates next pointer.

3. Final Answer:

   tail = tail->prev; delete tail->next; tail->next = nullptr; -> Option C

4. Quick Check:

   Move tail first, then delete old node [OK]

Hint: Move tail before deleting old node to avoid dangling pointer [OK]

Common Mistakes:

• Deleting tail before moving pointer causes invalid access
• Not setting tail->next to nullptr after deletion
• Incorrect pointer assignments causing memory leaks

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

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

Node* head = new Node{1, nullptr, nullptr};
Node* second = new Node{2, head, nullptr};
head->next = second;
Node* tail = second;

// Delete last node
Node* temp = tail;
tail = tail->prev;
tail->next = nullptr;
delete temp;

// Print list
Node* curr = head;
while(curr) {
  std::cout << curr->data << " ";
  curr = curr->next;
}

medium

A. 1 2

B. No output (empty list)

C. 2

D. 1

Solution

1. Step 1: Understand list before deletion

   List has two nodes: 1 (head) -> 2 (tail).

2. Step 2: Delete last node and update pointers

   Last node (2) is deleted, tail moves to node 1, and tail->next is set to nullptr.

3. Step 3: Print remaining list

   Only node 1 remains, so output is "1".

4. Final Answer:

   1 -> Option D

5. Quick Check:

   Last node deleted, only head remains [OK]

Hint: After deleting tail, print from head to see remaining nodes [OK]

Common Mistakes:

• Printing deleted node due to not updating tail->next
• Assuming list is empty after deletion
• Confusing head and tail pointers

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

void deleteEnd(Node*& tail) {
  delete tail;
  tail = tail->prev;
  tail->next = nullptr;
}

medium

A. Not updating head pointer causes memory leak

B. Accessing tail->prev after deleting tail causes undefined behavior

C. tail->next should be set to tail->prev

D. No error, code is correct

Solution

1. Step 1: Analyze order of operations

   The code deletes tail first, then tries to access tail->prev which is invalid after deletion.

2. Step 2: Correct approach

   First move tail pointer to previous node, then delete the old tail node, and update pointers.

3. Final Answer:

   Accessing tail->prev after deleting tail causes undefined behavior -> Option B

4. Quick Check:

   Delete after pointer update to avoid invalid access [OK]

Hint: Move pointer before delete to avoid invalid memory access [OK]

Common Mistakes:

• Deleting node before updating pointers
• Not checking if tail is nullptr before access
• Confusing order of pointer updates and deletion

5. Given a doubly linked list with nodes containing values [10, 20, 30, 40], write the steps to delete the last node and update the list correctly. What will be the final list after deletion?

hard

A. [10, 20, 30]

B. [10, 20, 30, 40]

C. [20, 30, 40]

D. [10, 20, 40]

Solution

1. Step 1: Identify the last node and tail pointer

   The last node has value 40, tail points to it.

2. Step 2: Move tail pointer to previous node and delete last node

   Tail moves to node with value 30, delete node 40, set tail->next to nullptr.

3. Step 3: Final list after deletion

   List nodes are now 10 -> 20 -> 30 with tail at 30.

4. Final Answer:

   [10, 20, 30] -> Option A

5. Quick Check:

   Last node removed, tail updated, list shortened [OK]

Hint: Delete last node, move tail back, set tail->next null [OK]

Common Mistakes:

• Not updating tail->next to nullptr
• Deleting wrong node or skipping pointer updates
• Assuming list shrinks from head instead of tail