Pop Using Linked List Node in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after popping the head node?

Consider a singly linked list with nodes 10 -> 20 -> 30 -> null. After popping the head node, what is the new list?

DSA C++

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* pop(Node* head) {
    if (!head) return nullptr;
    Node* temp = head;
    head = head->next;
    delete temp;
    return head;
}

int main() {
    Node* head = new Node(10);
    head->next = new Node(20);
    head->next->next = new Node(30);
    head = pop(head);
    Node* curr = head;
    while (curr) {
        std::cout << curr->data << " -> ";
        curr = curr->next;
    }
    std::cout << "null";
    return 0;
}

A 20 -> 30 -> null

B 10 -> 20 -> 30 -> null

C 30 -> null

D null

💡 Hint

Popping removes the first node and returns the next node as the new head.

The pop function removes the head node (10) and returns the next node (20) as the new head. The list becomes 20 -> 30 -> null.

❓ Predict Output

intermediate

What happens when popping from an empty linked list?

Given an empty linked list (head is nullptr), what is the result after calling pop(head)?

DSA C++

Node* pop(Node* head) {
    if (!head) return nullptr;
    Node* temp = head;
    head = head->next;
    delete temp;
    return head;
}

int main() {
    Node* head = nullptr;
    head = pop(head);
    if (!head) std::cout << "List is empty";
    else std::cout << "List is not empty";
    return 0;
}

A List contains one node

B List is not empty

C Runtime error

D List is empty

💡 Hint

Check the condition when head is nullptr before popping.

The pop function returns nullptr immediately if the list is empty. So the list remains empty and prints 'List is empty'.

🔧 Debug

advanced

Identify the error in this pop function implementation

What error will this pop function cause when called on a non-empty list?

DSA C++

Node* pop(Node* head) {
    Node* temp = head->next;
    delete head;
    return temp;
}

int main() {
    Node* head = new Node(5);
    head->next = new Node(10);
    head = pop(head);
    std::cout << head->data << " -> null";
    return 0;
}

A Runtime error: dereferencing null pointer

B No error, outputs: 10 -> null

C Memory leak error

D Compilation error

💡 Hint

Check if head is nullptr before accessing head->next.

The function assumes head is not nullptr and deletes it safely. It returns head->next as new head. The output is 10 -> null.

❓ Predict Output

advanced

What is the output after popping all nodes one by one?

Given a linked list 1 -> 2 -> 3 -> null, pop nodes until the list is empty. What is printed after each pop?

DSA C++

Node* pop(Node* head) {
    if (!head) return nullptr;
    Node* temp = head;
    head = head->next;
    delete temp;
    return head;
}

int main() {
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    while (head) {
        std::cout << head->data << " -> ";
        head = pop(head);
    }
    std::cout << "null";
    return 0;
}

A 1 -> 2 -> 3 -> null

B 1 -> 2 -> null

C 1 -> null

D null

💡 Hint

The code prints the head data before popping it in each iteration.

Each iteration prints the current head's data, then pops it. The output is 1 -> 2 -> 3 -> null.

🧠 Conceptual

expert

What is the time complexity of popping the head node from a singly linked list?

Consider a singly linked list with n nodes. What is the time complexity of removing the head node (pop operation)?

A O(n) - linear time

B O(log n) - logarithmic time

C O(1) - constant time

D O(n^2) - quadratic time

💡 Hint

Popping the head node only changes a few pointers.

Removing the head node requires updating the head pointer and deleting one node, which takes constant time regardless of list size.

Practice

1. What does the pop() operation do in a linked list when implemented using nodes?

easy

A. Removes the first node and returns its value

B. Adds a new node at the end

C. Deletes all nodes in the list

D. Finds a node with a specific value

Solution

1. Step 1: Understand pop operation in linked list context

   Pop means removing the top or first element from the list, similar to popping from a stack.

2. Step 2: Identify what happens to the first node

   The first node is removed and its value is returned, and the head pointer moves to the next node.

3. Final Answer:

   Removes the first node and returns its value -> Option A

4. Quick Check:

   Pop = Remove first node [OK]

Hint: Pop removes the first node like taking the top card [OK]

Common Mistakes:

• Thinking pop adds a node
• Confusing pop with delete all
• Assuming pop searches for a value

2. Which of the following is the correct syntax to remove the first node in a singly linked list in C++?

easy

A. delete head->next;

B. head->next = head;

C. head = head->next;

D. head = NULL;

Solution

1. Step 1: Understand how to move head pointer

   To remove the first node, head must point to the second node, which is head->next.

2. Step 2: Check each option

   head = head->next; correctly assigns head to head->next. head->next = head; reverses pointers incorrectly. delete head->next; deletes the second node, not the first. head = NULL; sets head to NULL, deleting entire list.

3. Final Answer:

   head = head->next; -> Option C

4. Quick Check:

   Move head to next node = head = head->next; [OK]

Hint: Pop means head points to next node [OK]

Common Mistakes:

• Deleting wrong node
• Setting head to NULL incorrectly
• Reversing pointer assignment

3. What will be the output after executing the following code?

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

Node* head = new Node{10, nullptr};
head->next = new Node{20, nullptr};

int popValue = head->data;
head = head->next;

std::cout << popValue << " -> " << (head ? std::to_string(head->data) : "null") << std::endl;

medium

A. 20 -> 10

B. 10 -> 20

C. 10 -> null

D. 20 -> null

Solution

1. Step 1: Trace initial list nodes

   head points to node with data 10, which points to node with data 20.

2. Step 2: Execute pop operation

   popValue stores 10 (head's data), then head moves to next node (20).

3. Step 3: Print values

   Output prints popValue (10) and head's data (20), separated by ' -> '.

4. Final Answer:

   10 -> 20 -> Option B

5. Quick Check:

   Pop value = 10, new head = 20 [OK]

Hint: Pop returns old head data, head moves to next node [OK]

Common Mistakes:

• Printing new head data as pop value
• Confusing order of output
• Not updating head pointer

4. Identify the error in the following pop function for a singly linked list:

int pop(Node*& head) {
  int val = head->data;
  delete head;
  head = head->next;
  return val;
}

medium

A. Accessing head->next after deleting head

B. Not checking if head is NULL before pop

C. Returning value before deleting node

D. Function should return void

Solution

1. Step 1: Analyze order of operations

   The code deletes head node first, then tries to access head->next, which is invalid after deletion.

2. Step 2: Identify correct order

   Must save head->next before deleting head to avoid accessing freed memory.

3. Final Answer:

   Accessing head->next after deleting head -> Option A

4. Quick Check:

   Delete before accessing next = error [OK]

Hint: Save next pointer before deleting node [OK]

Common Mistakes:

• Deleting node before saving next
• Ignoring NULL head check
• Wrong return type assumption

5. Given a linked list with nodes containing values [5, 10, 15], write the correct sequence of operations to pop all nodes and print their values in order using C++ pointers.
Which code snippet correctly implements this?

hard

A. while(head) { int val = head->data; head = head->next; std::cout << val << " "; }

B. while(head) { std::cout << head->data << " "; delete head; }

C. while(head) { delete head; head = head->next; }

D. while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; }

Solution

1. Step 1: Understand safe pop and print

   We must print the current node's data, save the current node pointer, move head to next, then delete the saved node.

2. Step 2: Check each option

   while(head) { int val = head->data; head = head->next; std::cout << val << " "; } does not delete nodes, causing memory leak. while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; } correctly saves node in temp, moves head, deletes temp, and prints value. while(head) { delete head; head = head->next; } deletes head before moving, causing invalid access. while(head) { std::cout << head->data << " "; delete head; } deletes head but does not move head pointer, causing infinite loop or crash.

3. Final Answer:

   while(head) { int val = head->data; Node* temp = head; head = head->next; delete temp; std::cout << val << " "; } -> Option D

4. Quick Check:

   Save node, move head, delete node, print value [OK]

Hint: Save node before delete, move head after [OK]

Common Mistakes:

• Deleting node before moving head
• Not deleting nodes causing leaks
• Not moving head pointer