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 from the linked list?

Consider a singly linked list with nodes containing values 10 -> 20 -> 30 -> null. The pop operation removes the head node. What is the state of the list after one pop?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    Node* head = malloc(sizeof(Node));
    head->data = 10;
    head->next = malloc(sizeof(Node));
    head->next->data = 20;
    head->next->next = malloc(sizeof(Node));
    head->next->next->data = 30;
    head->next->next->next = NULL;

    Node* popped = pop(&head);
    printList(head);
    free(popped);
    // Free remaining nodes omitted for brevity
    return 0;
}

A 30 -> null

B 20 -> 30 -> null

C 10 -> 20 -> 30 -> null

D null

💡 Hint

Pop removes the first node and moves the head pointer to the next node.

The pop function removes the head node (value 10) and updates the head pointer to the next node (value 20). The list after popping is 20 -> 30 -> null.

❓ Predict Output

intermediate

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

Given a linked list 5 -> 15 -> 25 -> null, pop nodes until the list is empty. What is printed after each pop?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    Node* head = malloc(sizeof(Node));
    head->data = 5;
    head->next = malloc(sizeof(Node));
    head->next->data = 15;
    head->next->next = malloc(sizeof(Node));
    head->next->next->data = 25;
    head->next->next->next = NULL;

    for (int i = 0; i < 3; i++) {
        Node* popped = pop(&head);
        printList(head);
        free(popped);
    }
    return 0;
}

A

15 -> 25 -> null
25 -> null
null

B

5 -> 15 -> 25 -> null
15 -> 25 -> null
25 -> null

C

15 -> 25 -> null
null
null

D

null
null
null

💡 Hint

Each pop removes the head node and updates the list.

After first pop, list is 15 -> 25 -> null; after second pop, 25 -> null; after third pop, list is empty (null).

❓ Predict Output

advanced

What error occurs when popping from an empty linked list?

What happens if you call pop on a linked list where the head pointer is NULL?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

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

int main() {
    Node* head = NULL;
    Node* popped = pop(&head);
    if (popped == NULL) {
        printf("List is empty, nothing to pop.\n");
    }
    return 0;
}

A Memory leak error

B Segmentation fault

C Null pointer dereference error

D List is empty, nothing to pop.

💡 Hint

Check if head is NULL before popping.

The pop function checks if the head is NULL and returns NULL safely, so no crash occurs. The program prints the message indicating the list is empty.

❓ Predict Output

advanced

What is the output after popping the last node from the list?

Given a linked list with a single node 42 -> null, what is the list state after popping once?

DSA C

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

#include <stdio.h>
#include <stdlib.h>

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

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

int main() {
    Node* head = malloc(sizeof(Node));
    head->data = 42;
    head->next = NULL;

    Node* popped = pop(&head);
    printList(head);
    free(popped);
    return 0;
}

A Segmentation fault

B 42 -> null

C null

D 42

💡 Hint

Popping the only node empties the list.

After popping the only node, head becomes NULL, so printing the list shows 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 the pop operation that removes the head node?

A O(1) - constant time

B O(log n) - logarithmic time

C O(n) - linear time

D O(n^2) - quadratic time

💡 Hint

Pop only changes the head pointer and frees one node.

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

Practice

1. What does the pop operation do in a linked list implemented as a stack?

easy

A. Deletes the entire linked list

B. Adds a new node at the end of the linked list

C. Removes and returns the first node (top) of the linked list

D. Returns the last node without removing it

Solution

1. Step 1: Understand pop operation in stack

   Pop removes the top element from the stack, which is the first node in a linked list stack.

2. Step 2: Apply to linked list

   In a linked list used as a stack, the first node is removed and returned during pop.

3. Final Answer:

   Removes and returns the first node (top) of the linked list -> Option C

4. Quick Check:

   Pop = remove first node [OK]

Hint: Pop removes the first node in linked list stack [OK]

Common Mistakes:

• Thinking pop removes last node
• Confusing pop with push
• Assuming pop deletes entire list

2. Which of the following is the correct syntax to pop a node from a linked list stack in C?

easy

A. node* pop(node** head) { node* temp = *head; *head = (*head)->next; return temp; }

B. node pop(node head) { head = head->next; return head; }

C. void pop(node* head) { head = head->next; }

D. node* pop(node* head) { head = head->next; return head; }

Solution

1. Step 1: Check parameter passing for head pointer

   To modify the head pointer, we pass a pointer to pointer (node** head).

2. Step 2: Verify pop logic

   Store current head in temp, move head to next node, return temp node.

3. Final Answer:

   node* pop(node** head) { node* temp = *head; *head = (*head)->next; return temp; } -> Option A

4. Quick Check:

   Pointer to pointer needed to update head [OK]

Hint: Use node** to update head pointer in pop [OK]

Common Mistakes:

• Passing head by value instead of pointer to pointer
• Returning wrong node or void
• Not updating head pointer correctly

3. What will be the output of the following code snippet?

typedef struct node { int data; struct node* next; } node;

node* pop(node** head) {
  if (*head == NULL) return NULL;
  node* temp = *head;
  *head = (*head)->next;
  return temp;
}

int main() {
  node n3 = {3, NULL};
  node n2 = {2, &n3};
  node n1 = {1, &n2};
  node* head = &n1;
  node* popped = pop(&head);
  printf("%d -> %d -> %d -> NULL", head->data, head->next->data, head->next->next->data);
  return 0;
}

medium

A. 2 -> 3 -> 0 -> NULL

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

C. 2 -> 3 -> NULL

D. 2 -> 3 -> (segmentation fault)

Solution

1. Step 1: Trace pop operation

   Pop removes node with data 1, head now points to node with data 2.

2. Step 2: Trace printf arguments

   head->data = 2, head->next->data = 3, head->next->next = NULL, accessing ->data causes segmentation fault after printing "2 -> 3 -> " .

3. Final Answer:

   2 -> 3 -> (segmentation fault) -> Option D

4. Quick Check:

   Accessing NULL->data causes crash [OK]

Hint: Accessing next of last node causes crash [OK]

Common Mistakes:

• Assuming NULL prints as 'NULL' string
• Ignoring segmentation fault on NULL dereference
• Misreading linked list after pop

4. Identify the bug in this pop function and select the correct fix:

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

medium

A. Pass head as node** and update *head inside function

B. Return head instead of temp

C. Set temp->next = NULL before return

D. No bug, function is correct

Solution

1. Step 1: Analyze head pointer update

   Function updates local copy of head, original head outside remains unchanged.

2. Step 2: Fix pointer update

   Pass pointer to pointer (node**) to modify original head pointer.

3. Final Answer:

   Pass head as node** and update *head inside function -> Option A

4. Quick Check:

   Use node** to update head pointer [OK]

Hint: Use node** to update head pointer in pop [OK]

Common Mistakes:

• Updating local copy of head only
• Not passing pointer to pointer
• Returning wrong node

5. Given a linked list stack with nodes containing integers, write the logic to pop all nodes and print their data in order. Which approach is correct?

hard

A. Traverse list without popping and print data

B. Use a loop: while head != NULL, pop node, print data, free node

C. Pop once, print data, then stop

D. Recursively call pop until head is NULL, then print all data after recursion

Solution

1. Step 1: Understand popping all nodes

   Pop removes top node each time, so loop until list is empty.

2. Step 2: Print and free nodes in loop

   Print data of popped node immediately, then free memory to avoid leaks.

3. Final Answer:

   Use a loop: while head != NULL, pop node, print data, free node -> Option B

4. Quick Check:

   Loop pop-print-free until empty [OK]

Hint: Loop pop until head NULL, print and free each node [OK]

Common Mistakes:

• Recursing without printing in correct order
• Not freeing nodes causing memory leaks
• Stopping after one pop