Pop Using Linked List Node in DSA C - Time & Space Complexity

Time Complexity: Pop Using Linked List Node

O(1)

Understanding Time Complexity

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

How does the time needed change when the list grows bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Remove the first node from a linked list
Node* pop(Node** head) {
    if (*head == NULL) return NULL;
    Node* temp = *head;
    *head = (*head)->next;
    temp->next = NULL;
    return temp;
}
    

This code removes the first node from the linked list and returns it.

Identify Repeating Operations

• Primary operation: Accessing and updating the head pointer once.
• How many times: Exactly once per pop call, no loops or recursion.

How Execution Grows With Input

Removing the first node always takes the same steps, no matter how big the list is.

Input Size (n)	Approx. Operations
10	5 steps
100	5 steps
1000	5 steps

Pattern observation: The number of steps stays the same as the list grows.

Final Time Complexity

Time Complexity: O(1)

This means removing the first node takes the same short time no matter how big the list is.

Common Mistake

[X] Wrong: "Removing a node takes longer if the list is bigger because you have to look through all nodes."

[OK] Correct: We only remove the first node by changing the head pointer, no need to check other nodes.

Interview Connect

Knowing this helps you explain why linked lists are good for quick removals at the front, a common question in interviews.

Self-Check

"What if we wanted to remove the last node instead? How would the time complexity change?"

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