Stack Implementation Using Linked List in DSA C++ - Time & Space Complexity

Time Complexity: Stack Implementation Using Linked List

O(1)

Understanding Time Complexity

We want to understand how fast stack operations run when using a linked list.

How does the time to push or pop change as the stack grows?

Scenario Under Consideration

Analyze the time complexity of the push operation in this stack implementation.

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

class Stack {
  Node* top = nullptr;
public:
  void push(int val) {
    Node* newNode = new Node{val, top};
    top = newNode;
  }
};
    

This code adds a new element to the top of the stack using a linked list.

Identify Repeating Operations

Look for loops or repeated steps inside push.

• Primary operation: Creating a new node and linking it to the current top.
• How many times: Exactly once per push call, no loops involved.

How Execution Grows With Input

Each push does the same small work no matter how big the stack is.

Input Size (n)	Approx. Operations
10	1 push = 1 operation
100	1 push = 1 operation
1000	1 push = 1 operation

Pattern observation: The work stays the same no matter how many items are in the stack.

Final Time Complexity

Time Complexity: O(1)

This means each push operation takes the same small amount of time regardless of stack size.

Common Mistake

[X] Wrong: "Push takes longer as the stack grows because it has to move all elements."

[OK] Correct: The linked list lets us add a new node at the front instantly without moving others.

Interview Connect

Knowing that stack push and pop are fast and constant time helps you explain efficient data handling in interviews.

Self-Check

"What if we used an array instead of a linked list for the stack? How would the time complexity of push change?"

Practice

1. What is the main advantage of implementing a stack using a linked list instead of an array?

easy

A. It uses less memory than an array

B. The stack can grow dynamically without a fixed size limit

C. It allows random access to elements

D. It is easier to sort the stack elements

Solution

1. Step 1: Understand stack size limitation in arrays

   Arrays have fixed size, so stack implemented with arrays has a maximum capacity.

2. Step 2: Linked list dynamic size property

   Linked lists can grow or shrink dynamically by adding or removing nodes, so stack size is not fixed.

3. Final Answer:

   The stack can grow dynamically without a fixed size limit -> Option B

4. Quick Check:

   Linked list stack size = dynamic growth [OK]

Hint: Linked list stacks grow dynamically, arrays have fixed size [OK]

Common Mistakes:

• Thinking linked list uses less memory always
• Assuming stack allows random access
• Confusing sorting with stack operations

2. Which of the following is the correct syntax to create a new node in a linked list stack in C++?

easy

A. Node* newNode = new Node(value);

B. Node newNode = Node(value);

C. Node newNode = new Node(value);

D. Node* newNode = Node(value);

Solution

1. Step 1: Understand dynamic memory allocation in C++

   To create a node dynamically, use 'new' keyword and pointer type.

2. Step 2: Check syntax correctness

   Node* newNode = new Node(value); uses pointer and 'new' correctly: Node* newNode = new Node(value);

3. Final Answer:

   Node* newNode = new Node(value); -> Option A

4. Quick Check:

   Dynamic node creation = new + pointer [OK]

Hint: Use 'new' with pointer to create nodes dynamically [OK]

Common Mistakes:

• Omitting pointer (*) when using new
• Using stack allocation instead of dynamic
• Assigning new object to non-pointer variable

3. What will be the output after executing the following code snippet for a stack implemented using linked list?

Stack s;
s.push(10);
s.push(20);
s.push(30);
s.pop();
s.push(40);
s.display();

Assuming display prints elements from top to bottom.

medium

A. 40 -> 30 -> 20 -> 10 -> null

B. 30 -> 20 -> 10 -> null

C. 20 -> 10 -> null

D. 40 -> 20 -> 10 -> null

Solution

1. Step 1: Trace push and pop operations

   Push 10, stack: 10; push 20, stack: 20->10; push 30, stack: 30->20->10; pop removes 30, stack: 20->10; push 40, stack: 40->20->10.

2. Step 2: Determine display output

   Display prints from top to bottom: 40 -> 20 -> 10 -> null.

3. Final Answer:

   40 -> 20 -> 10 -> null -> Option D

4. Quick Check:

   Stack after ops = 40->20->10 [OK]

Hint: Track push/pop stepwise to find top element [OK]

Common Mistakes:

• Forgetting pop removes top element
• Assuming push adds at bottom
• Mixing order of elements in display

4. Identify the error in the following pop function of a linked list stack:

int pop() {
  if (top == NULL) {
    cout << "Stack Underflow";
    return -1;
  }
  int val = top->data;
  delete top;
  top = top->next;
  return val;
}

medium

A. Returning -1 on empty stack is incorrect

B. Missing check for stack overflow

C. Accessing top->next after deleting top causes undefined behavior

D. Not updating top pointer after pop

Solution

1. Step 1: Analyze order of operations in pop

   Code deletes top node before accessing top->next, which is unsafe.

2. Step 2: Identify correct order

   Must save top->next in a temp pointer before deleting top, then update top.

3. Final Answer:

   Accessing top->next after deleting top causes undefined behavior -> Option C

4. Quick Check:

   Delete before next access = error [OK]

Hint: Save next pointer before deleting current node [OK]

Common Mistakes:

• Deleting node before updating pointer
• Ignoring stack underflow check
• Returning wrong values on empty stack

5. You want to implement a function to reverse a stack implemented using a linked list without using extra data structures. Which approach is correct?

hard

A. Use recursion to pop all elements and push them back in reverse order

B. Traverse the linked list and swap data of nodes pairwise

C. Create a new linked list and copy elements in reverse order

D. Use a queue to reorder elements and then push back to stack

Solution

1. Step 1: Understand constraints

   No extra data structures allowed, so cannot use queue or new list.

2. Step 2: Use recursion to reverse stack

   Recursion can pop all elements, then push them back in reverse order naturally.

3. Step 3: Why other options fail

   Swapping data pairwise does not reverse entire stack; copying needs extra list; queue is extra structure.

4. Final Answer:

   Use recursion to pop all elements and push them back in reverse order -> Option A

5. Quick Check:

   Recursion reverses stack without extra structures [OK]

Hint: Recursion can reverse stack without extra memory [OK]

Common Mistakes:

• Using extra data structures despite constraints
• Trying to swap data instead of nodes
• Not understanding recursion stack behavior