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DSA Pythonprogramming~10 mins

Stack Implementation Using Linked List in DSA Python - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the top of the stack as empty.

DSA Python
class Stack:
    def __init__(self):
        self.top = [1]
Drag options to blanks, or click blank then click option'
A[]
BNode()
C0
DNone
Attempts:
3 left
💡 Hint
Common Mistakes
Using 0 or an empty list instead of None for top.
2fill in blank
medium

Complete the code to create a new node with the given data.

DSA Python
class Node:
    def __init__(self, data):
        self.data = data
        self.next = [1]
Drag options to blanks, or click blank then click option'
Aself
Bdata
CNone
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Setting next to data or 0 instead of None.
3fill in blank
hard

Fix the error in the push method to correctly add a new node on top of the stack.

DSA Python
def push(self, data):
    new_node = Node(data)
    new_node.next = [1]
    self.top = new_node
Drag options to blanks, or click blank then click option'
Aself.top.next
Bself.top
CNone
Dnew_node
Attempts:
3 left
💡 Hint
Common Mistakes
Linking new_node.next to None or new_node itself.
4fill in blank
hard

Fill both blanks to correctly remove the top node and update the stack.

DSA Python
def pop(self):
    if self.top is None:
        return None
    popped_node = self.top
    self.top = [1]
    popped_node.next = [2]
    return popped_node.data
Drag options to blanks, or click blank then click option'
Aself.top.next
BNone
Cself.top
Dpopped_node
Attempts:
3 left
💡 Hint
Common Mistakes
Setting top to popped_node or None incorrectly.
5fill in blank
hard

Fill all three blanks to check if the stack is empty and return the top element's data.

DSA Python
def peek(self):
    if self.top == [1]:
        return [2]
    return self.top.[3]
Drag options to blanks, or click blank then click option'
ANone
Cdata
Dnext
Attempts:
3 left
💡 Hint
Common Mistakes
Returning top or next instead of data, or wrong empty check.

Practice

(1/5)
1. What is the main advantage of implementing a stack using a linked list instead of an array?
easy
A. The stack can grow dynamically without a fixed size limit
B. It uses less memory than an array
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 a fixed size, so stacks implemented with arrays have a maximum capacity.
  2. Step 2: Linked list dynamic size advantage

    Linked lists allocate memory dynamically, so the stack can grow as needed without size limits.
  3. Final Answer:

    The stack can grow dynamically without a fixed size limit -> Option A
  4. Quick Check:

    Linked list stacks grow dynamically [OK]
Hint: Linked list stacks grow dynamically, arrays do not [OK]
Common Mistakes:
  • Thinking linked lists use less memory always
  • Confusing stack with queue behavior
  • Assuming random access is possible in stacks
2. Which of the following is the correct way to create a new node class for a linked list stack in Python?
easy
A. class Node:\n def __init__(self):\n self.data = None\n self.next = None
B. class Node:\n def __init__(self, data, next):\n self.data = data\n self.next = next
C. class Node:\n def __init__(self, data):\n self.data = data\n self.next = None
D. class Node:\n def __init__(self, data):\n self.next = data\n self.data = None

Solution

  1. Step 1: Check Node class constructor parameters

    The node should store data and a pointer to next node, initialized to None.
  2. Step 2: Verify correct attribute assignment

    class Node:\n def __init__(self, data):\n self.data = data\n self.next = None correctly assigns data and sets next to None for a new node.
  3. Final Answer:

    class Node:\n def __init__(self, data):\n self.data = data\n self.next = None -> Option C
  4. Quick Check:

    Node stores data and next pointer [OK]
Hint: Node needs data and next pointer initialized to None [OK]
Common Mistakes:
  • Swapping data and next assignments
  • Forgetting to initialize next to None
  • Adding unnecessary parameters in constructor
3. What will be the output of the following code snippet implementing push and print operations on a linked list stack?
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None

class Stack:
    def __init__(self):
        self.top = None

    def push(self, data):
        new_node = Node(data)
        new_node.next = self.top
        self.top = new_node

    def print_stack(self):
        current = self.top
        while current:
            print(current.data, end=' -> ')
            current = current.next
        print('null')

stack = Stack()
stack.push(10)
stack.push(20)
stack.push(30)
stack.print_stack()
medium
A. 30 -> 20 -> 10 -> null
B. 10 -> 20 -> 30 -> null
C. null -> 10 -> 20 -> 30
D. 30 -> null -> 20 -> 10

Solution

  1. Step 1: Trace push operations

    Push 10: top points to node with 10. Push 20: new node with 20 points to 10, top updated to 20. Push 30: new node with 30 points to 20, top updated to 30.
  2. Step 2: Trace print_stack output

    Print starts at top (30), then 20, then 10, then null.
  3. Final Answer:

    30 -> 20 -> 10 -> null -> Option A
  4. Quick Check:

    Stack top is last pushed [OK]
Hint: Stack prints from top to bottom, last pushed first [OK]
Common Mistakes:
  • Printing in insertion order instead of stack order
  • Confusing next pointer direction
  • Forgetting to print null at end
4. Identify the error in the following pop method of a linked list stack and select the correct fix:
def pop(self):
    if self.top is None:
        return None
    popped_node = self.top
    self.top = popped_node.next
    return popped_node
medium
A. Raise exception instead of returning null when stack is empty
B. Check if popped_node.next is null before assigning
C. Set popped_node.next to null before returning
D. Return popped_node.data instead of popped_node

Solution

  1. Step 1: Understand pop method return value

    Pop should return the data of the popped node, not the node object itself.
  2. Step 2: Fix return statement

    Change return popped_node to return popped_node.data to return the value.
  3. Final Answer:

    Return popped_node.data instead of popped_node -> Option D
  4. Quick Check:

    Pop returns data, not node object [OK]
Hint: Pop returns data, not node object [OK]
Common Mistakes:
  • Returning node object instead of data
  • Not updating top pointer
  • Not handling empty stack case
5. You want to implement a function reverse_stack that reverses the elements of a stack implemented using a linked list without using extra data structures. Which approach correctly achieves this?
hard
A. Traverse the linked list and swap data of nodes pairwise from start to end
B. Recursively pop all elements, then push them back in reverse order
C. Use a temporary array to store elements, then push back in reverse
D. Create a new stack and push elements from the original stack to it

Solution

  1. Step 1: Understand constraints - no extra data structures

    Options C and D use extra structures (array or new stack), so they are invalid.
  2. Step 2: Recursive approach to reverse stack

    Recursively pop elements until empty, then push them back in reverse order using call stack, reversing the stack without extra storage.
  3. Final Answer:

    Recursively pop all elements, then push them back in reverse order -> Option B
  4. Quick Check:

    Recursion reverses stack without extra data structures [OK]
Hint: Use recursion to reverse stack without extra storage [OK]
Common Mistakes:
  • Using extra arrays or stacks despite constraints
  • Trying to swap node data incorrectly
  • Not handling base case in recursion
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10 questions~5 mineasy hard