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Stack 'Em Up!
TIP102 Unit 6 Session 2 Advanced (Click for link to problem statements)
- 💡 Difficulty: Easy
- ⏰ Time to complete: 10-20 mins
- 🛠️ Topics: Stack, Linked List, LIFO
Understand what the interviewer is asking for by using test cases and questions about the problem.
- Q: What does the problem ask for?
- A: The problem asks to implement a stack using a singly linked list.
- A: The stack should support standard stack operations like
push
,pop
,peek
, andis_empty
.
- Q: What should be returned?
- A: The
pop
method should return the value at the top of the stack and remove it. - A: The
peek
method should return the value at the top of the stack without removing it. - A: The
is_empty
method should returnTrue
if the stack is empty andFalse
otherwise.
- A: The
HAPPY CASE
Input:
- stack = Stack()
- stack.push(('Educated', 'Tara Westover'))
- stack.push(('Gone Girl', 'Gillian Flynn'))
- stack.push(('Dune', 'Frank Herbert'))
Output:
- Dune -> Gone Girl -> Educated
- Peek: ('Dune', 'Frank Herbert')
- Pop: ('Dune', 'Frank Herbert')
- Pop: ('Gone Girl', 'Gillian Flynn')
- Is Empty: False
- Pop: ('Educated', 'Tara Westover')
- Is Empty: True
EDGE CASE
Input:
- stack = Stack()
- Pop: None
- Peek: None
- Is Empty: True
Output:
- None
Explanation:
- The stack is empty, so `pop` and `peek` return `None`, and `is_empty` returns `True`.
Match what this problem looks like to known categories of problems, e.g. Stack, Linked List, etc.
For Stack problems involving Linked List, we want to consider the following approaches:
-
Node Management: Manage the nodes of the linked list such that
push
adds a node to the front,pop
removes the front node, andpeek
returns the value of the front node. -
Pointer Handling: Ensure that the
front
pointer always points to the top of the stack.
Plan the solution with appropriate visualizations and pseudocode.
General Idea: We will implement the stack using a singly linked list where the front
pointer always points to the top of the stack.
1) Implement the `Node` class to create nodes for the linked list.
2) Implement the `Stack` class with the following methods:
- `__init__`: Initialize an empty stack with `front` set to `None`.
- `is_empty`: Return `True` if `front` is `None`, otherwise return `False`.
- `push`: Create a new node, link it to the current `front`, and update `front`.
- `pop`: Remove and return the value of the node at `front`. Update `front` to point to the next node.
- `peek`: Return the value of the node at `front` without removing it.
- Forgetting to update the
front
pointer correctly afterpush
andpop
. - Incorrectly handling the empty stack case in
pop
andpeek
.
Implement the code to solve the algorithm.
class Node:
def __init__(self, value, next=None):
self.value = value
self.next = next
class Stack:
def __init__(self):
self.front = None
def is_empty(self):
return self.front is None
def push(self, value):
new_node = Node(value)
new_node.next = self.front
self.front = new_node
def pop(self):
if self.is_empty():
return None
popped_node = self.front
self.front = self.front.next
return popped_node.value
def peek(self):
if self.is_empty():
return None
return self.front.value
Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
- Example: Use the provided stack operations to verify that the class methods behave as expected.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N
represents the number of elements in the stack.
-
Time Complexity:
O(1)
for all operations (push
,pop
,peek
, andis_empty
) because they all involve only a constant number of operations. -
Space Complexity:
O(N)
because the space used by the stack is proportional to the number of elements in the stack.