What if you could always know the smallest item instantly, no matter how many you add or remove?
Why Min Stack Design in DSA Python?
Imagine you have a stack of books and you want to quickly find the smallest book without taking all of them out.
Doing this manually means checking each book every time you want the smallest one.
Manually searching for the smallest book every time is slow and tiring.
It wastes time and can cause mistakes if you lose track.
A Min Stack keeps track of the smallest book as you add or remove books.
This way, you always know the smallest book instantly without searching.
stack = [] # To find min, loop through stack each time min_value = min(stack) if stack else None
class MinStack: def __init__(self): self.stack = [] self.min_stack = []
You can get the smallest item in constant time while still using a stack.
Tracking the lowest price of stocks while adding and removing daily prices quickly.
Manual search for minimum is slow and error-prone.
Min Stack keeps track of minimum efficiently.
Allows instant access to minimum value during stack operations.
Practice
Min Stack data structure?Solution
Step 1: Understand the Min Stack concept
A Min Stack is designed to support normal stack operations and also return the minimum element quickly.Step 2: Identify the key feature
The key feature is retrieving the minimum element in constant time, not sorting or reversing.Final Answer:
To retrieve the minimum element in constant time along with stack operations -> Option CQuick Check:
Min Stack = fast minimum retrieval [OK]
- Thinking Min Stack sorts elements
- Assuming it stores only minimum elements
- Confusing with reversing stack order
Solution
Step 1: Check correct attribute initialization
Attributes must be initialized withselfto be instance variables.Step 2: Verify list initialization
Bothstackandmin_stackshould be empty lists initialized withself.Final Answer:
class MinStack:\n def __init__(self):\n self.stack = []\n self.min_stack = [] -> Option DQuick Check:
Use self for instance variables [OK]
- Not using self for attributes
- Initializing attributes as None instead of lists
- Assigning self to a list directly
min_stack = MinStack() min_stack.push(3) min_stack.push(5) min_stack.push(2) min_stack.push(1) min_stack.pop() print(min_stack.get_min())
Solution
Step 1: Trace push operations and min tracking
Push 3 (min=3), push 5 (min=3), push 2 (min=2), push 1 (min=1).Step 2: Pop removes 1, min updates to 2
After popping 1, the minimum is the next smallest, which is 2.Final Answer:
2 -> Option AQuick Check:
Pop removes min 1, next min = 2 [OK]
- Assuming min stays 1 after pop
- Confusing push order with min updates
- Ignoring min stack updates on pop
pop method and choose the correct fix:
def pop(self):
if self.stack:
self.stack.pop()
self.min_stack.pop()Solution
Step 1: Understand pop logic in Min Stack
Pop from min_stack only if the popped element is the current minimum to keep min_stack accurate.Step 2: Identify bug and fix
Current code always pops min_stack, which can cause errors if min_stack is out of sync.Final Answer:
Pop from min_stack only if popped element equals current min -> Option BQuick Check:
Pop min_stack conditionally to keep min correct [OK]
- Always popping min_stack without condition
- Not checking min_stack emptiness
- Changing pop order incorrectly
push, pop, top, and get_min all in O(1) time. Which approach is best?Solution
Step 1: Analyze time complexity requirements
All operations must be O(1), so scanning or traversing is not allowed.Step 2: Choose data structure approach
Using two stacks, one for elements and one for minimums, allows constant time min retrieval.Final Answer:
Use two stacks: one for all elements, one for current minimums -> Option AQuick Check:
Two stacks = O(1) min retrieval [OK]
- Using scanning which is O(n)
- Using queue which doesn't support stack operations
- Using linked list traversal for min
Ready to prove it?
Pass the test and this pattern moves unknown
