LC155. Min Stack¶
Problem Description¶
LeetCode Problem 155: Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
Implement the MinStack
class:
MinStack()
initializes the stack object.void push(int val)
pushes the elementval
onto the stack.void pop()
removes the element on the top of the stack.int top()
gets the top element of the stack.int getMin()
retrieves the minimum element in the stack.
You must implement a solution with O(1)
time complexity for each function.
Clarification¶
- Normal stack operation + retrieve the minimum
- O(1) time complexity
- What value should return for invalid cases: pop(), getMin, top() if no values in stack
Assumption¶
Solution¶
Approach - Stack of Value/Min Paris¶
Use one stack to store (value, min)
pair.
from collections import deque
class MinStack:
def __init__(self):
self.stack = deque()
def push(self, val: int) -> None:
# Push (val, min) pair
# If the stack is empty, the min value is the first value
if not self.stack:
self.stack.append((val, val))
return
current_min = self.stack[-1][1]
self.stack.append((val, min(val, current_min)))
def pop(self) -> None:
self.stack.pop()
def top(self) -> int:
return self.stack[-1][0]
def getMin(self) -> int:
return self.stack[-1][1]
Complexity Analysis¶
- Time complexity: \(O(1)\) for all operations
Similar to stand stack operations. - Space complexity: \(O(n)\)
The worse case to save all (value, min) pairs. The space used is \(O(2 n) = O(n)\).