Min Stack
Created: October 14, 2018 by [lek-tin]
Last updated: October 14, 2018
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
- push(x) – Push element x onto stack.
- pop() – Removes the element on top of the stack.
- top() – Get the top element.
- getMin() – Retrieve the minimum element in the stack.
Example
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
Solution
class MinStack:
def __init__(self):
"""
initialize your data structure here.
"""
self.stack = []
self.min_stack = []
def push(self, x):
"""
:type x: int
:rtype: void
"""
self.stack.append(x)
if len(self.min_stack) == 0 or self.min_stack[-1] >= x:
self.min_stack.append(x)
def pop(self):
"""
:rtype: void
"""
pop = self.stack.pop()
if pop == self.min_stack[-1]:
self.min_stack.pop()
def top(self):
"""
:rtype: int
"""
return self.stack[-1]
def getMin(self):
"""
:rtype: int
"""
return self.min_stack[-1]
# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.getMin()