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Min-Stackpy.py
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Min-Stackpy.py
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# Min Stack
# 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.
# Hide Hint #1
# Consider each node in the stack having a minimum value.
class MinStack:
def __init__(self):
self.s = []
def push(self, x):
cur_min = self.getMin()
if cur_min == None or x < cur_min:
cur_min = x
self.s.append((x,cur_min))
def pop(self):
self.s.pop()
def top(self):
if len(self.s) == 0:
return None
return self.s[-1][0]
def getMin(self):
if len(self.s) == 0:
return None
return self.s[-1][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()
#More Optimal Solutions
# Solution 1
# class MinStack(object):
# def __init__(self):
# self.stack = []
# self.mins = []
# def push(self, x):
# """
# :type x: int
# :rtype: None
# """
# self.stack.append(x)
# if not self.mins:
# self.mins.append(x)
# else:
# if x <= self.mins[-1]:
# self.mins.append(x)
# def pop(self):
# """
# :rtype: None
# """
# temp = self.stack.pop()
# if temp == self.mins[-1]:
# self.mins.pop()
# def top(self):
# """
# :rtype: int
# """
# return self.stack[-1]
# def getMin(self):
# """
# :rtype: int
# """
# return self.mins[-1]