max-stack.py

# Time:  push:    O(1)
#        pop:     O(n), there is no built-in SortedDict in python. If applied, it could be reduced to O(logn)
#        popMax:  O(n)
#        top:     O(1)
#        peekMax: O(1)
# Space: O(n), n is the number of values in the current stack

import collections


class MaxStack(object):

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.__idx_to_val = collections.defaultdict(int)
        self.__val_to_idxs = collections.defaultdict(list)
        self.__top = None
        self.__max = None


    def push(self, x):
        """
        :type x: int
        :rtype: void
        """
        idx = self.__val_to_idxs[self.__top][-1]+1 if self.__val_to_idxs else 0
        self.__idx_to_val[idx] = x
        self.__val_to_idxs[x].append(idx)
        self.__top = x
        self.__max = max(self.__max, x)


    def pop(self):
        """
        :rtype: int
        """
        val = self.__top
        self.__remove(val)
        return val


    def top(self):
        """
        :rtype: int
        """
        return self.__top


    def peekMax(self):
        """
        :rtype: int
        """
        return self.__max


    def popMax(self):
        """
        :rtype: int
        """
        val = self.__max
        self.__remove(val)
        return val


    def __remove(self, val):
        idx = self.__val_to_idxs[val][-1]
        self.__val_to_idxs[val].pop();
        if not self.__val_to_idxs[val]:
            del self.__val_to_idxs[val]
        del self.__idx_to_val[idx]
        if val == self.__top:
            self.__top = self.__idx_to_val[max(self.__idx_to_val.keys())] if self.__idx_to_val else None
        if val == self.__max:
            self.__max = max(self.__val_to_idxs.keys()) if self.__val_to_idxs else None