foo.py

class Solution(object):
    def __init__(self):
        """
        h_items is a hash (key = item, value = number of items purchased (counts))
        items_sorted is the list of items sorted by the counts
        space cost: O(n*n) ~ O(n)
        """
        self.h_items = {}
        self.items_sorted = []

    def itemPurchased(self, i_name):
        """
        Since getRanking will be called more often, let's do the sorting of
        the data right here.
        time cost: O(nlog(n)) (Since python uses quicksort like internally)
        """
        if i_name not in self.h_items:
            self.h_items[i_name] = 0
        self.h_items[i_name] += 1
        self.__sort()

    def getRanking(self):
        """
        time cost: O(1)
        """
        return self.items_sorted

    def getTopTenProducts(self):
        """
        time cost: O(1)
        """
        return self.items_sorted[0:10]

    def __sort(self):
        """
        Convert the hash to a list and order by the counts.
        Finally, update items_sorted by only keeping the items names
        Time cost: O(nlog(n))
        """
        self.items_sorted = []
        for item, count in self.h_items.iteritems():
            self.items_sorted.append([item, count])
        self.items_sorted.sort(key=lambda i: i[1], reverse=True)
        self.items_sorted = [e[0] for e in self.items_sorted]

solution_enable_basic_test = True
if solution_enable_basic_test:
    s = Solution()

    for i in range(0,20):
        s.itemPurchased('i' + str(i))

    s.itemPurchased('i1')
    s.itemPurchased('i1')
    assert(s.getRanking()[0] == 'i1')

    s.itemPurchased('i2')
    assert(s.getRanking()[1] == 'i2')

    assert(len(s.getRanking()) == 20)
    assert(len(s.getTopTenProducts()) == 10)