# Time:  O(n)
# Space: O(n)

# Given a chemical formula (given as a string), return the count of each atom.
#
# An atomic element always starts with an uppercase character,
# then zero or more lowercase letters, representing the name.
#
# 1 or more digits representing the count of that element may follow if the count is greater than 1.
# If the count is 1, no digits will follow. For example, H2O and H2O2 are possible, but H1O2 is impossible.
#
# Two formulas concatenated together produce another formula. For example, H2O2He3Mg4 is also a formula.
#
# A formula placed in parentheses, and a count (optionally added) is also a formula.
# For example, (H2O2) and (H2O2)3 are formulas.
#
# Given a formula, output the count of all elements as a string in the following form:
# the first name (in sorted order), followed by its count (if that count is more than 1),
# followed by the second name (in sorted order),
# followed by its count (if that count is more than 1), and so on.
#
# Example 1:
# Input: 
# formula = "H2O"
# Output: "H2O"
# Explanation: 
# The count of elements are {'H': 2, 'O': 1}.
#
# Example 2:
# Input: 
# formula = "Mg(OH)2"
# Output: "H2MgO2"
# Explanation: 
# The count of elements are {'H': 2, 'Mg': 1, 'O': 2}.
#
# Example 3:
# Input: 
# formula = "K4(ON(SO3)2)2"
# Output: "K4N2O14S4"
# Explanation: 
# The count of elements are {'K': 4, 'N': 2, 'O': 14, 'S': 4}.
# Note:
#
# All atom names consist of lowercase letters, except for the first character which is uppercase.
# The length of formula will be in the range [1, 1000].
# formula will only consist of letters, digits, and round parentheses,
# and is a valid formula as defined in the problem.

class Solution(object):
    def countOfAtoms(self, formula):
        """
        :type formula: str
        :rtype: str
        """
        parse = re.findall(r"([A-Z][a-z]*)(\d*)|(\()|(\))(\d*)", formula)
        stk = [collections.Counter()]
        for name, m1, left_open, right_open, m2 in parse:
            if name:
              stk[-1][name] += int(m1 or 1)
            if left_open:
              stk.append(collections.Counter())
            if right_open:
                top = stk.pop()
                for k, v in top.iteritems():
                  stk[-1][k] += v * int(m2 or 1)

        return "".join(name + (str(stk[-1][name]) if stk[-1][name] > 1 else '') \
                       for name in sorted(stk[-1]))

Create split-linked-list-in-parts.py