Longest Common Subsequence in Python (version 2.0)

In one of my previous posts I was implemented the Longest Common Subsequence algorithm using string shifting. Now I was enhanced the algorithm, simplified the code, and fixed some issues. Following the SC:

#!/usr/bin/env python
# -*- coding: utf-8 -*-
#
# Longest Common Subsequence, Shift algorithm.
# Copyright (C) 2012  Gonzalo Exequiel Pedone
#
# This program is free software: you can redistribute it and/or modify
# it under the terms of the GNU General Public License as published by
# the Free Software Foundation, either version 3 of the License, or
# (at your option) any later version.
#
# This program is distributed in the hope that it will be useful,
# but WITHOUT ANY WARRANTY; without even the implied warranty of
# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
# GNU General Public License for more details.
#
# You should have received a copy of the GNU General Public License
# along with This program. If not, see <http://www.gnu.org/licenses/>.
#
# Email   : hipersayan DOT x AT gmail DOT com
# Web-Site: http://hipersayanx.blogspot.com/
#
# The shift algorithm consist on finding the Longest Common Subsequence between
# a sequence 'a' and 'b' by comparing the last element of 'a' with the first
# element of 'b', and finding the longest sequence between the common elements,
# then shifting 'a' to the right, and comparing the next two elements, and so
# on, until reach the right wall, then shifting 'b' to the left until reach the
# left wall.
# This is a non-recursive algorithm, and the maximum number of comparissons are:
# (len(a) + len(b) - 1)**2.
#
#                 ===>
#             __ __ __ __
#          a |__|__|__|__|__ __ ) right wall
# left wall (         |__|__|__| b
#                        <===

def lcs(a=[], b=[]):
    if a == [] or b == []:
        return []

    l = len(a) + len(b) - 1

    # Fill non-comparable elements with null spaces.
    sa = a + (len(b) - 1) * ['']
    sb = (len(a) - 1) * [''] + b

    longest = []

    for k in range(l):
        cur = []

        for c in range(l):
            if sa[c] != '' and sb[c] != '' and sa[c] == sb[c]:
                cur.append(sa[c])
            else:
                if len(cur) > len(longest):
                    longest = cur

                cur = []

        if len(cur) > len(longest):
            longest = cur

        if sa[len(sa) - 1] == '':
            # Shift 'a' to the right.
            sa = [''] + sa[: len(sa) - 1]
        else:
            # Shift 'b' to the left.
            sb = sb[1:] + ['']

    return longest


a = 'Andrea always wears black cat to all places she go.'
b = 'She likes her black cat because it is very mischievous and playful.'

print(a)
print(b)
print()
print('What carry Andrea always?')
print()
print(''.join(lcs(list(a), list(b))))