'''
Note from ola:
The code below was done in class in
trying to understand the AnagramFree APT
'''

>>> x = ['tea', 'ate', 'eat', 'ales', 'sale']
>>> x
['tea', 'ate', 'eat', 'ales', 'sale']
>>> [sorted(e) for e in x]
[['a', 'e', 't'], ['a', 'e', 't'], ['a', 'e', 't'], ['a', 'e', 'l',
's'], ['a', 'e', 'l', 's']]
>>> sorted("apple")
['a', 'e', 'l', 'p', 'p']
>>> sorted("rainmaker")
['a', 'a', 'e', 'i', 'k', 'm', 'n', 'r', 'r']
>>> y = ["aet", "aet", "aet", "aels", "aels"]
>>> y
['aet', 'aet', 'aet', 'aels', 'aels']
>>> y = ['aels', 'aet', 'aels', 'aet', 'aet']
>>> y
['aels', 'aet', 'aels', 'aet', 'aet']
>>> for i, word in enumerate(y):
...     print word, y[:i-1].count(word)
... 
aels 2
aet 0
aels 1
aet 1
aet 1
>>> for i, word in enumerate(y):
...     print word, y[:(i-1)], y[:(i-1)].count(word)
... 
aels ['aels', 'aet', 'aels', 'aet'] 2
aet [] 0
aels ['aels'] 1
aet ['aels', 'aet'] 1
aet ['aels', 'aet', 'aels'] 1

>>> for i, word in enumerate(y):
...     print i,word, y[:i].count(word)
... 
0 aels 0
1 aet 0
2 aels 1
3 aet 1
4 aet 2
>>> y
['aels', 'aet', 'aels', 'aet', 'aet']
>>> x
['tea', 'ate', 'eat', 'ales', 'sale']
>>> y
['aels', 'aet', 'aels', 'aet', 'aet']
>>> sorted('eat')
['a', 'e', 't']
>>> ''.join(['a','e','t'])
'aet'
>>>