In [27]:

    

# from IPython.core.debugger import set_trace

def has_unique_characters(str=''):
    '''
    Checks if a string has all unique characters
    
    >>> has_unique_characters('abc')
    True
    >>> has_unique_characters()
    True
    >>> has_unique_characters('abac')
    False
    '''
#     1/ time: 1.74 µs ± 14.5 ns
#     unique_chars = dict.fromkeys(str,1)
#     return len(unique_chars)==len(str)

#     2/ time: 1.04 µs ± 7.39 ns using dictionary
#     unique_chars ={}
#     for char in str:
#         if char in unique_chars:
#             return False
#         unique_chars[char]=1
#     return True

#     3/ 1.4 µs ± 19.1 ns using a set
#     char_set =set()
#     for char in str:
#         if char in char_set:
#             return False
#         char_set.add(char)
#     return True

#     4/ 1.38 µs ± 8 ns using a bit vector
    bit_vector =0b0
    pos_a=ord('a')
#     set_trace()
    
    for char in str:
        pos_char=ord(char)-pos_a
#         print('vec:{:0>35b}'.format(bit_vector))
#         print('pos:{:0>35b}'.format(1 << pos_char))
        if bit_vector & (1 << pos_char):
            return False
        bit_vector |= (1 << pos_char)
    return True


# print(has_unique_characters('qwertyuuidfsdgsg'))


%timeit has_unique_characters('qwertyuuidfsdgsg')
# import doctest; doctest.testmod()


    

    




3.06 µs ± 49.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [37]:

    

# def isPermutations(s1,s2):
#     len1 = len(s1)
#     len2 = len(s2)
    
#     if len1 != len2:
#         return False
    
#     ss1=sorted(s1)
#     ss2=sorted(s2)
    
#     for i in range(0,len1):
#         if ss1[i] != ss2[i]:
#             return False
#     return True


def isPermutations(s1,s2):
    if len(s1) != len(s2):
        return False

    char_counts={}

    
    # fill it    
    for c in s1:
        char_counts[c] = char_counts.get(c,0) +1

    # empty it
    for c in s2:
        char_counts[c] = char_counts.get(c,0) -1

    # check if all are 0
    for c in char_counts:
        if char_counts[c] != 0:
            return False

    return True




        
print( isPermutations('abcdefffgh', 'dcabefghff') )


    

    




True

In [52]:

    




    

    
Out[52]:





97