Checking if a string's characters are ascending alphabetically and its ascent is evenly spaced python

Question

So need to check if a string\'s characters are ascending alphabetically and if that ascent is evenly spaced.

a = \"abc\"
b = \"ceg\"

So a i

Answer 1

without itertools

>>> a = 'abc'
>>> ords = [ord(c) for c in a]
>>> ords == sorted(ords)
True

>>> diffs = set()
>>> for i in range(len(ords) -1): diffs.add(ords[i] - ords[i+1])
>>> len(diffs) == 1
True

Answer 2

>>> from itertools import product
>>> from string import lowercase
>>> a="abc"
>>> any(a in lowercase[i::j+1] for i,j in product(range(26),repeat=2))
True
>>> b="ceg"
>>> any(b in lowercase[i::j+1] for i,j in product(range(26),repeat=2))
True
>>> c="longer"
>>> any(c in string.lowercase[i::j+1] for i,j in product(range(26),repeat=2))
False
>>> d="bdfhj"
>>> any(d in string.lowercase[i::j+1] for i,j in product(range(26),repeat=2))
True

It's not necessary to use product, and a little more efficient to do it this way

>>> any(a in string.lowercase[i::j+1] for i in range(26) for j in range(26-i))
True

Answer 3

A non intuitive approach in handling this

>>> somestring = "aceg"
>>> len(set([ord(y)-ord(x) for (x,y) in zip(*(iter(somestring),) * 2) if y > x]))==1
True

The concept is to create the difference of subsequent element and see if the difference is consistent. For that I create a set and determine if the length is 1 in which case the order is preserved. Also to ensure that the series

Answer 4

only hints for home work,,,

you can try to make use of something from this

In [100]: z = 'abc'
In [101]: [ord(x) for x in z]
Out[101]: [97, 98, 99]

then there can be several logics to check if the elements are evenly spaced :)