Getting all combinations of a string and its substrings

This is a fun exercise. I think other answers may use itertools.product or itertools.combinations. But just for fun, you can also do this recursively with something like

def subs(string, ret=['']):
    if len(string) == 0:
        return ret
    head, tail = string[0], string[1:]
    ret = ret + list(map(lambda x: x+head, ret))
    return subs(tail, ret)

subs('abc')
# returns ['', 'a', 'b', 'ab', 'c', 'ac', 'bc', 'abc']

You can do this easily using itertools.combinations

>>> from itertools import combinations
>>> x = 'abc'
>>> [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
['a', 'b', 'c', 'ab', 'ac', 'bc', 'abc']

If you want it in the reversed order, you can make the range function return its sequence in reversed order

>>> [''.join(l) for i in range(len(x),0,-1) for l in combinations(x, i)]
['abc', 'ab', 'ac', 'bc', 'a', 'b', 'c']

@Sunitha answer provided the right tool to use. I will just go and suggest an improved way while using your return_substrings method. Basically, my solution will take care of duplicates.

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I will use "ABCA" in order to prove validity of my solution. Note that it would include a duplicate 'A' in the returned list of the accepted answer.

Python 3.7+ solution,

x= "ABCA"
def return_substrings(x):
    all_combnations = [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
    return list(reversed(list(dict.fromkeys(all_combnations))))
    # return list(dict.fromkeys(all_combnations)) for none-reversed ordering

print(return_substrings(x))
>>>>['ABCA', 'BCA', 'ACA', 'ABA', 'ABC', 'CA', 'BA', 'BC', 'AA', 'AC', 'AB', 'C', 'B', 'A']

Python 2.7 solution,

You'll have to use OrderedDict instead of a normal dict. Therefore,

 return list(reversed(list(dict.fromkeys(all_combnations))))

becomes

return list(reversed(list(OrderedDict.fromkeys(all_combnations))))

Order is irrelevant for you ?

You can reduce code complexity if order is not relevant,

x= "ABCA"
def return_substrings(x):
    all_combnations = [''.join(l) for i in range(len(x)) for l in combinations(x, i+1)]
    return list(set(all_combnations))