The code below is my solution in python 3 to exercise 1.5 in Cracking the Coding Interview.  I would appreciate feedback on both the algorithm (improvements to space and time complexity) and/or coding style.  I think the time and space complexity of the code below is \$O(n^{2})\$ and \$O(n)\$ respectively. The exercise statement is as follows:Given 2 Strings write a function to check if they are 1 edit away.  There are three type of edits1) Insert a character2) Remove a character3) Replace a characterI wrote the code in Python 3.5 and confirmed that it passed a small unit test.  For this problem I am particularly interested in feedback on where in my code (if at all) I should include more comments.import unittestdef is_one_away(first: str, other: str) -> bool:    Given two strings, check if they are one edit away. An edit can be any one of the following.    1) Inserting a character    2) Removing a character    3) Replacing a character    if len(first) < len(other):        first, other = other, first    if len(first) - len(other) > 1:        return False    elif len(first) - len(other) == 1:        for pos in range(len(first)):            if first[:pos] + first[pos+1:] == other:                return True        return False    else:        num_different_chars = sum(1 for pos in range(len(first)) if first[pos] != other[pos])        return num_different_chars < 2class MyTest(unittest.TestCase):    def test_is_one_away(self):        self.assertEqual(is_one_away('pale', 'ale'), True)        self.assertEqual(is_one_away('pales', 'pale'), True)        self.assertEqual(is_one_away('pale', 'bale'), True)        self.assertEqual(is_one_away('pale', 'bake'), False)        self.assertEqual(is_one_away('ale', 'pale'), True)        self.assertEqual(is_one_away('aale', 'ale'), True)        self.assertEqual(is_one_away('aael', 'ale'), False)        self.assertEqual(is_one_away('motherinlaw', 'womanhitler'), False)        self.assertEqual(is_one_away('motherinlaw','motherinlow'), True)
Check if two strings are one 'edit' apart using Python
python;strings;python 3.x;complexity;edit distance
All three cases are the same: you iterate over both string until there is a difference, you skip that difference and you check that the remaining of the strings are the same.The only difference being how you skip the difference: you can store that in a dictionnary to also help short circuit in cases the length difference is 2 or more:def is_one_away(first: str, other: str) -> bool:    Given two strings, check if they are one edit away. An edit can be any one of the following.    1) Inserting a character    2) Removing a character    3) Replacing a character    skip_difference = {        -1: lambda i: (i, i+1),  # Delete        1: lambda i: (i+1, i),  # Add        0: lambda i: (i+1, i+1),  # Modify    }    try:        skip = skip_difference[len(first) - len(other)]    except KeyError:        return False  # More than 2 letters of difference    for i, (l1, l2) in enumerate(zip(first, other)):        if l1 != l2:            i -= 1  # Go back to the previous couple of identical letters            break    # At this point, either there was no differences and we exhausted one word    # and `i` indicates the last common letter or we found a difference and    # got back to the last common letter. Skip that common letter and handle    # the difference properly.    remain_first, remain_other = skip(i + 1)    return first[remain_first:] == other[remain_other:]