"""

One Edit Distance

Given two strings, determine if they are exactly one edit distance apart.

An edit is an insertion, deletion, or replacement of a single character.

Reference: https://leetcode.com/problems/one-edit-distance/

Complexity:

Time: O(n) where n is the length of the shorter string

Space: O(n) for string slicing

"""

from __future__ import annotations

def is_one_edit(source: str, target: str) -> bool:

"""Check if two strings are exactly one edit apart using slicing.

Args:

source: The first string.

target: The second string.

Returns:

True if the strings are exactly one edit apart, False otherwise.

Examples:

>>> is_one_edit("abc", "abd")

True

"""

if len(source) > len(target):

return is_one_edit(target, source)

if len(target) - len(source) > 1 or target == source:

return False

for index in range(len(source)):

if source[index] != target[index]:

return (

source[index + 1 :] == target[index + 1 :]

or source[index:] == target[index + 1 :]

)

return True

def is_one_edit2(source: str, target: str) -> bool:

"""Check if two strings are exactly one edit apart using modification.

Args:

source: The first string.

target: The second string.

Returns:

True if the strings are exactly one edit apart, False otherwise.

Examples:

>>> is_one_edit2("abc", "abd")

True

"""

source_length, target_length = len(source), len(target)

if source_length > target_length:

return is_one_edit2(target, source)

if len(target) - len(source) > 1 or target == source:

return False

for index in range(len(source)):

if source[index] != target[index]:

if source_length == target_length:

source = source[:index] + target[index] + source[index + 1 :]

else:

source = source[:index] + target[index] + source[index:]

break

return source == target or source == target[:-1]