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"""
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]