# 5 Best Ways to Check if Edit Distance Between Two Strings Is One in Python

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π‘ Problem Formulation: The ‘edit distance’ between two strings refers to the minimum number of operations required to transform one string into the other, with the allowable operations being insertion, deletion, or substitution of a single character. In Python, it’s a common problem to check if the edit distance between two strings is exactly one. For instance, the input strings might be `"cat"` and `"cut"`, and the desired output is `True`, as only one substitution is needed to convert one string into the other.

## Method 1: Brute-Force Comparison

This method involves manually iterating over the characters of both strings and comparing them to check for insertions, deletions, or substitutions. If more than one discrepancy is found, the function returns False; otherwise, it returns True if exactly one difference is detected.

Here’s an example:

```def is_edit_distance_one(str1, str2):
m, n = len(str1), len(str2)
if abs(m-n) > 1:
return False
count_diff = 0
i = j = 0
while i < m and j < n:
if str1[i] != str2[j]:
if count_diff == 1:
return False
if m > n:
i += 1
elif m < n:
j += 1
else:
i += 1
j += 1
count_diff += 1
else:
i += 1
j += 1
if i < m or j < n:
count_diff += 1

return count_diff == 1

print(is_edit_distance_one("cat", "cut"))```

Output:

`True`

This code snippet defines a function `is_edit_distance_one()` that takes two strings and computes the edit distance. It counts differences as it iterates through both strings. If differences exceed one, it returns False. Otherwise, True is returned if there is exactly one edit distance.

## Method 2: Dynamic Programming

Dynamic Programming (DP) is a method for efficiently solving a broad range of search and optimization problems which exhibit the property of overlapping subproblems. This method calculates distances at all pairs of prefixes and stores them in a DP table to avoid redundant calculations.

Here’s an example:

```def is_edit_distance_one_dp(str1, str2):
m, n = len(str1), len(str2)
if abs(m-n) > 1:
return False
dp = [[0] * (n+1) for _ in range(m+1)]
for i in range(m+1):
for j in range(n+1):
if i == 0:
dp[i][j] = j
elif j == 0:
dp[i][j] = i
elif str1[i-1] == str2[j-1]:
dp[i][j] = dp[i-1][j-1]
else:
dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1])
return dp[m][n] == 1

print(is_edit_distance_one_dp("cat", "cut"))```

Output:

`True`

The function `is_edit_distance_one_dp()` constructs a 2D array representing the edit distances between all prefixes of the input strings. It returns True if the edit distance between the entire strings is 1. This method is thorough but less efficient for our specific problem since it calculates more than what’s necessary.

## Method 3: Using Itertools

This approach utilizes the Python itertools library to generate possible combinations by adding, removing, or replacing characters. It checks these combinations to see if any have an edit distance of one from the original string.

Here’s an example:

```import itertools

def is_edit_distance_one_itertools(str1, str2):
if abs(len(str1) - len(str2)) > 1:
return False
edit_distance = sum((c1 != c2) for c1, c2 in itertools.zip_longest(str1, str2))
return edit_distance == 1

print(is_edit_distance_one_itertools("cat", "cut"))```

Output:

`True`

In this snippet, the `is_edit_distance_one_itertools()` function takes advantage of `itertools.zip_longest` to pair characters from both strings and count non-matching pairs without manually managing the indexes. It’s more concise but could be less readable for some users.

## Method 4: Early Exit

This technique optimizes the brute-force approach by adding an early exit clause. If strings vary significantly in length or if more than one edit is found early, the function short circuits to return False.

Here’s an example:

```def is_edit_distance_one_early_exit(str1, str2):
len1, len2 = len(str1), len(str2)
if abs(len1 - len2) > 1:
return False
differences = 0
i, j = 0, 0
while i < len1 and j < len2:
if str1[i] != str2[j]:
if differences:
return False
differences += 1
if len1 > len2:
i += 1
elif len1 < len2:
j += 1
else:
i += 1
j += 1
else:
i += 1
j += 1
if i < len1 or j < len2:
differences += 1

return differences == 1

print(is_edit_distance_one_early_exit("cat", "cut"))```

Output:

`True`

The function `is_edit_distance_one_early_exit()` immediately returns False if a second discrepancy is detected between the two strings or if their lengths differ by more than one. It is an efficient approach since it stops processing as soon as more than one edit is found.

## Bonus One-Liner Method 5: Pythonic Approach

A Pythonic approach often involves using built-in functions and idiomatic expressions. The following one-liner combines conditions and concise expressions to ascertain if the edit distance is one.

Here’s an example:

```is_edit_distance_one_pythonic = lambda s1, s2: len(s1) != len(s2) and any(s1[i] == s2[j] for i in range(len(s1)) for j in range(len(s2)) if abs(i-j) <= 1) or sum(a != b for a, b in zip(s1, s2)) == 1
print(is_edit_distance_one_pythonic("cat", "cut"))```

Output:

`True`

The one-liner lambda function `is_edit_distance_one_pythonic()` uses a combination of conditional statements and generator expressions with `zip()` to check for equal length strings, while it uses a nested loop condition for checking different length strings. It’s very concise but could be hard to understand and debug.

## Summary/Discussion

• Method 1: Brute-Force Comparison. Easy to understand. It may not be the most efficient for long strings.
• Method 2: Dynamic Programming. Exhaustively thorough. Overkill for checking distance of 1. More suitable for larger edit distances.
• Method 3: Using Itertools. Concise and leverages standard library. Less intuitive for those unfamiliar with itertools.
• Method 4: Early Exit. Best performance for our specific problem. Simple and efficient with an early stopping condition.
• Method 5: Pythonic Approach. Very concise. Can be difficult to read and may hide the logic’s complexity.