# Levenshtein distance

Levenshtein distance is an edit distance that may be used to compare how dissimilar two pieces of text are. For the Levenshtein distance, each addition, deletion, or substitution add 1 to the total distance. When considering Levenshtein distance, we are always interested on the smallest possible number of transformations required to make a string into another. Therefore, “john” and “jon” have a Levenshtein distance of 1, while “john” and “jose” have a Levenshtein distance of 2 (and not 4, as some would suggest two deletions and two additions).

Here is my implementation of Levenshtein distance using Python, just to show how it can be done in a scenario on which you cannot depend on any libraries other than the ones shipped with the language. Be aware that my implementation has space-complexity and time-complexity O(m·n), where m and n are the size of the input strings.

def levenshtein(a, b):
"""
Evaluates the Levenshtein distance between two strings.
"""
if len(a) < len(b):
a, b = b, a
if len(b) == 0:  # len(a) >= len(b)
return len(a)
a = ' ' + a
b = ' ' + b
d = {}
for i in range(len(a)):
d[i, 0] = i
for j in range(len(b)):
d[0, j] = j
for i in range(1, len(a)):
for j in range(1, len(b)):
if a[i] == b[j]:
# Got the same character
# Just use the answer to the prefix
d[i, j] = d[i - 1, j - 1]
else:
# Not the same character
# Use the smallest of diagonal, above or left
# And add 1 to account for the extra modification needed
d[i, j] = min(d[i - 1, j - 1], d[i, j - 1], d[i - 1, j]) + 1
return d[len(a) - 1, len(b) - 1]