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]
```