chore(Python): add Levenshtein distance (#685)

algorithms/Python/README.md

@@ -60,6 +60,7 @@
 - [N-th Term Of Fibonacci Series](dynamic_programming/fibonacci_series_nth_term.py)
 - [Catalan Sequence](dynamic_programming/catalan_sequence.py)
 - [0/1 Knapsack Problem](dynamic_programming/knapsack.py)
+- [Levenshtein distance](dynamic_programming/levenshtein_distance.py)
 
 ## Graphs
 - [Simple Graph](graphs/graph.py)

algorithms/Python/dynamic_programming/levenshtein_distance.py

@@ -0,0 +1,41 @@
+# The Levenshtein distance (Edit distance) Problem
+
+# Informally, the Levenshtein distance between two words is
+# the minimum number of single-character edits (insertions, deletions or substitutions)
+# required to change one word into the other.
+
+# For example, the Levenshtein distance between kitten and sitting is 3.
+# The minimal edit script that transforms the former into the latter is:
+
+# kitten —> sitten (substitution of s for k)
+# sitten —> sittin (substitution of i for e)
+# sittin —> sitting (insertion of g at the end)
+
+def levenshtein_distance(word_1, chars_1, word_2, chars_2):
+    # base case if the strings are empty
+    if chars_1 == 0:
+        return chars_2
+    if chars_2 == 0:
+        return chars_1
+
+    # if last characters of the string match, the cost of
+    # operations is 0, i.e. no changes are made
+    if word_1[chars_1 - 1] == word_2[chars_2 - 1]:
+        cost = 0
+    else:
+        cost = 1
+
+    # calculating the numbers of operations recursively
+    deletion =  levenshtein_distance(word_1, chars_1 - 1, word_2, chars_2) + 1
+    insertion = levenshtein_distance(word_1, chars_1, word_2, chars_2 - 1) + 1
+    substitution = levenshtein_distance(word_1, chars_1 - 1, word_2, chars_2 - 1) + cost
+
+    return min(deletion, insertion, substitution)
+
+# driving script
+if __name__ == '__main__':
+    word_1 = 'plain'
+    word_2 = 'plane'
+
+    print('The Levenshtein distance is:')
+    print(levenshtein_distance(word_1, len(word_1), word_2, len(word_2)))