Added Edit Distance Algorithm

Author: s-sanyal

pygorithm/dynamic_programming/edit_distance.py

@@ -0,0 +1,60 @@
+"""
+Given two strings, A and B. Find minimum number of edits required to convert 
+A to B. Following are the allowed operations(edits) :
+
+1. Insert(Insert a new character before or after any index i of A)
+2. Remove(Remove the charater at the ith index of A)
+3. Replace(Replace the character at ith indedx of A to any other character
+            (possibly same))
+
+Cost of all the operations are equal.
+
+Time Complexity : O(M X N)
+Space Complexity : O(M X N), where M and N are the lengths of A and B
+                    respectively.
+"""
+ 
+def edit_distance(str1, str2): 
+    """
+    :param str1: string
+    :param str2: string
+    :return: int
+    """
+    m = len(str1)
+    n = len(str2)
+
+	# Create a table to store results of subproblems 
+    dp = [ [0 for x in range(n + 1)] for x in range(m + 1) ]
+	
+    """
+    dp[i][j] : contains minimum number of edits to convert str1[0...i] to str2[0...j]
+    """
+
+	# Fill d[][] in bottom up manner 
+    for i in range(m + 1): 
+        for j in range(n + 1): 
+
+            # If first string is empty, only option is to 
+            # insert all characters of second string 
+            if i == 0: 
+                dp[i][j] = j # Min. operations = j 
+
+            # If second string is empty, only option is to 
+            # remove all characters of second string 
+            elif j == 0: 
+                dp[i][j] = i # Min. operations = i 
+
+            # If last characters are same, ignore last char 
+            # and recur for remaining string 
+            elif str1[i-1] == str2[j-1]: 
+                dp[i][j] = dp[i-1][j-1] 
+
+            # If last character are different, consider all 
+            # possibilities and find minimum 
+            else: 
+                dp[i][j] = 1 + min(dp[i][j-1],	 # Insert 
+                                    dp[i-1][j],	 # Remove 
+                                    dp[i-1][j-1]) # Replace 
+
+
+    return dp[m][n]