human
Edit Distance | DP | .
If any string is empty , return the remaining characters of other string.
To check if the recursive tree for given n & m has already been executed.
If characters are equal , execute recursive function for n - 1 , m - 1.
If characters are nt equal , we need to find the minimum cost out of all 3 operations ..
temp variables.
Driver code"
Solution in Python:

Write me a code to edit Distance and the other steps required?
assistant
#Here is code to edit Distance and the other steps required.
 def minDis ( s1 , s2 , n , m , dp ) :
	if ( n == 0 ) :
		return m
	if ( m == 0 ) :
		return n
	if ( dp [ n ] [ m ] != - 1 ) :
		return dp [ n ] [ m ] ;
	if ( s1 [ n - 1 ] == s2 [ m - 1 ] ) :
		if ( dp [ n - 1 ] [ m - 1 ] == - 1 ) :
			dp [ n ] [ m ] = minDis ( s1 , s2 , n - 1 , m - 1 , dp )
			return dp [ n ] [ m ]
		else :
			dp [ n ] [ m ] = dp [ n - 1 ] [ m - 1 ]
			return dp [ n ] [ m ]
	else :
		if ( dp [ n - 1 ] [ m ] != - 1 ) :
		m1 = dp [ n - 1 ] [ m ]
		else :
		m1 = minDis ( s1 , s2 , n - 1 , m , dp )
		if ( dp [ n ] [ m - 1 ] != - 1 ) :
		m2 = dp [ n ] [ m - 1 ]
		else :
		m2 = minDis ( s1 , s2 , n , m - 1 , dp )
		if ( dp [ n - 1 ] [ m - 1 ] != - 1 ) :
		m3 = dp [ n - 1 ] [ m - 1 ]
		else :
		m3 = minDis ( s1 , s2 , n - 1 , m - 1 , dp )
		dp [ n ] [ m ] = 1 + min ( m1 , min ( m2 , m3 ) )
		return dp [ n ] [ m ]
str1 = " voldemort "
str2 = " dumbledore "
n = len ( str1 )
m = len ( str2 )
dp = [ [ - 1 for i in range ( m + 1 ) ] for j in range ( n + 1 ) ]
print ( minDis ( str1 , str2 , n , m , dp ) )