human
Minimum Cost To Make Two Strings Identical | Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ].
Following steps build L [ m + 1 ] [ n + 1 ] in bottom up fashion . Note that L [ i ] [ j ] contains length of LCS of X [ 0. . i - 1 ] and Y [ 0. . j - 1 ].
L [ m ] [ n ] contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ].
Returns cost of making X [ ] and Y [ ] identical . costX is cost of removing a character from X [ ] and costY is cost of removing a character from Y [ ].
Find LCS of X [ ] and Y [ ].
Cost of making two strings identical is SUM of following two 1 ) Cost of removing extra characters from first string 2 ) Cost of removing extra characters from second string.
Driver Code"
How can the above be solved in Python?

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What is a code to minimum Cost To Make Two Strings Identical and the other steps required?
assistant
#Here is code to minimum Cost To Make Two Strings Identical and the other steps required.
 def lcs ( X , Y , m , n ) :
	L = [ [ 0 for i in range ( n + 1 ) ] for i in range ( m + 1 ) ]
	for i in range ( m + 1 ) :
		for j in range ( n + 1 ) :
			if i == 0 or j == 0 :
				L [ i ] [ j ] = 0
			elif X [ i - 1 ] == Y [ j - 1 ] :
				L [ i ] [ j ] = L [ i - 1 ] [ j - 1 ] + 1
			else :
				L [ i ] [ j ] = max ( L [ i - 1 ] [ j ] , L [ i ] [ j - 1 ] )
	return L [ m ] [ n ]
def findMinCost ( X , Y , costX , costY ) :
	m = len ( X )
	n = len ( Y )
	len_LCS = lcs ( X , Y , m , n )
	return ( costX * ( m - len_LCS ) + costY * ( n - len_LCS ) )
X = " ef "
Y = " gh "
print ( ' Minimum ▁ Cost ▁ to ▁ make ▁ two ▁ strings ▁ ' , end = ' ' )
print ( ' identical ▁ is ▁ = ▁ ' , findMinCost ( X , Y , 10 , 20 ) )