human
Length of longest common subsequence containing vowels | function to check whether ' ch ' is a vowel or not.
function to find the length of longest common subsequence which contains all vowel characters.
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 ] which contains all vowel characters.
Driver Code"
Solution in Python:
 What is a code to length of longest common subsequence containing vowels and the other steps required?
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#Here is code to length of longest common subsequence containing vowels and the other steps required.
 def isVowel ( ch ) :
	if ( ch == ' a ' or ch == ' e ' or ch == ' i ' or ch == ' o ' or ch == ' u ' ) :
		return True
	return False
def lcs ( X , Y , m , n ) :
	L = [ [ 0 for i in range ( n + 1 ) ] for j in range ( m + 1 ) ]
	i , j = 0 , 0
	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 ] ) and isVowel ( X [ i - 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 ]
X = " aieef "
Y = " klaief "
m = len ( X )
n = len ( Y )
print ( " Length ▁ of ▁ LCS ▁ = " , lcs ( X , Y , m , n ) )