human
LCS ( Longest Common Subsequence ) of three strings | Returns length of LCS for X [ 0. . m - 1 ] , Y [ 0. . n - 1 ] and Z [ 0. . o - 1 ].
Following steps build L [ m + 1 ] [ n + 1 ] [ o + 1 ] in bottom up fashion . Note that L [ i ] [ j ] [ k ] contains length of LCS of X [ 0. . i - 1 ] and Y [ 0. . j - 1 ] and Z [ 0. . ... k - 1 ].
L [ m ] [ n ] [ o ] contains length of LCS for X [ 0. . n - 1 ] and Y [ 0. . m - 1 ] and Z [ 0. . o - 1 ].
Driver program to test above function"
Solution in Python:
assistant
#Here is code to lCS ( Longest Common Subsequence ) of three strings and the other steps required.
 def lcsOf3 ( X , Y , Z , m , n , o ) :
	L = [ [ [ 0 for i in range ( o + 1 ) ] for j in range ( n + 1 ) ] for k in range ( m + 1 ) ]
	for i in range ( m + 1 ) :
		for j in range ( n + 1 ) :
			for k in range ( o + 1 ) :
				if ( i == 0 or j == 0 or k == 0 ) :
					L [ i ] [ j ] [ k ] = 0
				elif ( X [ i - 1 ] == Y [ j - 1 ] and X [ i - 1 ] == Z [ k - 1 ] ) :
					L [ i ] [ j ] [ k ] = L [ i - 1 ] [ j - 1 ] [ k - 1 ] + 1
				else :
					L [ i ] [ j ] [ k ] = max ( max ( L [ i - 1 ] [ j ] [ k ] , L [ i ] [ j - 1 ] [ k ] ) , L [ i ] [ j ] [ k - 1 ] )
	return L [ m ] [ n ] [ o ]
X = ' AGGT12'
Y = '12TXAYB '
Z = '12XBA '
m = len ( X )
n = len ( Y )
o = len ( Z )
print ( ' Length ▁ of ▁ LCS ▁ is ' , lcsOf3 ( X , Y , Z , m , n , o ) )